Brain Teasers
Brain Teasers Trivia Mentalrobics Games Community
Personal Links
Submit a Teaser
Your Favorites
Your Watchlist
Browse Teasers
All

Cryptography
Group
Language
Letter-Equations
Logic
Logic-Grid
Math
Mystery
Optical-Illusions
Other
Probability
Rebus
Riddle
Science
Series
Situation
Trick
Trivia

Random
Daily Teasers
Search Teasers

Advanced Search
Add to Google Add to del.icio.us

More ways to get Braingle...

Red and Black Cards

Probability puzzles require you to weigh all the possibilities and pick the most likely outcome.

 

Puzzle ID:#279
Fun:*** (2.47)
Difficulty:** (1.83)
Category:Probability
Submitted By:jerry1***
Corrected By:absy

 

 

 



You have two cards, one red on both sides, one red on one side and black on the other. You reach into a hat and pull out one card, viewing only the top. It is red. What`s the probability that the other side is red?




What Next?

  
  

See another brain teaser just like this one...

Or, just get a random brain teaser

If you become a registered user you can vote on this brain teaser, keep track of
which ones you have seen, and even make your own.

 



Comments

fatlingon
Nov 26, 2001

Unfortunally the answer is not true. it would be true if there was 3 red and one black dot that wasnīt attached to any cards. But never then less, when you do the probability maths, you canīt focus on the dots when they are attached to 2 cards. Cause there is only a 50%/50% chance to get either card. Therefor the amount of red dots does not matter.
(user deleted)
Dec 11, 2001

i was thinking the same thing fatingon. only the initial probability governs it.
tommo
Jan 14, 2002

There are two cards in the problem but it’s actually irrelevant.
When you pick a card out there are 4 alternatives you will see:
1)Side A of red/red card=red
2)Side B of red/red card=red
3)Side A of red/black card=red
4)Side B of red/black card =black
So when you pick a card out the chance that the other side is red is 3/4 just the same as the side you see is 3/4 but the problem has a condition that the side you see is always red.
The three alternatives you now see are:
1)Side A of red/red card=red
2)Side B of red /red card=red
3)Side A of red/black card=red
The chance that the other side is black is 1/3 because if 1)or 2)the other side is red, if 3)the other side is black.
lamentor79
Feb 10, 2002

Here we are considering two separate events; as per the first comment: to clarify the argument, let us consider that event A has been determined (according to the problem), therefore the probability of event B (a separate event) is calculated given the results of event A, ==>> P: B|A, or "the probability of B, given A is 50%."
tommo
Feb 20, 2002

I really underestimated how good this one is. I’d like to change my vote to 4/4.
I’ve tried it out on a number of people and even when they are given the answer they still don’t get it.
OJee
Apr 09, 2002

lamentor79: your comment is correct, but then again false. The events are divided in two: the drawing of the card, and the flipping of it to se wheter the other side is red /black. When we have drawed the card, we either se one of the two red side of the red/red card, or the red side of the black card. (3 posible outcomes) only of these gives us a black side. The answer is 2/3. as said before
BabyB*
Jul 09, 2002

Let A = you see a red side when pulling out a card if we can only see one side.
Let B = you pull out the two-sided red card.

Probability of A is 3/4. I think we can all agree on that. There are 3 red sides and 1 black side.
Probability of B is 1/2. We can all agree on that too: we pull out either the red/red OR the red/black.

The question is asking for the conditional probability of B given A. If we suppose that event 1 is pulling out a side and event 2 is determining the nature of the card, the question asks us about event 2 given that the results of event 1 are already known: What is the probability of the card being the red/red given we already pulled out a red side?

Bayes' Rule governs conditional probabilities. It says P(B given A) = P(B and A)/P(A).
In this case, P(B and A) is 1/2. (see *below if you don't understand why)
P(A) is 3/4.

By Bayes' Rule (1/2)/(3/4) = 2/3.

*Before we pull out a card, we have these options:
pull out red/red, see red (front)
pull out red/red, see red (back)
pull out red/black, see red
pull out red/black, see black

Thus in 2 out of 4 cases we pull out a red side AND pull out the red/red card.

Now, we have ALREADY pulled out a red side, which is what the question calls for.
Our options have been reduced to:
pull out red/red, see red (front)
pull out red/red, see red (back)
pull out red/black, see red

It's easy to see that our chance of getting the red/red is now 2/3.
electronjohn*us*
Dec 03, 2002

I have to agree with the answer. It is what I came up with also. Very good teaser.
curtiss82**
Dec 28, 2003

I say the answer is wrong and I agree witht he first comment. The correct answer is 1/2. Lets call the red/red card card A and the red/black card card B. If card A is drawn, then there is a 100% chance the other side is red. If card B is drawn, then there is a 0% change the other side is red. Since there are only 2 cards, this averages out to a probability of 1/2. Another way to think of it is that since you can see one side of the card, there are only two options. It can be red if it is card A or it can be black if it is card B. Only 1 of the 2 options is red, so you have a probability of 1/2. The probability is not the number of red sides over the total number of sides. 3/4 would be the probability of drawing a red face, not the probability of the option side of a drawn card being red. Nor is the answer 2 out of 3. The probability is the number of red sides over the number of options. Only 1 of the 2 cards has a red back.
curtiss82**
Dec 28, 2003

I ment to say opposite side not option side in the second to the last line in my previous comment. I would also like to explain why it is not 2/3. You can only draw the red/red card once, so just because it has 2 red sides, doesn't mean it gets counted twice. It would only be 2/3 if there were 2 red/red cards and 1 red/black cards.
cnmne*us*
Jan 14, 2004

Ditto to Curtiss82. There are only two cards. If both cards are laid on a table with the red sides showing, then you would only have a 1 in 2 chance of drawing the card with black on the other side.
fatlingon
Feb 01, 2004

Itīs like you have 4 colored marbles, 3 of them beeing red and one beeing black. If you take away two red ones there is a 50/50 chance of the next marble beeing either red or black, the answere in this teaser is based on the flawed assumption that it would take away only one red side. When the card in the teaser is drawn the other cardīs red side(one of them or the only red side) loses itīs value in the final equation and therefor itīs still only a 50/50 chance.
fatlingon
Feb 01, 2004

To put it more breafly.. Just because you can flip one of the cards over(and still beeing the same color) doesnīt mean the odds increases. Itīs still only ONE or TWO cards.
fatlingon
Feb 01, 2004

Typo. one OF two cards... not one OR two cards.
jmcleod*
Jun 02, 2004

fatlingon, you are missing a key element. If you ask what is the probability of chosing the card with tow red dots, it is 1/2. But in this problem you ask what is the probability that you choose the card with one black side pluss what is the probability that you will see the red side if you look at only one side, 1/4. Given that you are looking at one red dot now you know that the chance that thire is a black dot on the other side has to be less than 50%.
krishnanAin*
Jun 08, 2004

Hmm.. Interesting discussion but I have to agree with the original answer of 2/3. In answering 1/2 you are making the assumption that picking each of the cards was equally likely which is wrong in this case. Consider this question...You have two cards, one red on both sides, one red on one side and black on the other. You reach into a hat and pull out one card, viewing only the top. It is red. What`s the probability that you picked the first card? In this case, it is clearly more likely that you picked the first card and hence the answer cannot be 1/2.
curtiss82**
Jun 09, 2004

Krishnan, you are wrong. It does not matter what the color of the card is in your example, there are only 2 cards, so each card has a probability of 1/2 of being drawn. The color of the sides has a different story depending on what you know about the card drawn. If you can't see the card that was drawn, there is a 3/4 chance that the top is red. Another way of stating that is that there is a 3/4 chance of drawing card and having the top being red. Once you see that the top is red, there are only two choices for the opposiste side (it can either be red because one card has two red sides or it can be black because the other card has different colored sides.) Since there are only two options for the opposite side, there is a 1/2 chance that the opposite side is red. Likewise, there is also a 1/2 chance that it is black.
curtiss82**
Jun 10, 2004

If you want to do this in a more mathematical way, there are three sides that are red, so there is a 3/4 chance of drawing a card with the top being red. After that, there are 3 unknown sides, 2 of which are red thus the probability is (2/3). The total probability of drawing a card with the top being red and the opposite side being red is equal to (3/4)*(2/3) which is 1/2. The people who think that it is 2/3 are only looking at the second half of the total probability, which is not correct. The 2/3 probability is before the card is even drawn, when the top of the card is unknown. The 2/3 probability, and thus the total probability shown above, is an approach that views each side of the card individually (it excludes the fact that each card has an opposite side.) The answer given in the teaser is the equivalent of having 4 one sided cards and asking what is the probability that the 2nd draw is red. In that situation, the answer is 2/3. But since the cards are 2 sided, you have to look at the total probability, the probability of drawing 2 red cards.
krishnanAin*
Jun 11, 2004

No, Curtis, I disagree. The correct answer is 2/3 only. This is a problem of conditional probability where the condition is that the side viewed is red. You have calculated the total probability correctly to be 1/2. To find the conditional probability, you have to divide the total probability by the probability of the conditioning event. In this case, the probability of the conditioning event is 3/4 which is the probability that the side viewed will come out to be red. Hence the required answer is 0.5/0.75 = 2/3.
curtiss82**
Jun 12, 2004

You contradict yourself. The conditional probability is 1/2 because there are only two options (it can either be red or black, not red, red, or black.) The question eliminates the red, red, black options by saying the cards are two sided and that you draw one card with the top red. The card can only be the red/red card, or the red/black card, thus the answer is 1/2. 2/3 is not an option since the cards are two sided, they are not single colored marbles that you draw out of a hat. There are 3 red sides, you know one of them, so you think that only leaves 2, but one of the sides is not independent, it is attached to another side, so it has no chance of being in play. You know the card you drew has a red top, but you also know that the other card in the hat also has a red top. So you actually only have 1 unknown side. It can be the card you drew, or it can be the card in the hat. So the answer is 1/2. It is only when the sides are independant that you get an answer of 2/3. If the cards were one sided and you had four cards in a hat, 3 red, 1 black, and your first draw was red, the conditional probility for the second draw to be red is 2/3. This is because the cards are independant and you have to draw twice. In this teaser, the sides are not independant and you only draw once.
beanie89**
Jun 13, 2004

At first I though the answer to this question was 1/2. Then I read these comments and got confused as to what the answer really was. So, I went to my math teacher and asked her about it. She explained it to me about exactly how BabyB explained it. The answer IS 2/3. My math teacher is the smartest person I know. So if she said it, it MUST be true!!
krishnanAin*
Jun 13, 2004

I don't know how to explain this more clearly but BabyB has given a perfect explanation. Let me try to explain one final time. You have two cards RR and RB. The question states that you pull out one card and then view one of its sides and you see it is red. This situation is equivalent to picking one out of 4 sides with 3 red and 1 black. This means that you could have seen each of the three red sides with the same probability of 1/3 (because you are given that the side you view is red). Out of these three, in two of them, the other side is also red and in one, th other side is black. Hence the probability that the other side is red is 2/3. Now, on to your reasoning that the answer is 1/2. You have made the assumption that both cards are equally likely to have been picked. That is true only when you have no conditioning. But, when you have the condition that you have seen a red face, it distorts the initial probability of picking the cards and the two cards are no longer equally likely to have been picked up. Clearly, if I said that I chose one card and saw one of its sides at random and found it to be red, you would find it more likely that I picked the RR card instead of the RB card. In fact, in two out of three ways of seeing a red, I would have pickked the RR card. So the answer is 2/3.
tsimkinAus*
Nov 26, 2004

The answer is most definitely 2/3. Krishnan is absolutely right (making curtis wrong...) The fact that there are two possibilities does not make it 50%. Today I might win the lottery, or I might not, but that doesn't mean that I am 50% to do so. This is a Bayes' Theorem problem, one of conditional probability. Pr(double red|red side) = Pr(red side and double)/Pr(red side) = (1/2)/(3/4) = 2/3. Curtis, your set up of the answer was a little off on the math (you multiplied these numbers), but this is indeed the right answer. Good question.
nkatha23
Nov 09, 2005

Love math but never had a thing for probability...Gud one though
hidentreasure**
Nov 29, 2005

Easy.
stephiesd**
Dec 09, 2005

You people make it sound so hard. here's my explanation:

you have three cards. they all have at least one red side. two of them have two red sides. so out of the 6 sides, 5 are red. which really doesn't make a difference.

since the probablity that you have the red/black is 1/3 since one third of the cards is red/black. therefore, 2/3 of the cards is red/red. since you know that one side of the card is red, chances that the other side is red is 2/3 sicne that is the value of the red/red cards.
mr_brainiac*
Jan 01, 2006

I'm gonna go test this out by actually doing it at least a thousand times and come back with an answer.
mr_brainiac*
Jan 02, 2006

There are 2 cards and therefore a 1/2 chance of picking each card, but one of those cards has 2 ways to show a red face, the other has only 1. Therefore, 1/3 of the time that you see a red face first, it is on the card with a black face opposite. 2/3 of the time that you see a red face first, it is on the double red card and there is another red face opposite it.
sammy66*
Jan 29, 2006

acually, it would be 3/4.
YVAU
Mar 02, 2006

The answer is 1/2. Forget sides for a moment. The hat contains 2 cards: card A (r/r) and card B (r/b). You picked, so you are now holding 1 card in your hand. It could be card A (50% chance of that) or card B (50% chance of that). If it's card A, then the other side is red (100% chance other side is red if you picked card A). If it's card B that you picked, then other side is black (0% chance red). So the answer is 0.5*1+0.5*0=0.5.
The 3/4 answer is actually the answer to another question: if you pick a card in the hat and look at the top, what is the probability that the top will be red, i.e. what is the probability that you find yourself in this situation to start with. That would be the 3/4.
tsimkinAus*
Mar 03, 2006

YVAU -- While I agree that you have a 100% chance of seeing red if you picked the double-red card, and a 0% chance of seeing red if you picked the red-black card, I disagree that these two things are equally likely. Since the double red-card has two red sides, and the red-black card only one, once you see the red side, it is twice as likely that this is because you have the double-red. Expand the problem. Imagine you have two decks of cards -- one which is a normal 52-card deck, and the other which is the Ace of Spades 52 times. Now I remove the decks from your sight, shuffle one up, and turn over the top card. Lo and behold, it is the Ace of Spades. I would hope you would think that this deck is more likely to be the one that is only Aces of Spades, and I am pretty likely (52/53 chance, in fact) to turn over another Ace of Spades if I turn over another card.
YVAU
Mar 03, 2006

Hi tsimkin, thanks for your reply. I still dont' agree with you but thanks to your 2-deck example, I understand what I think is your confusion. You are indeed more likely to pick an ace of spade from deck2 than from deck1 (52 times more likely) in your 2-deck example. However, that is not the question asked by the problem. The confusion comes from the double use of the color red when in fact these are 2 separate incidents. Allow me offer an analogy that I hope will make sense: you must pick a ball from a bag: The bag contains 1,000 black balls and 1 red; If you pick a black ball, you'll be killed. If you pick the red ball, you'll be asked to go to another bag that contains 2 balls: yellow and blue. You find yourself in front of the 2nd bag; What is your probability of getting blue vs. yellow. It's 50%. Now, replace yellow and blue in the second bag by red and black. It's still 50%, of course, these are the same balls with just different colors. It does not matter that you had to defeat incredible odds and death itself to come to the second bag. You are in front of the second bag, so now the problem starts. Once you are there your probability of picking red is 50%. And that's what the problem is asking. If the question was: what is prob of picking red twice in a row, of course it would not be 50%.
YVAU
Mar 03, 2006

I realize at this time that we have at least 2 different strong beliefs on what the answer is. I cannot guarantee that I am right, even if I believe so. I think the only way to settle the argument is to actually run the experiment. Has anyone tried it? I will try and I will come back and report. And if I am wrong, so be it, I will have learned something. And if I am right, I'll gloat for days.
YVAU
Mar 03, 2006

I did the experiment (computer). I was wrong. Doing the actual experiment helped me understand where I screwed up; I must say that the explanation of the 2/3 crowd is not entirely compelling, although their result is certainly accurate. I used about 200 points of data, and the answer came back at 0.71; it's not 2/3 but it's pretty close. It's certainly not 0.5; doing the experiment, I realized why I was wrong. If you are looking at a red card, the other side might be another red (card 1) or a black (card 2). There is indeed 50% chance to pick card 1 or card 2, that part has not changed. However, you get 2x as many chances to pick 1 red side with card 1 than with card 2. So card 1 carries 2x as much weight, while card 2 carries only 1 (card 1 is red whether you are looking at the top or the bottom). If I wanted to super-picky I could say that the problem states that you are looking at the top of the card, therefore prob is 50%, not 2/3, but I think it;'s just the wording and I'll concede that if you pick a card, and you're looking at a red side, probability that other side is red is 2/3, not 50%. I stand corrected. Please note however, that the probability of picking card 1 or 2 is 50%; it's because you can pick card 1 2 ways (while you can only pick card 2 1 way) that card 1 carries 2x as much weight.
jj_is_coolAus*
Mar 29, 2006

why did this get accepted
brainjuice**
Mar 31, 2006

i dunt understand the question..
Swordoffury1392*us*
May 27, 2006

I agree, the answer is definately wrong!
Jimbo*au*
May 28, 2006

This teaser got accepted because the answer is "definitely" (you can check the spellling on an on-line dictionary) correct. It should be rather obvious when many of the objectors change their minds and agree after some rational debate. Personally I love these conditional probability teasers. They certainly stir up some passionate debate but it would be a tragedy to censor them (remove them from the site) on the basis of one or two uninformed opinions.
tonjawithaj*us*
Jun 07, 2006

Wow! I have learned something today!
xdbtcpAus*
Jun 22, 2006

cool teaser - didn't read the comments, but obviously some people have a lot of spare time....
rrn0rrnrrnY*us*
Sep 04, 2006

As soon as I read this and the answer I knew there would be a bevvy of debate.
ciotogca*
Oct 26, 2006

The answer is wrong because the question was not worded properly. As soon as you say "You reach into a hat and pull out one card, viewing only the top. It is red." you are applying a deterministic event that invalidates the question as a question of probability.

To phrase it properly you should say "suppose you reach into a hat..." - this negates the cases where the face is black without invalidating the question.
Eshootzi_scrs
Apr 24, 2007

Ok I realize no one will read this and it will probably be removedbut...You reach into a hat and pic a card the face is red. what are the odds the other side is blue?
Or you picked and orange.
Its an impossible question wihtout given facts. Are there any black black cards? Are we dealing with only two colors? How many cards in the hat and how many are red/red vs red/black?
I f all the cards are either red/red or red/black then it is 50/50 that the other side is red. It can only be red or black its that simple.
Eshootzi_scrs
Apr 24, 2007

I know it says you have two cards but it doesnt say they are the two in the hat. Still assuming that they are...50% you picked the red/red card. 50% you picked the black.
4 sides..3 red 1 black. You see red then only 3 sides left...2 red 1 black. 2:1 or 2/3.
But if you think about it like they are standard playing cards you have on red card one black card and the back of all the cards is red.
You would then know if it was the back you were seeing so it would then be 50 50. Love the ones that are debatable.
Eshootzi_scrs
Apr 24, 2007

One more thing. If anybody has any doubts about this we could meet and run the experiment 100 times for $100 per draw, you take black each time I'll take red. BTW, you have to stay with black even if I draw the black card black up. Ok you can win that one (once) (25% win rate and up to 33% on red up(75% of the time) =25 +(33%x75%) =25%+25% =50% )
So even if I conceded the black up draws its even, If we play only red up I think I win alot. $6666.67.
leftclickAau*
Jul 29, 2007

First of all, great teaser, and the fact that people are arguing about the answer shows that it has inspired thought, which is obviously a good thing

Secondly, the answer is correct. I had to think about it for quite a while before coming to that conclusion.

The question asks "what is the chance that the other side is red". What that really means is "what is the chance that both sides will be red, given that the first side is red." If you think the answer is 1/2, then you are forgetting about the last part.

Since we know that the first face we saw is red, then we know that it is twice as likely to be the red/red card as the red/black card. (This is because every second time we draw the red/black card, it is black face up and therefore doesn't count). Therefore the chance that both sides are red (i.e. that the other side is red) is 2/3 (which is twice as likely as 1/3).

Hmmmm that was really hard to explain and I don't think I did a very good job...
AKMark
Sep 08, 2007

This really chewed me up. I had to go to pencil and paper to get this to work.

Two cards, four sides. R1, R2, R3,B.

If the top is R1, the reverse is R2
If the top is R2, the reverse is R1
If the top is R3, the reverse is B

Since ONLY when the top is R3 is the other side Black, two of three times the other side will be red.

I don't know why I couldn't think this through, but I couldn't. I had to write it down.
LilLadyPunkAca*
Oct 21, 2007

I don't get it... if the red face you have now is R1 , then the other face would either be blackj or red..thus 50% or 1/2 chance.
bbbz*
Oct 10, 2008

answer 2/3 is correct.
I got a new way of explaining it for anyone new stumbling across this one and still thinks its 1/2:

Lets use an example where we change the red/black card to a black/black card. The other is still red/red. You pick a card and see a red side. What is probability that the other side is also red? Wouldn't everyone agree that its 100%? However, using the logic of the 1/2ers would tell you that its 1/2, because "there are two cards and you're just picking one of them"

In my example the other side would NEVER be black becasue everytime you see a black side, the conditions aren't met, the draw is voided, you return the card (the black/black card) and you must pick again until you get a red side! Just as the red/black gets returned half the time its picked in the original problem. This reduces the times the red/black card is "legally" drawn to just once for every two times the red/red card is drawn.

Why can't people see the implications of the condition that the side you see has to be red to begin with?

Trying to think of different ways to explain it. Best i can do. anyone else?
ShadowsAca*
Nov 22, 2008

The answer is correct. Whoever says it's wrong doesn't appreciate the wonderfully confusing world of probability. Top-notch mathematicians say that the answer is 2/3, and I personally agree.
Gale
Dec 31, 2008

Suppose you had four cards in the hat. One with both sides black, one with both sides red and two each with one side red and one side black. Then you randomly pull out one card and notice that it's red on the exposed side.

What then is the probability both sides are red?
tsimkinAus*
Dec 31, 2008

Gale -- it would be 1/2
Gale
Dec 31, 2008

Yes. I agree that it's 1/2.

But, if you structure the puzzle just a little differently you can get a 1/3 result. To do this you have an assistant pick the card at random, look at both sides and then answer the question, "is at least one side red"? , affirmitively
tsimkinAus*
Dec 31, 2008

I agree.
RomanG417Aus*
Jul 28, 2009

Anyone who still thinks it is one-half look at it like this.

The probability of the first side being red and the second side being red(probability of picking out the all red card)=1/2

Probability of the first side being red=3/4
Probability of the second side being red after assuring the first side is red=x
3/4*x=1/2
x=2/3
opqpop
Sep 28, 2010

lol how is there 55 comments on this!

You have 3 possible red faces you're staring at. 2 of them have a red on the other side. Answer is 2/3.
lamentor79
Mar 03, 2011

...revisited 9 years later....

If the question asked the probability of drawing the 2nd card from the bag (prior to seeing the back side of held card) and seeing a red face, then the probability would be 2/3. READ THE QUESTION, PLEASE. READ IT AGAIN, PLEASE!

I can't imagine that there has been so much debate over this question.

Event "A" has already occured, leaving event "B" (asking the color on the other side of the drawn card) to have only two options. How interesting that I have accidentally rediscovered this teaser 9 years later and yet have gained great amusement from it!
lamentor79
Mar 03, 2011

To explain further:

Probability that chosen card is RR is 50% and probability that chosen card is RB is 50%.

Case 1. Red/red, viewing top:

Probability of remaining card showing black is 50%, hence red is 50% (0.5).

Case 2. Red/red, viewing bottom:
Probability of remaining card showing black is 50%, hence red is 50% (0.5).

Case 3. Red/black, viewing RED side:
Probability of remaining card showing red is 100% (1.00).

Probability of remaining card being red equals P(1) + P(2) + P(3) all divided by 3 = (0.5 +0 .5 + 1.0)/3 = 2/3

TA DA! Is it magic? Or, is it misinterpretation of the question?



Back to Top
   



Users in Chat : None 

Online Now: 14 users and 448 guests

Copyright © 1999-2014 | Updates | FAQ | RSS | Widgets | Links | Green | Subscribe | Contact | Privacy | Conditions | Advertise

Custom Search





Sign In A Create a free account