Brain Teasers
Dogs
Probability
Probability puzzles require you to weigh all the possibilities and pick the most likely outcome.Probability
There are two dogs, a black one and a white one.
One is male.
What is the probability they are both male?
One is male.
What is the probability they are both male?
Answer
1/3There are 3 combinations:
Black is male, white is male,
Black is male, white is female,
Black is female, white is male.
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Comments
First off, probability is not my strong suit. Now, that being said, if one of them is a male, isn't there a 50/50 chance of the second dog being a mail? Isn't this just like a coin tossed into the air and landing heads 5 times in a row.....the 6th toss has a 50/50 chance of being heads. Color is irrelevent....if you don't know the sex of the dog, it has a 50% chance of being male
If you said the black on is male, then there is a 1/2 chance the other is Male/Female, but when you are only told 1 is Male, that can be either the black or the white.
Its like a 1/2 chance the first is male, and 1/2 chance the second is male, that would give 1/4 chance they are both male. (There are 4 combinations, mf, mm, fm, ff)
But at least one is male, so that elliminates one of the possible combinations giving 1/3
Its like a 1/2 chance the first is male, and 1/2 chance the second is male, that would give 1/4 chance they are both male. (There are 4 combinations, mf, mm, fm, ff)
But at least one is male, so that elliminates one of the possible combinations giving 1/3
sex of dogs is A & B. B or A is known to be male.
we have the four posibilities: mm, ff, mf & fm.
ff is out of boundry (one must be male). mm is the one we are seeking (1 condition fulfills our requirement. we have 3 possible solutions. the P(A|B) = 1/3.
we have the four posibilities: mm, ff, mf & fm.
ff is out of boundry (one must be male). mm is the one we are seeking (1 condition fulfills our requirement. we have 3 possible solutions. the P(A|B) = 1/3.
Probablity isn't my strong suit either, but it seems to me that there are only TWO possibilities for the second dog (and color makes no difference at all). It is either male or female. Also, order doesn't matter here, so mf is the same as fm (or, if you think order DOES matter, then fm isn't a possibility because the problem stated that the first one was male). It seems that the answer should be 1/2. I'm with pusandave.
In the last two examples in the answer, the gender is switched with a different color. Color is irrelevant in this problem; isn't it?
If you think about this, the way the problem is worded eliminates only ONE of the FOUR possible situations: that both the bears are female.
It would be like saying if you tossed a coin twice, and either the first time came out heads, or the second time, then what is the probability of both heads.
Try tossing a coin TWICE, lots of times. Count all the times where there is at least one head (ie any time there are two tails ignore them) If you do it 100 times, there should be about 25 tails-tails to ignore. (1/4 chance of tails-tails)
Then out of the rest (should be around 75) count how many times there are two heads - should be about 25 (1/4 chance)
This would show 25/75 chance of getting two heads if at least one of the coins is heads
It would be like saying if you tossed a coin twice, and either the first time came out heads, or the second time, then what is the probability of both heads.
Try tossing a coin TWICE, lots of times. Count all the times where there is at least one head (ie any time there are two tails ignore them) If you do it 100 times, there should be about 25 tails-tails to ignore. (1/4 chance of tails-tails)
Then out of the rest (should be around 75) count how many times there are two heads - should be about 25 (1/4 chance)
This would show 25/75 chance of getting two heads if at least one of the coins is heads
There are two dogs, a black one and a white one.
One is Male.
What is the Probability they are both male?
Answer:
1/3
There are 3 combinations:
Black is Male, White is Male,
Black is Male, White is Female,
Black is Female, White is Male.
There is a problem with your logic cathalmacabe, or not really with your logic, with your question. The question asks for a simple piece of information of which there are only two choice....male & female. The question is not whether the "2nd" dog is black or white or whether it has fleas. It asks what is the probability that that individual animal with no other hints or clues is a male or a female. Color is irrelelevant and any information regarding those facts is superfulous. --RAYN
One is Male.
What is the Probability they are both male?
Answer:
1/3
There are 3 combinations:
Black is Male, White is Male,
Black is Male, White is Female,
Black is Female, White is Male.
There is a problem with your logic cathalmacabe, or not really with your logic, with your question. The question asks for a simple piece of information of which there are only two choice....male & female. The question is not whether the "2nd" dog is black or white or whether it has fleas. It asks what is the probability that that individual animal with no other hints or clues is a male or a female. Color is irrelelevant and any information regarding those facts is superfulous. --RAYN
rayneeday : Nothing wrong with my logic. I thought it would be a little trickier if people were thrown off by the colour, its means they might have to think about it more and see if it is relavent or useful. Its not my fault if you were confused by this.
It is also easier to see in the answer. If you say on is male, the other is female, and then one is female and the other is male, some people might say whats the difference. I think it makes it slightly harder to get the answer, but easier to see when given the answer
e-mail me on [email protected] if you still have a problem
It is also easier to see in the answer. If you say on is male, the other is female, and then one is female and the other is male, some people might say whats the difference. I think it makes it slightly harder to get the answer, but easier to see when given the answer
e-mail me on [email protected] if you still have a problem
Alright, the gentleman from Texas yeilds to Mr. (or Ms.) McCabe, Upon further review, I must admit that there may have been a slight (very slight) flaw in my conclusions, but I have seen the light. And I have found the error in my ways. So with a tip of the hat and a kind nod, I'm going to stroll on out of here with my tail between my legs. BUT I'LL BE BACK. hehehe
Doesn't the statement
"One is male"
Make the probablity of the other being male Zero?
I.e. only one dog is male the other isn't?
"One is male"
Make the probablity of the other being male Zero?
I.e. only one dog is male the other isn't?
cathalmccabe, your logic IS screwed up and I hope you read this twice just to realize it. Listen...we were TOLD that the first one was male. That was a GIVEN. In your coin example, if we were TOLD that the first toss was heads, there would be NO CHANCE of a tails/tails situation. Only heads-heads and heads-tails. Think about it and quit disputing it. You are WRONG and the answer is 50%. I saw the same flaw in logic in the "muddy waters" puzzle. There answer in that probability puzzle is also 50%.
Cath, One more comment on this. Take two coins. Flip the first one and if it's tails, forget about flipping the second one and don't even count it as a throw (because if you go ahead and flip the second one, there is a chance it could be tails and there isn't supposed to be ANY chance of tails-tails. Now then, if you throw the first one and it's heads, count it as a throw and flip the second one. Of the throws you counted, you'll find that the second flip was 50% heads and 50% tails. Thank you, thank you very much. I hope you now see where I'm coming from and why the answer is in fact 50%.
Ok, last one I swear. I have two coins that I'm going to flip at the same time. If they happen to both be tails then I will not count it as a toss since there is no chance (not 25%) of that happening (there is no chance because remember, one of the dogs is male so there is absolutely NO chance of having two female dogs). Actually, to save time, why don't I just put one of the coins down on the table as heads. That way I'm SURE I won't get two tails. Isn't that the same as you saying that one of the dogs is male? Ok, now then I'm going to flip the other coin 100 times. It will be 50% heads and 50% tails. Ok, Sorry. I'm done now but I just can't believe you still don't see it.
Flip a coin, then flip another coin.
Do this 100 times.
Probability wise you will get:
50 heads on the first throw, and 50 tails.
The same for the second throw.
25 of the 50 heads on the first throw will have 25 heads on the second throw.
giving 25heads-heads.
The other 25 heads on the first throw will have will have tails giving 25heads-tails.
Same for the 50 tails on the first throw:
25tails-heads and 25 tails-tails
You are told there must be at least one heads (NOT THE FIRST ONE IS HEADS)
Therefore you elliminate the 25tails-tails from this example, leaving you with 25 heads-heads; 25 heads-tails; 25 tails-heads
Then you are asked for the chance that if one is heads, whats the chance the other is heads: sample group of 75, number of favourable outcomes is 25.
25/75 = 1/3
I hope you read this twice and stop pestering me. (I have e-mail notification on, and its a bit annoying to get 3 e-mails because of your ignorance)
Do this 100 times.
Probability wise you will get:
50 heads on the first throw, and 50 tails.
The same for the second throw.
25 of the 50 heads on the first throw will have 25 heads on the second throw.
giving 25heads-heads.
The other 25 heads on the first throw will have will have tails giving 25heads-tails.
Same for the 50 tails on the first throw:
25tails-heads and 25 tails-tails
You are told there must be at least one heads (NOT THE FIRST ONE IS HEADS)
Therefore you elliminate the 25tails-tails from this example, leaving you with 25 heads-heads; 25 heads-tails; 25 tails-heads
Then you are asked for the chance that if one is heads, whats the chance the other is heads: sample group of 75, number of favourable outcomes is 25.
25/75 = 1/3
I hope you read this twice and stop pestering me. (I have e-mail notification on, and its a bit annoying to get 3 e-mails because of your ignorance)
You just don't get it and I'm sorry. heads-tails and tails-heads is the same situation, not two different situations. You sound like a broken record but you've never disputed my argument about leaving one coin up as heads to save time. Doesn't matter if it came up heads first or second, the point is that it HAD to come up...so I left it up as heads and just flipped the other (not necessarily second) coin. And I got 50/50. It's simple binomial statistics. Open up any stats book and see for yourself.
That is complete stupidity to say that heads-tails is the same as tails heads. Its not. If you want to consider them the same, you must at least realise that it is two times more likely to have a head and a tail together than it is to have a head-head.
Tell me what is wrong with my example doing it 100 times. If you do it 100 times, statistically what happens? Its just like I described.
I didn't bother to "dispute the argument about leaving one coin up as heads to save time" because this is again a stupid thing to do. This changes the puzzle completely.
Wait for a moment, why bother save time here, first make sure that you are doing things correctly. Is tossing a coin twice, 100 times, the same as placing one on the table and tossing the other 100 times?
The whole point of this puzzle, (which seems simple to be beyond your level of comprehension) is that the heads-tails and tails-head are completely different. (Look that up in any stat book)
"Take two coins. Flip the first one and if it's tails, forget about flipping the second one and don't even count it as a throw (because if you go ahead and flip the second one, there is a chance it could be tails and there isn't supposed to be ANY chance of tails-tails."
By "forgetting the second throw" you are eliminating a possible outcome (namely tails-heads) which is possible, because one is heads.
Do you know anything about probability? Probability is number of favourable outcomes/number of possible outcomes.
You are "forgetting" about 1/3 of the possible outcomes.
I'm not going to apoligise for sounding like a broken record. I'm saying the same thing because it is right.
Tell me what is wrong with my example doing it 100 times. If you do it 100 times, statistically what happens? Its just like I described.
I didn't bother to "dispute the argument about leaving one coin up as heads to save time" because this is again a stupid thing to do. This changes the puzzle completely.
Wait for a moment, why bother save time here, first make sure that you are doing things correctly. Is tossing a coin twice, 100 times, the same as placing one on the table and tossing the other 100 times?
The whole point of this puzzle, (which seems simple to be beyond your level of comprehension) is that the heads-tails and tails-head are completely different. (Look that up in any stat book)
"Take two coins. Flip the first one and if it's tails, forget about flipping the second one and don't even count it as a throw (because if you go ahead and flip the second one, there is a chance it could be tails and there isn't supposed to be ANY chance of tails-tails."
By "forgetting the second throw" you are eliminating a possible outcome (namely tails-heads) which is possible, because one is heads.
Do you know anything about probability? Probability is number of favourable outcomes/number of possible outcomes.
You are "forgetting" about 1/3 of the possible outcomes.
I'm not going to apoligise for sounding like a broken record. I'm saying the same thing because it is right.
The only reason I'm saying that you should leave one coin up as heads is because it was given that one of them would be heads. That's it. What's wrong with that? Maybe the first coin came up heads, or maybe the second one did. Maybe the white dog was the male, or maybe the black one was male. Doesn't matter. Remeber that......"doesn't matter". I'll say that time and time again. Tails/heads and heads/tails is the same situation with the ORDER reversed...which remember, doesn't matter. But in consulation, I do know where you are coming from with your answer. You think that just because one of the dogs came up male, that would somehow decrease the chances that the other one would also be male. That's a misconception. You'd lose a fortune in Vegas thinking that there is more of a chance red would come up simply because the ball landed on black last time....or even the last five times. I may read your response, but I'm done replying. I think this thread is long enough anyway. Good puzzle. (obviously) Got me thinking. Ohya, and if you don't want comments/debates, turn your email notification off. ;-)
Whats wrong with leaving one coin up is that you elliminate 1/3 of the possible outcomes (this is my broken record speech because you keep asking for it.)
You are saying heads-tails is the same as tails-heads. In the context of this puzzle it is in the same group of outcomes, ie if it comes out heads-heads, that is one type of outcome (favourable) and the other two are un-favourable. You can say they are the same if you like, but that doesn't mean you can delete half of them.
if you have 4 and 8/2 would you just say 8/2 is the same so forget about it?
"You think that just because one of the dogs came up male, that would somehow decrease the chances that the other one would also be male. That's a misconception" It does decrease the chance. YOU are wrong. If I went to Vegas this is not the same as if the first ball comes up red. It would be *AT LEAST ONE BALL* has come up red. You don't know wether it is the first or second one. You can't bet on this directly. The red ball in Vegas is the same as you leaving one coin turned up. I have told you why it is wrong which you dispute. You tell me what is wrong with when you do it my way 100 times. Tell me why it does not come out the same as you leaving one coin up. I am interested in comments/debates, but not one person telling me I'm WRONG WRONG WRONG 3 times in a row.
I think you have seen that you are wrong, and won't answer in an acceptance of defeat. At least Ray earlier on could admit he made an error with this.
I've found this if you want some more info http://maa.truman.edu/probability.html
(By the way if I did see five black balls come up in a row, I probably would have a bet on the next being red - as long as I knew things weren't rigged I might lose, but the odds are in my favour just because you win or lose doesn't mean the odds are different)
You are saying heads-tails is the same as tails-heads. In the context of this puzzle it is in the same group of outcomes, ie if it comes out heads-heads, that is one type of outcome (favourable) and the other two are un-favourable. You can say they are the same if you like, but that doesn't mean you can delete half of them.
if you have 4 and 8/2 would you just say 8/2 is the same so forget about it?
"You think that just because one of the dogs came up male, that would somehow decrease the chances that the other one would also be male. That's a misconception" It does decrease the chance. YOU are wrong. If I went to Vegas this is not the same as if the first ball comes up red. It would be *AT LEAST ONE BALL* has come up red. You don't know wether it is the first or second one. You can't bet on this directly. The red ball in Vegas is the same as you leaving one coin turned up. I have told you why it is wrong which you dispute. You tell me what is wrong with when you do it my way 100 times. Tell me why it does not come out the same as you leaving one coin up. I am interested in comments/debates, but not one person telling me I'm WRONG WRONG WRONG 3 times in a row.
I think you have seen that you are wrong, and won't answer in an acceptance of defeat. At least Ray earlier on could admit he made an error with this.
I've found this if you want some more info http://maa.truman.edu/probability.html
(By the way if I did see five black balls come up in a row, I probably would have a bet on the next being red - as long as I knew things weren't rigged I might lose, but the odds are in my favour just because you win or lose doesn't mean the odds are different)
I have a headache, but think I get this now. (maybe not) My gut reaction was to go for the 50% option. i.e. What sex is the second dog? We are not asked what is the sex of the second dog though, if we were, I think it WOULD be 50/50. We are asked for the chances of the OTHER dog being female. Given that if you have a 50/50 chance of any given dog being male or female (which is probably not quite right, but near enough for the sake of argument) we have 4 total possible outcomes. MM, MF, FM, or FF. Remember, the question does NOT ask for the chance of one dog being M or F, but of BOTH dogs in a random couple being MM. Eliminating the FF option (as this is the only one we can, given that AT LEAST 1 is male, but not the first) we have the 3 possibilities that the questioner gave in the answer (although it's taken me about an hour to see that!!) Only 1 is MM (the MF or FM is irrelavent - remember - not asking for sex of second dog SEQUENTIALLY!) so I think 1/3 is right. maybe. Get me another drink.
I think chamber has finally realised he was wrong. Bit of a cop out, eh I think this has gone on long enough so I'm not going to reply any more! I think at least he has gotten an education out of it even if he won't admit it! ...
"He who laughs last ...."
"He who laughs last ...."
Alright Cathal, I'm not going down like that. I saw Andy's reply and it got me going again. Listen, in your muddy waters puzzle, Wasn't the 1E picked up FIRST? There is order there. I believe the answer is 50% in that one too. You are failing to realize the givens in the problem. In the muddy waters teaser there is a chance of 1E/1E or 1E/2E (first one picked/next one you're gonna pick). Right? Isn't that all the possibilies? I'll tell you one thing, 2E/1E and 2E/2E isn't possible (which is why I don't count tails as a throw in one of my previous arguments) because it's GIVEN that the FIRST one picked was a 1E. Do you follow me so far? Andy said something about "first". Well, what was first and what was second in that dog puzzle? Nothing and it doesn't matter. He doesn't understand that MF and FM are the same because...IN BOTH CASES THERE IS ONE MALE which WAS GIVEN IN THE PROBLEM. The problem doesn't mention anything about order. It doesn't need to. All it says is that one is male. So when you say MF is different than FM, how can that be? That reduces the selection to MM and MF. One you know/one you don't 50% Simple Binomial Statistics. Wow. 21 comments on this thread. Very good.
Come off it. Number one, the muddfy waters is a different puzzle if it was the same I wouldn't have bothered rewriting it. Number two. You keep saying MF and FM is the same. Lets just for the sake of argument call MM outcome 1, MF outcome 2 FM outcome 3. Would you not agree the three outcomes are all equally likely? lets go back to basics.
A simple question: You have 2 dogs (but you know they are definitely not both female), whats the chance of the first being female? 1/3 OK? whats the chance of the second being female? 1/3 OK? Whats the chance of them both being Male? 1/3? OK? Whats the chance of them both being male? 1/3 again.
I know you (wrongly) think MF and FM are the same. If you want to say they are the same, lets call all the outcomes 2 and 3 Outcome B So we have Outcome 1 and outcome B. You have to see now that outcome B is twice as likely as outcome 1. (Because 2 and 3 and 1 are equally likely)
Going back to the other one, Money in Water, the answer would be 50/50 if there was a 50/50 chance of either coin. Because you have already taken out one €1 coin there is a higher chance that the second one is also €1. So far its DEFINITELY not €2, but there is a chance that you pulled out a 2nd €1. That increases the chances of finding another €1.
Do you not agree with what happens if you actually did this 100 times and counted the results? Do you not think that if you say MF and FM are the same that that whole outcome is twice more likely? If you can't understand this, you are missing one of the fundamentals of probability, and I'm afraid all this will just be above you. Some people see it themselves, most people can see it when its explained to them, but some people just aren't able. I've told you lots of times now, if you want to look back over my previous posts, or try here: http://maa.truman.edu/probability.html
but if you can't understand what I'm telling you you won't be able to understand this page either. At first I thought you didn't get this because I have not explained properly. I think you have seen it, but don't want to admit it because you can't admit you were wrong. You tell me whats wrong with my explainations, especially when you do it 100 times.
A simple question: You have 2 dogs (but you know they are definitely not both female), whats the chance of the first being female? 1/3 OK? whats the chance of the second being female? 1/3 OK? Whats the chance of them both being Male? 1/3? OK? Whats the chance of them both being male? 1/3 again.
I know you (wrongly) think MF and FM are the same. If you want to say they are the same, lets call all the outcomes 2 and 3 Outcome B So we have Outcome 1 and outcome B. You have to see now that outcome B is twice as likely as outcome 1. (Because 2 and 3 and 1 are equally likely)
Going back to the other one, Money in Water, the answer would be 50/50 if there was a 50/50 chance of either coin. Because you have already taken out one €1 coin there is a higher chance that the second one is also €1. So far its DEFINITELY not €2, but there is a chance that you pulled out a 2nd €1. That increases the chances of finding another €1.
Do you not agree with what happens if you actually did this 100 times and counted the results? Do you not think that if you say MF and FM are the same that that whole outcome is twice more likely? If you can't understand this, you are missing one of the fundamentals of probability, and I'm afraid all this will just be above you. Some people see it themselves, most people can see it when its explained to them, but some people just aren't able. I've told you lots of times now, if you want to look back over my previous posts, or try here: http://maa.truman.edu/probability.html
but if you can't understand what I'm telling you you won't be able to understand this page either. At first I thought you didn't get this because I have not explained properly. I think you have seen it, but don't want to admit it because you can't admit you were wrong. You tell me whats wrong with my explainations, especially when you do it 100 times.
I thought this was appropriate: "For those who understand no explanation is needed, ...For those who don't none will do".
"whats the chance of the first being female? 1/3 OK? whats the chance of the second being female?..."
What are you talking about? First ?!? Second ?!? Do you mean to say black? White? By "first", do you mean "the one you know is male"? I don't think that's what you mean because that would completely contradict your argument. I KNOW what "first" is in the muddy water teaser...it was the first coin picked up. What is first in this one? What is second? DOESN'T MATTER. The answer is 50%.
What are you talking about? First ?!? Second ?!? Do you mean to say black? White? By "first", do you mean "the one you know is male"? I don't think that's what you mean because that would completely contradict your argument. I KNOW what "first" is in the muddy water teaser...it was the first coin picked up. What is first in this one? What is second? DOESN'T MATTER. The answer is 50%.
Can I start at the top?
There are two dogs, a black one and a white one. One is male. (I'm assuming that means at least 1 is male, or the probability of them both being male is 0.)
I think the black/white thing is not irrelevant.
4 possibilies (ignoring the 1 is male for now.)
White - Male ; Black - Male.
White - Male ; Black - Female.
White - Female ; Black - Male.
White - Female ; Black - Female.
I think that blows Chambers - MF is the same as FM thoery pretty sky high.
Now we know the last cannot be possible (bringing back the 'one is male' thing.)
Of the 3 remaining. one is that they are both male. QED. The end. Goodnight.
There are two dogs, a black one and a white one. One is male. (I'm assuming that means at least 1 is male, or the probability of them both being male is 0.)
I think the black/white thing is not irrelevant.
4 possibilies (ignoring the 1 is male for now.)
White - Male ; Black - Male.
White - Male ; Black - Female.
White - Female ; Black - Male.
White - Female ; Black - Female.
I think that blows Chambers - MF is the same as FM thoery pretty sky high.
Now we know the last cannot be possible (bringing back the 'one is male' thing.)
Of the 3 remaining. one is that they are both male. QED. The end. Goodnight.
Andy, go back to school. Your comment makes no sense. We've already established that color is irrelevent. Maybe next time you should read all of the comments. "the end. Goodnight"
Chamber - I'm fed up trying to expain it too you. You must be about 2 years old. I know this, because my 3 year old understands it. Ho ho. By the way, I knew a kid like you at school. Tried to be a bully, and rather than explain arguments, and accept defeat gracefully, just got abusive. I kicked his behind all over school in the end. Maybe you know that feeling? By the way, I can still see why you don't agree, but you're not even trying to listen now, so I will not bother any more. We'll agree to disagree.
Yes I agree, can't accpet defeat gracefully. Didn't read my link, and didn't answer what was wrong with what I said. (ie 100 times; and everything I told him was wrong with his answer)
I thing this is finally finished. I can't agree to differ because he is very very wrong!
I thing this is finally finished. I can't agree to differ because he is very very wrong!
actually the probabilility is 1/2, in the puzzle he/she asked what gender and had nothing to do with the color, just because you can work extra info in the answer. You have to word it right. now say you said what's the probability of the other color being male THEN it would be right, but then again it would be to easy so you shouldn't have put this here in the first place!!!!
I meant work extra info in the question
There is nothing wrong with the wording of the question, and the answer is unambiguously 1/3. I'm surprised this question generated so much argument.
It's Because People With Big Egos Don't Like To Get Proven wrong And Try And Make It Look Like Their Right Even When They'ra Blatently Wrong. Nice Puzzle cathalmccabe And I Did Think it Was 50% But I Am Crap At Probability So I Believe You Are Right.
the answer is 50%! The colour is irrelevant so all we need to know is there are two dogs. One is a male. Then we move to the second dog it is either male or female, so theres are 50% chance of the dogs being male and female and a 50% chance of both dogs being male!
Ok Cathal. Lots of comments here. I have 2 cents to give and a question for you. What is the difference between your puzzle and the following:
Two dogs. One black and one white. One is male, other is unknown. What are the chances the other is male (both of them male)?
Be careful of your answer Cathal. I worded it a little different.
Two dogs. One black and one white. One is male, other is unknown. What are the chances the other is male (both of them male)?
Be careful of your answer Cathal. I worded it a little different.
that was !ntersting!! Anyways go to my website www.pinkspage.com
Love,
Alecia Moore
(P!NK)
Love,
Alecia Moore
(P!NK)
I see what you are getting at. But your puzzle is very different. By asking what the other dog is, implies you know which dog is male. So you could say the only possibilities possibly are MMor MF. (You can't have FM if you do it like this. You could have FM and MM but then not MF)
If you word it more explicitly you might say you have two dogs, number one and number two. Number one is male, whats the chance of number two male (whats the chance both of them are male)? (1/2)
What I'm saying is you know that one of the dogs is male. You don't know if its the first or second one. The probability will depend on that. Its conditional.
If you word it more explicitly you might say you have two dogs, number one and number two. Number one is male, whats the chance of number two male (whats the chance both of them are male)? (1/2)
What I'm saying is you know that one of the dogs is male. You don't know if its the first or second one. The probability will depend on that. Its conditional.
I believe my example problem is exactly the same as the original teaser. Keep in mind that niether one said which one was male and there should be no implying in these teasers. But I also think that it doesn't matter wheather or not you know which one is male because that knowledge won't statistically effect the chances of the other (unknown) dog being male (independent). Consider the following adaptation:
Two dogs. A black one and a white one. One is male. If you picked one, what are the chances it would be male?
In this (now you can say totally different) version, you are asked to choose. This is where they become dependent. The chances become 2/3 that the white one is male. Also 2/3 that the black one is male. 1/3 female for both black and white. As you can see, color isn't really relevant but I think that's been established already. Even knowing which dog was the male dog would not effect the outcome as long as you chose randomly. Please give me your feedback on this (let's keep it clean to) because I am intrigued by this teaser and I think you do good work.
Two dogs. A black one and a white one. One is male. If you picked one, what are the chances it would be male?
In this (now you can say totally different) version, you are asked to choose. This is where they become dependent. The chances become 2/3 that the white one is male. Also 2/3 that the black one is male. 1/3 female for both black and white. As you can see, color isn't really relevant but I think that's been established already. Even knowing which dog was the male dog would not effect the outcome as long as you chose randomly. Please give me your feedback on this (let's keep it clean to) because I am intrigued by this teaser and I think you do good work.
Re-reading your previous comment leaves me to add one more thought. In your critisizm of my example you say, "number one and number two. Number one is male." But that's not accurate. My example DID NOT say "Two dogs. Black one and white one. Black one is male..." as you suggest it did or implied. You used numbers to distinguish the two dogs. I used known and unknown to distinguish the two dogs. One could argue that you "implied" known and unknown to distinguish the two dogs of the original teaser. All I was trying to do was show the "more explicit" version of that original teaser by exposing the known and unknown distinguishers which, as you pointed out, would make the result 50%. I think chamber44 was on to something but needs to get off of his high horse.
Because you say one dog is known, what is the OTHER ONE, this means you know which dog is which. You can use numbers, colours, positions anything to differentiate between them, but it all comes down to you know which is which. I my puzzle you don't know which is definitely the male, which changes it completely.
By trying to show "the more explicit version" you changed the puzzle.
My version is explicit. One dog is male, you don't know which one, it could be first, could be second, could be black white etc. Some people just can't understand that concept, and they try to make the puzzle into one they can answer easily.
Your puzzle does say "Two dogs. Black one and white one. Black one is male"
It just swaps known and unknown for black and white. known for number one and number two. By asking "What is the other one?", you are saying you know one is known/Black/number one or whatever.
It changes it statistically because you must consider all possibilities, IF the first is male, whats the chances the next is male, AND IF the second is male, whats the chances the first is male.
You puzzle would just consider if either first or second is male whats the chances the other is male. You can see these are different?
Thanks for the comment, but I believe all before you have realised why they are wrong and have given up.
Look at what happens if you doi this 100 times as I've shown above.
By trying to show "the more explicit version" you changed the puzzle.
My version is explicit. One dog is male, you don't know which one, it could be first, could be second, could be black white etc. Some people just can't understand that concept, and they try to make the puzzle into one they can answer easily.
Your puzzle does say "Two dogs. Black one and white one. Black one is male"
It just swaps known and unknown for black and white. known for number one and number two. By asking "What is the other one?", you are saying you know one is known/Black/number one or whatever.
It changes it statistically because you must consider all possibilities, IF the first is male, whats the chances the next is male, AND IF the second is male, whats the chances the first is male.
You puzzle would just consider if either first or second is male whats the chances the other is male. You can see these are different?
Thanks for the comment, but I believe all before you have realised why they are wrong and have given up.
Look at what happens if you doi this 100 times as I've shown above.
I still believe the answer is 50%! Ok two dogs a black one and a white one. We know that ONE of the dogs in male. So we have a male dog and a mystery dog. The mystery dog can be male or the mystery dog can be female. Leaving us with either of these possible combination; M/M, M/F. As there are two possible combinations the probabilty of both dogs being male equals 50%! Can I ask a question, if I painted the white dog black would the answer to the question be different?
What you are missing is the fact that there is a third situation F/M, because you talk about "the other one" then this situation is elliminated from your puzzle. Because of the way I word this its still valid. M/F is not the same as F/M. Two completely different situations, which must be taken into account as possible outcomes. If you have read all the comments, this really should be clear by now.
Ok, i'm sorry, I'm a little slow, can i just ask, is colour irrelevant?
DO you really need me to answer that?
If one dog was brown and the other grey do you think you would get a different answer?
The reason I gave a colour and then made no reference to it is intentional. I say one dog is ... That means it could be either the white or the black, so you have to consider both possibilities.
When the other person says what is the other dog, they might as well say what is the sex of the black dog.
I don't mean to be nasty but if you think changing the colour of a dog is going to effect the probability, you will have no hope of understanding this, and I'm not surprised you think its 50/50.
If you add two yellow banannas to one red apple how many pieces of fruit do you have? If you take away the red apple and add a green apple...
If one dog was brown and the other grey do you think you would get a different answer?
The reason I gave a colour and then made no reference to it is intentional. I say one dog is ... That means it could be either the white or the black, so you have to consider both possibilities.
When the other person says what is the other dog, they might as well say what is the sex of the black dog.
I don't mean to be nasty but if you think changing the colour of a dog is going to effect the probability, you will have no hope of understanding this, and I'm not surprised you think its 50/50.
If you add two yellow banannas to one red apple how many pieces of fruit do you have? If you take away the red apple and add a green apple...
3 fruits! Ok, well I think everyone must think I'm really thick because I still don't get this! There are two dogs, we need to find out the probabilty of the dogs both being males. We are given a clue to help us! We are told that one dog is a male, that does half the work for us, now to the tricky part, we must find the sex of the other dog. There are two types of dogs, the male dog and the female dog. Lets call male dog M and female dog F. Ok, we know that one dog is M, thus the sex of this dog does not effect the sex of the other dog. So now the question states, there is one dog, what is the probability of it being male? It could be F or it could be M. Two variables. Which means the answer has to be 50%!
Let me ask a question. There are two chickens. One of these chickens is a female. What is the probabilty of both the chickens being female?
Let me ask a question. There are two chickens. One of these chickens is a female. What is the probabilty of both the chickens being female?
Are you having a laugh at me now? I've already explained this answer loads of times now and you keep asking the same question.
IT IS NOT A QUESTION OF WHAT SEX IS THE OTHER DOG. YOU DON'T KNOW THE SEX OF *EITHER* DOG. YOU ONLY KNOW THER IS DEFINITELY ONE MALE.
There are three possibilities:
Possibility number one:
The first Dog could be male (to satisfy the condition) and the second dog could be female.
Possibility number two:
The first Dog could be male (to satisfy the condition) and the second dog could also be male.
Possibility number three:
The first Dog could be female and the second dog could be male (to satisfy the condition).
The situation where the first dog is male and the second female IS NOT THE SAME as the first dog being female and the second male.
So there are three possibilities, because I only said ONE of the dogs is male not specifying which one. I DID NOT SAY the first dog is male, or what is the "other one" or anything else like that.
3 possibilities, 1 favourable outcome. It is that simple.
Call the dogs chickens call them different colours do whatever you like, but the numbers and wording stay the same. The answer is still 1/3
IT IS NOT A QUESTION OF WHAT SEX IS THE OTHER DOG. YOU DON'T KNOW THE SEX OF *EITHER* DOG. YOU ONLY KNOW THER IS DEFINITELY ONE MALE.
There are three possibilities:
Possibility number one:
The first Dog could be male (to satisfy the condition) and the second dog could be female.
Possibility number two:
The first Dog could be male (to satisfy the condition) and the second dog could also be male.
Possibility number three:
The first Dog could be female and the second dog could be male (to satisfy the condition).
The situation where the first dog is male and the second female IS NOT THE SAME as the first dog being female and the second male.
So there are three possibilities, because I only said ONE of the dogs is male not specifying which one. I DID NOT SAY the first dog is male, or what is the "other one" or anything else like that.
3 possibilities, 1 favourable outcome. It is that simple.
Call the dogs chickens call them different colours do whatever you like, but the numbers and wording stay the same. The answer is still 1/3
I'm not having a laugh at you, I still believe the answer is 50%. M/F and F/M is the same as for both combinations equal one male dog and one female dog. To me it seems simple there are two dogs. One is male. Which dog is male, is irrelevant to the question, all we are asked to do is find the chance of two dogs being male when one dog is already male. As we already know that ONE of the dogs is male we need to know what the other dog could be. Leaving us with M/M and M/F (or F/M which is equivilent to M/F). This is a simple question with a simple answer, I've looked at it over and over again and too me there seems only one possible answer!
You are right in saying there is only one possible answer.
You are wrong in saying M/F and F/M is the same. If you consider the possibility of two dogs being male when you are given no information, the answer is 1/4.
Going by your logic you would get 1/3.
You would have 4 combinations.
MM, FM, MF, FF.
Giving an answer 1/4
If you did this problem you would say FM and MF is the same so the answer is 1/3 which its easy to see is wrong.
If you say there is one female and one male, and it doesn't matter if you say F/M or M/F, you must realise this is wrong. If you call either MF or FM "condition 1" MM condition 2 and FF condition 3 you must see that condition 1 is twice as likely as the other two.
You are wrong in saying M/F and F/M is the same. If you consider the possibility of two dogs being male when you are given no information, the answer is 1/4.
Going by your logic you would get 1/3.
You would have 4 combinations.
MM, FM, MF, FF.
Giving an answer 1/4
If you did this problem you would say FM and MF is the same so the answer is 1/3 which its easy to see is wrong.
If you say there is one female and one male, and it doesn't matter if you say F/M or M/F, you must realise this is wrong. If you call either MF or FM "condition 1" MM condition 2 and FF condition 3 you must see that condition 1 is twice as likely as the other two.
Surely which dog is male is irrelevant. The question being asked states: what is the probability of both dogs being male leaving the options of M/F or M/M. If the question stated, how many possible combinations involving the sexes of the dogs are there, then the answer would be 1/3-GLENN
You have to be joking. GO BACK TO THE START AND THINK ABOUT THE PUZZLE RATHER THAN JUST REPEATING THE FIRST THING THAT COMES INTO YOUR HEAD. READ ALL MY OTHER COMMENTS.
I've already said this I don't know how many times. Will you please actually read what I'm saying? Think about it?
You are saying MF and FM don't matter because there are the same number of males and females. You conclude they represent one outcome
So what you then conclude is that there are TWO possible outcomes, FM and MM, and ONE of them (MM) is favourable so the odds are 50/50
You are wrong to say this.
If you say FM and MF are the same (Which is wrong) You must see that there is an equal chance of FM or MF happening, and its the same chance that MM will happen. The THREE possible outcomes are all equally likely.
Or if you insist on saying they are the same you must see that a situation with one female and one male is TWICE as likely as the one with MM That gives MM a 1/3 chance.
Look at my comment above.
OUT OF TWO DOGS, WHATS THE CHANCE OF HAVING TWO MALES? = 1/4
BECAUSE 4 SITUATIONS mf, fm, mm, ff,
1 FAVOURABLE OUTCOME MM = 1/4
YOU MUST BE ABLE TO SEE THAT.
HOWEVER YOU SAY mf, AND fm ARE THE SAME SO YOU WOULD GET 1/3 WHICH YOU MUST BE ABLE TO SEE IS WRONG.
I've already said this I don't know how many times. Will you please actually read what I'm saying? Think about it?
You are saying MF and FM don't matter because there are the same number of males and females. You conclude they represent one outcome
So what you then conclude is that there are TWO possible outcomes, FM and MM, and ONE of them (MM) is favourable so the odds are 50/50
You are wrong to say this.
If you say FM and MF are the same (Which is wrong) You must see that there is an equal chance of FM or MF happening, and its the same chance that MM will happen. The THREE possible outcomes are all equally likely.
Or if you insist on saying they are the same you must see that a situation with one female and one male is TWICE as likely as the one with MM That gives MM a 1/3 chance.
Look at my comment above.
OUT OF TWO DOGS, WHATS THE CHANCE OF HAVING TWO MALES? = 1/4
BECAUSE 4 SITUATIONS mf, fm, mm, ff,
1 FAVOURABLE OUTCOME MM = 1/4
YOU MUST BE ABLE TO SEE THAT.
HOWEVER YOU SAY mf, AND fm ARE THE SAME SO YOU WOULD GET 1/3 WHICH YOU MUST BE ABLE TO SEE IS WRONG.
I have read every comment, I have read the problem and I have thought about this! I am not wrong in saying that the answer is 50% as the dogs are not numbered and are both exactly the same! M/F and F/M must equal the same as in both cases there is one male dog and one female dog. It doesn't matter if the first dog is male in the first example and the second dog is male in the second example BECAUSE in both cases there is simply one male dog and one female dog. Lets do this in a process of elimination. We will leave out the black and white part as we have both agreed colour is irrelevant, note fruit example. SO we have two dogs. We know barely anything about these dogs EXCEPT that one dog is male. Ok, now that we know the first dog is male, lets toss that out of the picture as it is no longer relevant to the puzzle as it is 'given' information. Lets move onto the second dog. We know that there is now only TWO possible options for the second dog, male or female! The problem DOES NOT state that we need to find out all possible combinations, all we need to know is what is the chance of the two dogs being male OR in other words, using the process of elimination what is the probability of the second dog being male-GLENN
No you are wrong to say you know the first dog is male so toss out the rest of the information.
Tell me about the four situations MM, FF, MF and FM. when you are asked simply what is the chance to 2 males? 1/4 yes??
You can say MF and FM is the same if you want to call it that but you must be very careful, then you MUST say the 4 possibilities are MM, FF, FM and **FM** there will be two occurences of FM.
Its twice as likely as the others.
Your problem is you can't see past saying "the first dog is male whats the other".
You are only told there is definitely one dog male.
Three things can happen. The first dog is male, then the second dog is male.
Or the first dog is male and the second is female. But it is equally likely that the second dog is male. Then you could have the first dog is male (but we have already counted that situation)
then the only other possibility is that the first dog is female. You must count it.
YOU CANNOT SAY OK WE KNOW ONE DOG WHATS THE OTHER.
YOU DON'T KNOW *which* DOG IS DEFINITELY MALE.
Ask it first another way.
Answer all these questions:
You have two dogs,
1: What is the chance of getting 1 male?
2: What is the chance of getting 2 males?
3: What is the chance of getting 1 male and 1 female?
4: What is the chance of getting 2 males?
5: If you know both are definitely not Male, whats the chances of them both being female?
Tell me about the four situations MM, FF, MF and FM. when you are asked simply what is the chance to 2 males? 1/4 yes??
You can say MF and FM is the same if you want to call it that but you must be very careful, then you MUST say the 4 possibilities are MM, FF, FM and **FM** there will be two occurences of FM.
Its twice as likely as the others.
Your problem is you can't see past saying "the first dog is male whats the other".
You are only told there is definitely one dog male.
Three things can happen. The first dog is male, then the second dog is male.
Or the first dog is male and the second is female. But it is equally likely that the second dog is male. Then you could have the first dog is male (but we have already counted that situation)
then the only other possibility is that the first dog is female. You must count it.
YOU CANNOT SAY OK WE KNOW ONE DOG WHATS THE OTHER.
YOU DON'T KNOW *which* DOG IS DEFINITELY MALE.
Ask it first another way.
Answer all these questions:
You have two dogs,
1: What is the chance of getting 1 male?
2: What is the chance of getting 2 males?
3: What is the chance of getting 1 male and 1 female?
4: What is the chance of getting 2 males?
5: If you know both are definitely not Male, whats the chances of them both being female?
Ok, I'll show you what I have done to finish with my answer. There are two dogs. This leaves open the following possibilities: M/M, F/F and M/F(F/M). I am not numbering the dogs or specifying that a certain dog is male while the other is female, I am simply saying that a male dog and a female dog is the equivilent to a female dog and a male dog. For example; there are two different houses standing next to each other with two different families. Each house has two dogs a black one and a white one. In house 1 the female is black and the male white while in house 2 it is the opposite. House 1 and House 2 both have a male and a female dog and even though the dogs who are male are not the same they agree that the houses still share the fact that they both have two dogs a male one and a female one and that they are equal. A new person moves in, they tell the neighbours that they have two dogs and that one is male. Nothing else. The neighbours think carefully, and go through the possible options. They automatically knock out the F/F option as it can't exist because they know one of the dogs is male. Then they come across the M/M option and leave that in as it a valid option. Now they come across the M/F or F/M option and as they know that which dog is male has to be irrelevant through their own situation. They are left with two possible options having a male and female dog or having two female dogs!
In answer to your questions;
There are four options for each question: M/M, F/F, M/F, F/M
1. 2/4
2. 1/4
3. 2/4
4. 1/4
5. 1/1
NOTE: In this question no extra information has been given unlike the original teaser.
There are four options for each question: M/M, F/F, M/F, F/M
1. 2/4
2. 1/4
3. 2/4
4. 1/4
5. 1/1
NOTE: In this question no extra information has been given unlike the original teaser.
Your example of having two houses doesn't exactly say anything. I phrased my last question (number 5 poorly)
Go through the first four questions again, but for 5 :You know that BOTH dogs are not male, but one of them might be. Whats the chance of two females?
Go through the first four questions again, but for 5 :You know that BOTH dogs are not male, but one of them might be. Whats the chance of two females?
1/3
1/3 bingo. Its the same as the original teaser, I just worded it differently.
Ok, with that worked out, I'll leave quietly and apologise for any troubles caused. Good day to you!
Aug 18, 2002
It's all dependent on wording. In your last example, you said both are not female, but one of them "might be" female - which is not the same as one of them "definitely being" female. Although I actually agree with your logic of 1/3 as the correct answer, you have to admit that the riddle could have been phrased a little more clearly. Surely you see that if you altered the wording merely slightly, "the first one is male" - instead of "one of them is male", how people could get confused.
Saying they both are definitely not female is saying that one is definitely male. Think about it.
I also worded this problem corretcly. If I say like you suggest the first one is male, then that leaves a 50/50 chance of both males.
I also worded this problem corretcly. If I say like you suggest the first one is male, then that leaves a 50/50 chance of both males.
Well well,this teaser has ruffled a few feathers! sorry cat but your solution is bollox, You know that one dog IS male,whats the chance of them both being male? well they either are or they aren't!! thats 50/50!
Oh cat,on your second comment dated APR 16 where the hell did the bear come from? was that what they where when you copied the original teaser!!? And you say Mad-ade copies!!!ha!ha!
I agree.
Sep 01, 2002
What is the probability of Cathalmcabe getting at least one of his teasers right? Any one know the answer?
Don't know, but this one is wrong!
Sep 20, 2002
Color is irrelevant.
Two dogs, X and Y.
Male = 1. Female = 0.
X + Y >= 1
What is the probability that X + Y = 2?
50%
What is the difference between this problem and the one posed at the start of this thread? None that I can see.
Two dogs, X and Y.
Male = 1. Female = 0.
X + Y >= 1
What is the probability that X + Y = 2?
50%
What is the difference between this problem and the one posed at the start of this thread? None that I can see.
Even after reading all of the above comments, I'm not convinced that the answer is 1/3. I emailed cathalmccabe a number of days ago (and haven't received a reply) suggesting that the situation of MF and FM being possible outcomes is ONLY relevant when going through a selection process (eg. lift the first dog's tail, then lift the second dog's tail and note the sex of both dogs - then the MM, MF and FM outcomes are relevant and the answer would be 1/3). To have an answer of 1/3 the question would have to be something along the following lines: "There are two dogs. One is male. If I lifted the tails of both dogs and checked to see what sex they were, what is the probability of them both being male?" The way I see the wording of this teaser though, if you simply said that one dog is definitely M (but we don't know which one), then the other dog MUST be either M or F, giving an answer of 1/2. As soon as you introduce an order of selection into the question the probabilities will be different.
Well cath Probability is right up my alley, let me say that you are correct in everything you have said. The theory is known as conditional probability and those doubters who wish to learn something may look it up. As a simple explanation. Take this problem which is DIFFERENT to yours. "Two dogs, black and white. The white one is male. What is the probability that the black one is male?" The answer to THIS problem is 50%. Now in YOUR problem WE KNOW LESS INFORMATION! ;as you have already pointed out! We don't know which one is male! Phrase your teaser this way. You flip two coins and look at them (without showing them to your tormenters). If they come down 2 tails you flip them again. If not you can safely say that at least one of them is a head. Now you opponents are willing to bet money as a even money bet that 50% of the time they will be 2 heads. According to them, if they bet $1 every time the declaration is made that at least one is a head that they will come out even? I would give them $1.20 every time it came down 2 heads if they would give me $1 every time there was one of each. I could retire! One last comment, why don't you critics get 2 coins and a box of matches. Divide the matches in two and play the game I suggested. See how long it takes to lose your matches if you're betting on 2 heads!!
I disagree Jimbo.
A mathematician could make an argument that the colour of the dogs matters. In theory. But of course, a biologist will tell you that colour and gender are mutually exclusive. In the real world, therefore, the answer is incorrect, and the reference to colour is extraneous information. The type of information that math teachers like to throw into a word problem to make it appear more complicated than it really is.
A mathematician could make an argument that the colour of the dogs matters. In theory. But of course, a biologist will tell you that colour and gender are mutually exclusive. In the real world, therefore, the answer is incorrect, and the reference to colour is extraneous information. The type of information that math teachers like to throw into a word problem to make it appear more complicated than it really is.
The colour is entirely irrelevant. Even if 99.9% of dogs in the world were black, and 0.1% were white, the teaser states that "there are 2 dogs one is black and one is white." Regardless of how I came by these dogs or how unlikely it is that they are together or whatever is not the subject of the question. The only assumed knowledge (which I think is a fair assumption to make) is that randomly choosing any dog from the population of all dogs is just as likely to result in a male as a female. If you now want to argue that there is some genetical correlation between the colour of a dog and its gender then you are getting way out of my league. I think its a great teaser with an obvious logical choice between either one half or one third for the answer. Once you have modelled the problem with some cards or coins, it doesn't take long to realise that the odd pair comes up twice as often as the matching pair. The author of this teaser has even more interesting puzzles on this site.
It is outrageous that someone is trying to edit this teaser to change the answer to 0.5 which is entirely wrong. I teach probability for a living! The answer given here is entirely correct 1/3.
[1] Probability is a theoretical construct and cannot affect reality. Proability of a Head tossed on a coin is 0.5 but if you tossed a coin 6 times you might get 6 heads. It is irrelevant. So are all these comments about whether male and female dogs are 50% of the population.
[2] The definition of probability is the number of ways that the described event can happen divided by the total number of possible outcomes.
[3] Thus: Here is a picture of the described situation.
(a) Black dog = male, White dog = male
(b) Black dog = male, White dog = female.
(c) Black dog = female, white dog = male.
That's it. There are only 3 possible situations described by the sentence "I have a black and a white dog; one of them is male."
What breed of dogs they are (French poodles are mostly white) is irrelevant. The question is not asking what is the probability that I have 2 dogs, or what is the probability they might be black and white or what is the probability they are male and female. The FACTS of the case are describe above in (a) (b) and (c).
Now by the definition of Probability, how many of these result in the "other one" being also a male ie 2 males? Answer 1 case only has 2 males!
Probability = 1/3.
[1] Probability is a theoretical construct and cannot affect reality. Proability of a Head tossed on a coin is 0.5 but if you tossed a coin 6 times you might get 6 heads. It is irrelevant. So are all these comments about whether male and female dogs are 50% of the population.
[2] The definition of probability is the number of ways that the described event can happen divided by the total number of possible outcomes.
[3] Thus: Here is a picture of the described situation.
(a) Black dog = male, White dog = male
(b) Black dog = male, White dog = female.
(c) Black dog = female, white dog = male.
That's it. There are only 3 possible situations described by the sentence "I have a black and a white dog; one of them is male."
What breed of dogs they are (French poodles are mostly white) is irrelevant. The question is not asking what is the probability that I have 2 dogs, or what is the probability they might be black and white or what is the probability they are male and female. The FACTS of the case are describe above in (a) (b) and (c).
Now by the definition of Probability, how many of these result in the "other one" being also a male ie 2 males? Answer 1 case only has 2 males!
Probability = 1/3.
I agree with Jimbo here. The answer is indeed 1/3. As Gizzer said, the colours don't matter. If the question is reworded as follows, the answer is still 1/3. "I have 2 dogs. One of them is male. What is the probability that both are male?" The solution of 1/3 is based on the reasonable assumption that it is equally likely for a random dog to be male or female. If that were not the case, then the probability would not be 1/3.For example, if all dogs were only males, then the answer would be 1. Or if there existed only one male dog in the world, then the probability would be 0. But, since we are given no other information, we have to assume that male dogs and female dogs are equally likely to occur. Then, we get the answer of 1/3 from Jimbo's argument. (His cases a,b and c become equally likely).
You go to the kennel to pick up two dogs: 1 black and 1 white. The kennel keeper tells you that he has two black dogs (1 male and 1 female) in one cage, and two white dogs (1 male and 1 female) in a second cage. He then blindfolds you and tells you to grab one black dog and one white dog (by the collar, no fair reaching for the back underside). At this point, there are 4 possible outcomes: BM/WM, BM/WF, BF/WM, BF/WF. The keeper then tells you that at least 1 of the dogs is male. This eliminates the BF/WF combination. You are now left with a 1 in 3 chance that both dogs are male. If you had been told that the black dog was male, then there would have been a 1 in 2 chance of two male dogs. Although the color does not affect the outcome, I think that it does make it easier to visualize the answer. My fingers are now out of breath.
almost a year after the last comment but i have a perfect way to explain this!
? = mysterydog
M = male
theres one male and one mystery
M? or ?M (because apparently order DOES matter)
M? could be MF or MM
?M could be FM or MM
all possible outcomes:
MF
MM
FM
MM
50%
? = mysterydog
M = male
theres one male and one mystery
M? or ?M (because apparently order DOES matter)
M? could be MF or MM
?M could be FM or MM
all possible outcomes:
MF
MM
FM
MM
50%
almost a year after the last comment but i have a perfect way to explain this!
? = mysterydog
M = male
theres one male and one mystery
M? or ?M (because apparently order DOES matter)
M? could be MF or MM
?M could be FM or MM
all possible outcomes:
MF
MM
FM
MM
50%
? = mysterydog
M = male
theres one male and one mystery
M? or ?M (because apparently order DOES matter)
M? could be MF or MM
?M could be FM or MM
all possible outcomes:
MF
MM
FM
MM
50%
That's just silly! To see why, try not assuming that there are not two females and list the solutions:
MM
MM
MF
FM
FF
Clearly you can see the problem.
MM
MM
MF
FM
FF
Clearly you can see the problem.
but there cant be FF
That is beside the point. I clearly said to forget what had been said about "they can't be that" and include that possibility.
THE ANSWER IS 1/2. THE END
BM+WM=1/2
BM+WF=1/4
BF+WM=1/4
NOT
BM+WM=1/3
BM+WF=1/3
BF+WM=1/3
All you people who said it was a half give yourself a pat on the back, all you people who said it was a 1/3, go and stand in the corner.
BM+WM=1/2
BM+WF=1/4
BF+WM=1/4
NOT
BM+WM=1/3
BM+WF=1/3
BF+WM=1/3
All you people who said it was a half give yourself a pat on the back, all you people who said it was a 1/3, go and stand in the corner.
What poker is pointing out is that if you can have MM and another MM then why can't you have MF and another MF. The point being that some people seem to be suggestin that there are 2 possibilities [1] A white male and a black male and [2] A black male and a white male ? Huh?
the probability is 1/2. if your mom has two child. the first child is boy. what second child's gender? the possibilities are boy or girl (50%). if it is 1/3, then there must be one more gender, it is HERMAPHRODITE!
I will use coins in this comment because that makes more sense to me.
If you just flip 1 coin, it is definitely 1/2 chance of being heads and 1/2 of being tails. Now:
If you flip 2 coins (coin A and coin B) that are identical to you and only you (so someone else can tell) without knowing which one is A or B and just see the results of one of them and know that it is heads, then you have these outcomes: (results are written as 'coin A/coin B')
Either one of them could be the heads that you picked up, so:
H/? or
?/H
Referring to the second paragraph (of my comment), you know that the ? can be H or T, so now the possible outcomes are:
H/T (H/?)
H/H (H/?)
T/H (?/H)
H/H (?/H)
making there be 2/4 (1/2) that have 2 heads.
If you just flip 1 coin, it is definitely 1/2 chance of being heads and 1/2 of being tails. Now:
If you flip 2 coins (coin A and coin B) that are identical to you and only you (so someone else can tell) without knowing which one is A or B and just see the results of one of them and know that it is heads, then you have these outcomes: (results are written as 'coin A/coin B')
Either one of them could be the heads that you picked up, so:
H/? or
?/H
Referring to the second paragraph (of my comment), you know that the ? can be H or T, so now the possible outcomes are:
H/T (H/?)
H/H (H/?)
T/H (?/H)
H/H (?/H)
making there be 2/4 (1/2) that have 2 heads.
The Answer is 1/3. Everyone who disagrees should read (and try to understand!!!) cnmne's 'rewording' of the problem (posted on Sep 05, 2004
). However, all you guys who are completely confused about the answer this question; ***CONSIDER YOURSELVES NORMAL***
). However, all you guys who are completely confused about the answer this question; ***CONSIDER YOURSELVES NORMAL***
I remember a quote made by my probability teacher once : "Where there is probability, there is an argument"
Guys the answer is indeed 1/3 as per the wordings of the teaser. In probability its the wordings which changes the answer and not the data.
The child example given by brain juice says that first child is male. But this teaser says that one is male (not the 1st one).
Comparing this with the child example, a mother has two kids and one is male. What is the probability that the other child is male.
Now there are 4 combinations for the two
Older Younger
Male male
Male Female
Female Male
Female Female
But we know that 4th possibility is not possible in this case as we know that at least one is male.
With that we are left with 3 outcomes and answer as 1/3
Guys the answer is indeed 1/3 as per the wordings of the teaser. In probability its the wordings which changes the answer and not the data.
The child example given by brain juice says that first child is male. But this teaser says that one is male (not the 1st one).
Comparing this with the child example, a mother has two kids and one is male. What is the probability that the other child is male.
Now there are 4 combinations for the two
Older Younger
Male male
Male Female
Female Male
Female Female
But we know that 4th possibility is not possible in this case as we know that at least one is male.
With that we are left with 3 outcomes and answer as 1/3
this is absurd -- if "one is male," then there are only 2 possiblities for the sex of the other dog -- male or female. The color is completely irrelevant. If the black dog is male, then the other one is white and either male or female with a 50% chance of being male. If the white dog is male, then the other one is black with a 50% chance of being male or female -- color becomes irrelevant. I think the answer should be 1/2
Everyone, go to this link: http://en.wikipedia.org/wiki/Brain_teaser
Now read where it says "If we encounter someone with two children, given that at least one of them is a son, what is the probability that the other is also a son?"
According to conditional probability, the answer is in fact 1/3.
Now read where it says "If we encounter someone with two children, given that at least one of them is a son, what is the probability that the other is also a son?"
According to conditional probability, the answer is in fact 1/3.
There are two dogs, a black one and a white one.
One is male.
What is the probability they are both male?
zero. You said already said one is male. So they're not both male.
Oh wait, you meant at least one is male. oh, ok.
> There are two dogs, a black one and a white one.
ok, so the possibilities so far are:
1 black one is male, white one is male
2 black one is male, white one is female
3 black one is female, white one is male
4 black one is female, white one is female
> [at least] One is male.
ok, so cut out possibility #4.
> What is the probability they are both male?
Looks like 1 / 3. Is there any flaw in my logic?
One is male.
What is the probability they are both male?
zero. You said already said one is male. So they're not both male.
Oh wait, you meant at least one is male. oh, ok.
> There are two dogs, a black one and a white one.
ok, so the possibilities so far are:
1 black one is male, white one is male
2 black one is male, white one is female
3 black one is female, white one is male
4 black one is female, white one is female
> [at least] One is male.
ok, so cut out possibility #4.
> What is the probability they are both male?
Looks like 1 / 3. Is there any flaw in my logic?
1/2. I can see your reasoning for 1/3, but I truely don't believe that is right.
Let's use coins to demonstrate this.
OK... we'll make 'male' to 'heads' and 'female' to 'tails'.
Alright. From these we are told one is heads.
So...
Turn one coin over on to heads. OK? You know at least one is heads, so you may as well turn one over. If there were none, then you wouldn't need to, hey? But since you know there is at least 1, why not turn the first one over.
Now, flip the second coin. It has a 50% chance of landing on heads, and a 50% chance of landing on tails.
I really can't see why there is so much of an argument... it is pretty simple, really...
Let's use coins to demonstrate this.
OK... we'll make 'male' to 'heads' and 'female' to 'tails'.
Alright. From these we are told one is heads.
So...
Turn one coin over on to heads. OK? You know at least one is heads, so you may as well turn one over. If there were none, then you wouldn't need to, hey? But since you know there is at least 1, why not turn the first one over.
Now, flip the second coin. It has a 50% chance of landing on heads, and a 50% chance of landing on tails.
I really can't see why there is so much of an argument... it is pretty simple, really...
The answer is 1/3, even though this answer seems counterintuitive to me.
The reason why is because the dogs are different, one is black and one is white. Therefore, there are 4 different possibilities for the gender "combinations" of the dogs, as pointed out in comments above are: mm, mf, fm, and ff, because the "order" of the gender matters. Now, if the dogs were the "same", except for the fact that they may be different genders, then the answer would be 1/2, since the possible gender "combinations" would then be: mm, mf, ff (notice that mf = fm, order doesn't matter). Then, since we know that at least one of the dogs is male, there are only 2 possibilities: mm or mf, and therefore the probability of both dogs being male is 1/2.
Clever teaser. Had me confused.
The reason why is because the dogs are different, one is black and one is white. Therefore, there are 4 different possibilities for the gender "combinations" of the dogs, as pointed out in comments above are: mm, mf, fm, and ff, because the "order" of the gender matters. Now, if the dogs were the "same", except for the fact that they may be different genders, then the answer would be 1/2, since the possible gender "combinations" would then be: mm, mf, ff (notice that mf = fm, order doesn't matter). Then, since we know that at least one of the dogs is male, there are only 2 possibilities: mm or mf, and therefore the probability of both dogs being male is 1/2.
Clever teaser. Had me confused.
Okay you know what....
Two years ago I said 1/2 now i say 1/3, and heres why
If you flip two coins, 2 heads, will come up 1/4 of the time, and eliminating the possibilty of two tails, that makes 1/3.
Seriously, do an experiment with coins, it's 1/3
Two years ago I said 1/2 now i say 1/3, and heres why
If you flip two coins, 2 heads, will come up 1/4 of the time, and eliminating the possibilty of two tails, that makes 1/3.
Seriously, do an experiment with coins, it's 1/3
The answer is 1/3 but not because one is black and one is white. Color actually has nothing to do with it. Cath was just throwing that in as a red herring. If you didn't know anything, the combinations would be MM, MF, FM, FF..but you know one is male so that excludes FF. This leaves only 3. Of the three, only one has the other dog as male. But, in defense of everyone who disagrees...
IF he had said that the WHITE one was male, then it would be 50/50 that the Black one was also male.
IF he had said that the WHITE one was male, then it would be 50/50 that the Black one was also male.
Actually, the correct answer is zero. It says that one dog is male. It doesn't say "we know that at least one dog is male" or anything like that.
If I told you that I have two kids, one is a boy, isn't it clear that the other is a girl?
Oh, yeah, if we assume that the speaker is not a native speaker of English, and he really meant to say "at least" then I'm on board with the 1/3 answer.
If I told you that I have two kids, one is a boy, isn't it clear that the other is a girl?
Oh, yeah, if we assume that the speaker is not a native speaker of English, and he really meant to say "at least" then I'm on board with the 1/3 answer.
I take that back. It's 50%. I like the comment about X+Y. Let me repost it in my own way:
Male=1 : Female=0
X=a dog (nevermind color)
Y=a dog (nevermind color)
we need X+Y=2
we know that X or Y is 1 so we have:
1+Y=2
X+1=2
for the first equation Y can be 0 or 1 and still satisfy the teaser givens of one dog being male. For the second equation X can be 0 or 1 and still satisfy the teaser.
for the first equation:
1+0=1
1+1=2
for the second equation:
0+1=1
1+1=2
50/50 chance of the answer being 2. Sorry Cath, you had me as a believer but now I see the true light lol.
Male=1 : Female=0
X=a dog (nevermind color)
Y=a dog (nevermind color)
we need X+Y=2
we know that X or Y is 1 so we have:
1+Y=2
X+1=2
for the first equation Y can be 0 or 1 and still satisfy the teaser givens of one dog being male. For the second equation X can be 0 or 1 and still satisfy the teaser.
for the first equation:
1+0=1
1+1=2
for the second equation:
0+1=1
1+1=2
50/50 chance of the answer being 2. Sorry Cath, you had me as a believer but now I see the true light lol.
I wrote a computer simulation and now I'm getting 1/3 again so in conclusion, I suck at probability.
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