### Brain Teasers

# Money in Water

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

You drop two coins into a muddy river by mistake. You saw one falling in, a 1 Euro coin. You know the other one had to be either a 1 Euro coin, or a 2 Euro coin.

After searching and pulling out one, and discovering it's a 1 Euro coin, what is the chance that the other is also a 1 Euro coin?

After searching and pulling out one, and discovering it's a 1 Euro coin, what is the chance that the other is also a 1 Euro coin?

### Hint

It's not 50/50!### Answer

2/3There are 3 possible combinations after you pull out a 1 Euro:

A 1 Euro was in the river initially, and you pull out the second one you saw fall in; therefore a 1 Euro is in the river still.

A 1 Euro was in the river initially, but you pull it out; therefore a 1 Euro is still in the river.

A 2 Euro fell in first, you pull out the 1 Euro you saw fall in; therefore a 2 Euro is still in the river.

These three states are all equally likely, so there is a 2 in 3 chance of pulling out a 1 Euro.

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## Comments

combinations:

1€, 2€

1€, 1€ - you pick up the one you saw first

1€, 1€ - you pick up the one you didn't see first.

2/3 combinations give the satisfactory(?) result of getting two 1€ coins. =)P(2x1€)=2/3

1€, 2€

1€, 1€ - you pick up the one you saw first

1€, 1€ - you pick up the one you didn't see first.

2/3 combinations give the satisfactory(?) result of getting two 1€ coins. =)P(2x1€)=2/3

Nice!

It's not 66% people. I understand why you think that. It's a 50% chance. Because you already picked out the 1 "1 euro" coin, that leaves 1 of 2 possibilities for the other coin. That's the bottom line. It's either a 1 euro or a 2 euro. 50% chance of it being either/or.

chamberpot you are wrong. List all combinations, like my friend did here in the first comment. Find the ones that fit. Put that number over the first number and you will get 2/3

How about you think of doing this 100 times. (remember all the numbers are statistics)

Each time throw if 1 €1. Then randomly throw in €1 or €2

Pick out a coin.*This is the crucial part*

IF YOU PULL OUT A €2 *FIRST* THEN YOU DON'T COUNT THAT ONE.

Count the amount of times you do this, and then count the amount of times you pull out another €1

In 100 times, you will use 150 €1 and 50 €2

On the first go you will pull out €1 3/4 times (75) and €2 1/4 of 100 times. (25) On the second go you will pull out €1 3/4 of the times (75) and €2 1/4 of the time (25)

Now the odds are going to be how many times €1 came out at the end.

At the moment, 75/100!!

Discount the 25 times when you pulled out €2 first. So lets say you only did the experiment 75 times really

Each time you pulled €2 the second coin you pulled out had to be €1 so if you don't count those 25 times (100-25) (BECAUSE YOU ARE TOLD IN THE PUZZLE €1 COMES OUT FIRST), thats 25 times LESS you pulled out €1 the second time (75-25=50)

SO the odds are now: 50/75 = 2/3

How about you think of doing this 100 times. (remember all the numbers are statistics)

Each time throw if 1 €1. Then randomly throw in €1 or €2

Pick out a coin.*This is the crucial part*

IF YOU PULL OUT A €2 *FIRST* THEN YOU DON'T COUNT THAT ONE.

Count the amount of times you do this, and then count the amount of times you pull out another €1

In 100 times, you will use 150 €1 and 50 €2

On the first go you will pull out €1 3/4 times (75) and €2 1/4 of 100 times. (25) On the second go you will pull out €1 3/4 of the times (75) and €2 1/4 of the time (25)

Now the odds are going to be how many times €1 came out at the end.

At the moment, 75/100!!

Discount the 25 times when you pulled out €2 first. So lets say you only did the experiment 75 times really

Each time you pulled €2 the second coin you pulled out had to be €1 so if you don't count those 25 times (100-25) (BECAUSE YOU ARE TOLD IN THE PUZZLE €1 COMES OUT FIRST), thats 25 times LESS you pulled out €1 the second time (75-25=50)

SO the odds are now: 50/75 = 2/3

Good one, but someone did this with cars too, I don't know who did it first.

I did a similar one with cars in a tunnel, but a different answer. The logic is similar. Is that the one you mean?

HUH?

Jun 17, 2002

wow I must be really dumb, I thought it had to be 50% chance, it sems so logical, but how you guys worked all that probability, and what ifs out is brilliant

cath take it from me... just because you show all your fancy math and crap your wrong... try taking a college level prob and stat course ... then you will learn about statistics my friend

The answer is 50%. The author of this "teaser", and those who agree with him are looking at the situation the wrong way. Their logic is flawed.

I have 2 coins; one is a 1Euro coin, and I don't know what the second is because I didn't get a good look at it. I know it has to be a 1E or 2E coin, though.

I'm walking along one day and I accidentally drop my 2 coins in the river. I bend over and pick one up and notice that it is a 1E coin. I now wonder what the second coin is. Is it a 1E or a 2E coin? Hmmmm...

The first thing we need to do is COMPLETELY forget about the first coin. You picked that up and you already know what it is. I want to know what the second coin is. It is either a 1E or a 2E coin. I could take a guess at it, and I would have a 50/50 chance of guessing correctly.

YOU MUST COMPLETELY FORGET ABOUT THE FIRST COIN!!!

If you feel that I am incorrect, please prove me wrong.

I have 2 coins; one is a 1Euro coin, and I don't know what the second is because I didn't get a good look at it. I know it has to be a 1E or 2E coin, though.

I'm walking along one day and I accidentally drop my 2 coins in the river. I bend over and pick one up and notice that it is a 1E coin. I now wonder what the second coin is. Is it a 1E or a 2E coin? Hmmmm...

The first thing we need to do is COMPLETELY forget about the first coin. You picked that up and you already know what it is. I want to know what the second coin is. It is either a 1E or a 2E coin. I could take a guess at it, and I would have a 50/50 chance of guessing correctly.

YOU MUST COMPLETELY FORGET ABOUT THE FIRST COIN!!!

If you feel that I am incorrect, please prove me wrong.

Assuming it's 50/50 the second coin is 1 or 2 Euro, and 50/50 that you pick the first or second coin initially...Scenario 1: Coin #1 is 1 Euro, coin #2 is 1 Euro, you picked coin #1. So the next coin should be 1 Euro.Sceanrio 2: Coin #1 is 1 Euro, coin #2 is 1 Euro, you picked coin #2. So the next coin should be 1 Euro.Scenario 3: Coin #1 is 1 Euro, coin #2 is 2 Euro, you picked coin #1. So the next coin should be 2 Euro.No more scenarios that will fit the story.Hence probability is 2/3.

Look:

1, 1

1, 2

You cannot, repeat, CANNOT draw the 2 first. Therefore, you pick one of the 1s first. There are three of them. Two of them are paired with another 1, one of them is paired with the 2. 2/3 probability.

However, I do understand your point about the 1/2 odds as well. FROM THAT POINT ON, it would be 1/2 odds. However, like the flipping of a coin, you can have cummulative odds. The odds are 1/4 that you'll get heads twice. After getting one head, there is a 1/2 chance of the second head. Still, it was 1/4 odds of getting two. Which number do you believe?

1, 1

1, 2

You cannot, repeat, CANNOT draw the 2 first. Therefore, you pick one of the 1s first. There are three of them. Two of them are paired with another 1, one of them is paired with the 2. 2/3 probability.

However, I do understand your point about the 1/2 odds as well. FROM THAT POINT ON, it would be 1/2 odds. However, like the flipping of a coin, you can have cummulative odds. The odds are 1/4 that you'll get heads twice. After getting one head, there is a 1/2 chance of the second head. Still, it was 1/4 odds of getting two. Which number do you believe?

1, 1

1, 2

You cannot, repeat, CANNOT draw the 2 first. Therefore, you pick one of the 1s first. There are three of them. Two of them are paired with another 1, one of them is paired with the 2. 2/3 probability.

However, I do understand your point about the 1/2 odds as well. FROM THAT POINT ON, it would be 1/2 odds. However, like the flipping of a coin, you can have cummulative odds. The odds are 1/4 that you'll get heads twice. After getting one head, there is a 1/2 chance of the second head. Still, it was 1/4 odds of getting two. Which number do you believe?

This is the same idea and answer as the red and black cards puzzle.

The answer is 50/50. If the question had asked what the chance of pulling a 1 euro coin was before the first pull then the answer would be 2/3(assuming no other coins in the river).

However since the first pull had already occured the only answer is 50/50. There is one coin in the river with 2 possible values, each equally likely (again assuming no other coins in the river).

If the first coin pulled had been the 2 euro coin the probability of pulling a 1 euro coin next would be 100%.

However since the first pull had already occured the only answer is 50/50. There is one coin in the river with 2 possible values, each equally likely (again assuming no other coins in the river).

If the first coin pulled had been the 2 euro coin the probability of pulling a 1 euro coin next would be 100%.

Okay, maybe this will clear things up a bit: use the "given" probability rule - P(A given B) = P(A and B) / P(B). In the problem, A is getting a 1 euro second, B is getting a 1 euro first. The equation simplifies to P(A given B) = P(A) since P(A and B) = P(A) * P(B). Therefore, all we need to consider is P(getting a 1 euro second). This is where we can use the general probability rule of (# of successes) / (# of outcomes) and examine scenarios. cathal has already done this for us, thus getting the 2/3 answer.

Now, for all you people struggling with the 50/50 deal, wolves' flaw is the easiest to spot and explain. you say that the second coin still in the river has an "equally likely" chance of being 1 or 2 euro's. it in fact does not, so the (# of successes) / (# of outcomes) formula does not apply in this situation.

Now, for all you people struggling with the 50/50 deal, wolves' flaw is the easiest to spot and explain. you say that the second coin still in the river has an "equally likely" chance of being 1 or 2 euro's. it in fact does not, so the (# of successes) / (# of outcomes) formula does not apply in this situation.

(# of successes) / (# of outcomes) Will always hold for Probability. I've been watching all the comments with interest! Basically, three possibilities for the second coin.

1 Its the two Euro,

2 its a second €1 coin you had but didn't know,

3 the second €1 coin you had and didn't know came out first, so you pull out the €1 coin you know you had.

2/3 of getting €1.

Reading back over my original solution, maybe its a bit confusing, sorry if it is.

1 Its the two Euro,

2 its a second €1 coin you had but didn't know,

3 the second €1 coin you had and didn't know came out first, so you pull out the €1 coin you know you had.

2/3 of getting €1.

Reading back over my original solution, maybe its a bit confusing, sorry if it is.

Cath is right- it's 2/3. There are two things the coin could be - 1E or 2E- but there are 3 possible outcomes.

Chamber you're on the money. All the statisticians in the world can't change the way the question was posed.

This getting ridiculous now. Come on guys. Read the question. Just take a moment and read it carefully. There is one coin in a river. It is either a 1E or 2E coin. It has a 50/50 chance of being either.

It is necessary to FORGET about the first coin.

Look at it this way. There are 2 dead bodies in a river. One is pulled out and it is a male. The other body is either a male or a female. What are the chances it is a male? What are the chances it is a female? - 50/50!

It is necessary to FORGET about the first coin.

Look at it this way. There are 2 dead bodies in a river. One is pulled out and it is a male. The other body is either a male or a female. What are the chances it is a male? What are the chances it is a female? - 50/50!

Sorry farrell but your logic is flawed. You cannot simply forget about the first event (the first coin) when determining probabilities for the second event (the second coin) as the two events are not independent. Suppose the first coin drawn from the river was a 2E coin. In this case the 2nd coin MUST be a 1E. Hence, the probability that the 2nd coin is a 1E is clearly influenced by the result of the 1st event. Your analogy with the bodies in the river is a different problem as you have no information about the sex of either body. Hence, in that situation the two events are independent. The sex of the first body has no influence over the sex of the second, so probabilities are 50/50.

Absolutely unbelievable.... My analogy with the two bodies in the water fits perfectly! Let's just assume, as we did in the teaser, that we know the first body is a boy. We pull him out of the water. O.K. great. He's a male. We can clearly see that there is a second body in there, but we do not know what sex it is. We find ourselves staring into the river wondering what the sex of the second body is. We have one object with 2 possibilities, male or female. We have a 50/50 chance of getting it correct. The fact that the first one was a male has no impact whatsoever on the sex of the other. Cut and dry, that's the bottom line.

Farrell2k, you have to consider that a. you already *know* one of the body is male, and b. you *do not know* that which body you fish out first. You have 50% chance of fishing out the first body, in which case you have 50/50 of finding the next body male or female. You *also* have a 50% change of fishing out the second body, in which case the next body is 100% male. Combine these two probabilities, and you get the 2/3 answer. Which is what others have repeated so many times: you cannot forget about the first event.

I think you all might be wrong, if you focus on the second part of the question i.e. ‘You know the other one had to be either a 1 euro coin, or a 2 euro coin’. The question that you must ask how do you know and what reference are you using. If your pocket is the reference, then you must have known what was in it originally. So if you picked up the 1 euro coin, you just compare back what is left in you pocket. You will then know that it is 100% chance that the other one in the river is a 1 euro coin, or a 0% chance that the other one in the river is not a 1 euro coin. (The only exception you might get with this approach is when somebody else already drop some 1 euro coins previously to this event into river)

which brings us back to my statement. Example:

1. everyone should agree that the probability of pulling a 1 unit coin from a river is 50/50 if only one of each is present.

2.with 2 1unit and 1 2unit coins in the river the probability that the second coin pulled is a 1 unit coin is 2/3.

If the question had been what is the probability that the second coin pulled would be a 1 unit coin the answer would be 2/3 but since the question was "now that you know the remaining coin is either a 1 or 2 unit coin what is the probability of pulling the 1 unit coin."

The answer to the specific question asked is 50/50.

1. everyone should agree that the probability of pulling a 1 unit coin from a river is 50/50 if only one of each is present.

2.with 2 1unit and 1 2unit coins in the river the probability that the second coin pulled is a 1 unit coin is 2/3.

If the question had been what is the probability that the second coin pulled would be a 1 unit coin the answer would be 2/3 but since the question was "now that you know the remaining coin is either a 1 or 2 unit coin what is the probability of pulling the 1 unit coin."

The answer to the specific question asked is 50/50.

At point A we have 2 coins in a river. At point B we have 1 coin in a river. Start from point B. The coin is either a 1E or a 2E. 50/50.

Myself, wolvesheart, and a few others are correct with 50/50.

However, I do commend the 66% percenters for their critical thinking.

Myself, wolvesheart, and a few others are correct with 50/50.

However, I do commend the 66% percenters for their critical thinking.

the 'correct' answer is 50/50 chance.

Think of this another way. When you flip a coin it has a 50/50 chance for heads or tails. If you flip it three times in a row and get heads three times, the 4th flip still has a 50/50 chance of being heads (not a 1 in 16)

CORRECT answer 50/50 chance.

Think of this another way. When you flip a coin it has a 50/50 chance for heads or tails. If you flip it three times in a row and get heads three times, the 4th flip still has a 50/50 chance of being heads (not a 1 in 16)

CORRECT answer 50/50 chance.

Hopefully the definitive comment:

The probability of taking two 1€ coins out is 2/3. This is the probability BEFORE you dip your hand in the first time.

Once you have the first 1€ coin, the probability you are calculating is then 'Of the two remaining coins, what is the likelyhood of you fetching another 1€ out of the water?'. The answer to that question is 50% as you are only dealing with two coins by now. You cannot calculate a probability of an already happened event (ie what are the chances of the first coin being 1€) - that is 100%. Unless it has already rolled down the riverbank and back into the water!

The probability of taking two 1€ coins out is 2/3. This is the probability BEFORE you dip your hand in the first time.

Once you have the first 1€ coin, the probability you are calculating is then 'Of the two remaining coins, what is the likelyhood of you fetching another 1€ out of the water?'. The answer to that question is 50% as you are only dealing with two coins by now. You cannot calculate a probability of an already happened event (ie what are the chances of the first coin being 1€) - that is 100%. Unless it has already rolled down the riverbank and back into the water!

I can't believe this is so hard for some people. Forget about 2/3 or 50/50 just think equally likely, less likely or more likely.

Because you know that you had a 1 euro, and because you didn't pull out 2 euro there is a higher chance the 2nd coin is 1 euro.

If you KNEW you pulled out the 1st 1 Euro THEN the 2nd event is 50/50, but because you MAY have pulled out a 2nd 1euro, this increases the chances for the 2nd event being a 1 euro.

Is it equally likely that the second coin is 1 or 2? THe answer is NO. Its not just about 2 coins. Its about *THREE*

The chance of pulling out 1 euro second depends on the first event (THIS IS WHAT EVERYONE WHO DOESN'T GET IT IS MISSING)

Because of the first event, there is still an equal probability that a 1 or 2 euro fell in (you don't know which you pulled out). But, there is a 1/3 that the 2nd 1 euro coin has been pulled out 1st. There is a 1/2 that the 1st 1 euro has been pulled out.

Because you know that you had a 1 euro, and because you didn't pull out 2 euro there is a higher chance the 2nd coin is 1 euro.

If you KNEW you pulled out the 1st 1 Euro THEN the 2nd event is 50/50, but because you MAY have pulled out a 2nd 1euro, this increases the chances for the 2nd event being a 1 euro.

Is it equally likely that the second coin is 1 or 2? THe answer is NO. Its not just about 2 coins. Its about *THREE*

The chance of pulling out 1 euro second depends on the first event (THIS IS WHAT EVERYONE WHO DOESN'T GET IT IS MISSING)

Because of the first event, there is still an equal probability that a 1 or 2 euro fell in (you don't know which you pulled out). But, there is a 1/3 that the 2nd 1 euro coin has been pulled out 1st. There is a 1/2 that the 1st 1 euro has been pulled out.

Cathal - re-read your question.

You drop two coins in one of which is a 1€, which you retrieve. Of the remaining coin, all you know is that it may be either a 1€ or 2€ coin. It is irrelevant what came out first as you are only certain of 50% of the coins. You have eliminated one of them, so the question is 'What is the probability that the next (and other) coin will be 1€, given that it can only be 1€ or 2€?' as these two are mutually exclusive then the answer, for the next delve can only 50%. It is not dependent on the first retrieval.

An easier, and drier solution would be to see which coin you have left in your pocket!!

You drop two coins in one of which is a 1€, which you retrieve. Of the remaining coin, all you know is that it may be either a 1€ or 2€ coin. It is irrelevant what came out first as you are only certain of 50% of the coins. You have eliminated one of them, so the question is 'What is the probability that the next (and other) coin will be 1€, given that it can only be 1€ or 2€?' as these two are mutually exclusive then the answer, for the next delve can only 50%. It is not dependent on the first retrieval.

An easier, and drier solution would be to see which coin you have left in your pocket!!

You re-read it. You drop a 1 euro in definitely. You definitely pull out a 1 euro but they are not necessarily the same 1 euro coin.

I give up.

I give up.

About time.

I missed this part: "as these two are mutually exclusive"

They are not, they are dependant. You have three options: you can pull out the first 1 euro, the other possible 1 euro, or a possible 2 euro. If you pull out the 2 euro first, the 2nd is certain to be a 1 euro. If you pull out a 1 euro (as happend) its UNcertain what the second coin is. Its not my fault you can't understand this. Go find a maths book

They are not, they are dependant. You have three options: you can pull out the first 1 euro, the other possible 1 euro, or a possible 2 euro. If you pull out the 2 euro first, the 2nd is certain to be a 1 euro. If you pull out a 1 euro (as happend) its UNcertain what the second coin is. Its not my fault you can't understand this. Go find a maths book

How silly of me! I thought this was an intellectual debate instead of a means to score points at a personal level. Never mind - to continue...

Substitute the first 1€ by a RED counter, the second 1€ by a BLUE counter and the 2€ by a YELLOW counter. (This is equivalent as you state there are two 1€ coins with equal probabilities of being present.) Re-phrase the question "You have three coloured counters you drop two of them. The first you retrieve is red. What is the probability that the next one is either red or blue?" Given that these are equivalent problems, what is the answer? To paraphrase - you pull out a 1€ (as happend)(sic) it is CERTAIN that the other is a 1€ or a 2€ ie 50%. QED

Oh - and I have lots of maths books, thanks - and I've read them, and I've understood them - shall I scan my certificates? No - let's not go down that route, why not email the whole page to Mensa instead?

Substitute the first 1€ by a RED counter, the second 1€ by a BLUE counter and the 2€ by a YELLOW counter. (This is equivalent as you state there are two 1€ coins with equal probabilities of being present.) Re-phrase the question "You have three coloured counters you drop two of them. The first you retrieve is red. What is the probability that the next one is either red or blue?" Given that these are equivalent problems, what is the answer? To paraphrase - you pull out a 1€ (as happend)(sic) it is CERTAIN that the other is a 1€ or a 2€ ie 50%. QED

Oh - and I have lots of maths books, thanks - and I've read them, and I've understood them - shall I scan my certificates? No - let's not go down that route, why not email the whole page to Mensa instead?

OK fair enough maybe that comment was approaching the belt line. I apoligise, but most of the other comments don't know what they are talking about. And to be honest yours sounds the same. If it is an intelluctual debate, then why did you say just look in your pocket for the other coin. (for a start there is no other coin in your pocket) Whats wrong with your example is: you don't know you've pulled out the first thing. if you call them red blue and yellow, then you pull out either red OR blue, you don't know which. You want to know the chance of pulling out red or blue again. Have you read all the comments? I think if you do this 100 times is the best. if you drop one 1euro coin in followed by a 1 euro, 50 times and then 2 euro 50 times, and pull out the two, obviously out of the first pairing you will pull a 1 euro out first every time (50) and then you also pull out a 1 euro 50 times. the second pairing you pull out 1 euro 25 times first and 2 euro 25 times first (Conversly you pull out second 2 euro 25 times and 1 euro 25 times)

Because I tell you you pull out 1 euro first it elliminates the 25 times in this experiment the 2 euro comes out first. you have 75 samples where a 1 euor came out 50 times. If you notice when the penny drops (eventually with most people they won't acknoledge they were wrong they just stop replying which is annoying)

Because I tell you you pull out 1 euro first it elliminates the 25 times in this experiment the 2 euro comes out first. you have 75 samples where a 1 euor came out 50 times. If you notice when the penny drops (eventually with most people they won't acknoledge they were wrong they just stop replying which is annoying)

Tell you what, why not rephrase the conumdrum so we can all agree with your answer?

What do you mean rephrase it, I'm only going to say the exact same thing again. That is how I want and meant to phrase it. Its very clear. If you don't understand something, ask me. Tell me whats wrong with my assumption of doing it 100 times. Stastically 1 euro will come out 2/3 times.

As you know one coin is 1€ and you pull a 1€ out with your first delve, discard this occurence, as by definition implicit within the problem,this is certain, 100%. What you are left with is the fact that you may have dropped a 1€ or a 2€ coin in. You imply there is no other possibility. Your question now is "Is the next coin going to be 1€ or 2€, given the fact that it could only be one or the other?". The answer to that has to be 50/50 as there are only two outcomes, each equally likely.

The question, as phrased, only concerns the second dip into the river.

Thats not really true. Thats why you can't understand this. The second pick is related to the first. You see one coing fall in a 1 euro. YOU DON"T NECESSARILY PICK THE SAME 1 EURO OUT. That is why there is a higher chance than 50-50 of picking another one.

The second pick depends on the first one. Before you say anything else tell me whats wrong with what I said about doing it 100 times. You are missing the point, you might not pull out the same coin you saw fall in.

The second pick depends on the first one. Before you say anything else tell me whats wrong with what I said about doing it 100 times. You are missing the point, you might not pull out the same coin you saw fall in.

That was simple!!

Love,

Alecia Moore

(P!NK)

Love,

Alecia Moore

(P!NK)

People struggle more with this fundamental probability problem than any other. The answer is definitely 2/3. It is exactly the same as the red and black card question and many others you will come across. In terms of the mathematics, it's all been covered above. P(A given B) = P(A and B)/P(B) is called Bayes' Rule, and while not at all intuitive, it is a staple of probability and statistics. The only thing I can say is this: Before you pull out a coin, the outcomes are these:

Drop 1E(I)1E(II), Pull 1E(I)=Left 1E(II). Drop 1E(I)1E(II), Pull 1E(II)=Left 1E(I). This is where the hang up occurs. The question says, "After searching and pulling out one, and discovering its [*AN*] 1 euro coin..." You know you have dropped 1E(I) in, but when you pull a 1E out, you absolutely do not know if you pulled out 1E(I) or 1E(II), or even if there is a 1E(II). Your other options of course are Drop 1E(I)2E, Pull 1E(I)= Left 2E. Drop 1E(I)2E, Pull 2E=Left 1E(I). NOW, we pull out *AN* 1E. It could be 1E(I) or 1E(II), if there is one, and we have absolutely no way of KNOWING which it is. Once we have pulled out *AN* 1E, we really have only reduced our possible outcomes by one, not two. If the question had been "You drop in 1E(I) and either an 1E or a 2E. You pull out *THE* 1E(I), what is the probability you pull out a 1E next?" then 1/2 is the answer. However, the question as posed does not say that you pulled out the same 1E(I) you dropped in and therefore the answer is 2/3.

Drop 1E(I)1E(II), Pull 1E(I)=Left 1E(II). Drop 1E(I)1E(II), Pull 1E(II)=Left 1E(I). This is where the hang up occurs. The question says, "After searching and pulling out one, and discovering its [*AN*] 1 euro coin..." You know you have dropped 1E(I) in, but when you pull a 1E out, you absolutely do not know if you pulled out 1E(I) or 1E(II), or even if there is a 1E(II). Your other options of course are Drop 1E(I)2E, Pull 1E(I)= Left 2E. Drop 1E(I)2E, Pull 2E=Left 1E(I). NOW, we pull out *AN* 1E. It could be 1E(I) or 1E(II), if there is one, and we have absolutely no way of KNOWING which it is. Once we have pulled out *AN* 1E, we really have only reduced our possible outcomes by one, not two. If the question had been "You drop in 1E(I) and either an 1E or a 2E. You pull out *THE* 1E(I), what is the probability you pull out a 1E next?" then 1/2 is the answer. However, the question as posed does not say that you pulled out the same 1E(I) you dropped in and therefore the answer is 2/3.

Jul 26, 2002

Okay here is another answer to get everyone riled up again. After fishing out the second coin, you will know 100% what it is.

The answer is 2/3 for the reasons explained by Baby B.

The set of outcomes can only be:

1,1

1,2

(pull1, pull2)

This is clearly explained in the problem. If you did not pull the first Euro, then the situation would be entirely different.

Since you've already taken the first euro, then you have 2 1s left out of 3 total: 2/3.

The set of outcomes can only be:

1,1

1,2

(pull1, pull2)

This is clearly explained in the problem. If you did not pull the first Euro, then the situation would be entirely different.

Since you've already taken the first euro, then you have 2 1s left out of 3 total: 2/3.

Aug 09, 2002

The 2/3 answer is only valid BEFORE the

first coin is retrieved. After that

point, it is 1/2. Without considereation

of the fist coin, the logic behind the

2/3 answer would say that if the first

coin drawn was a 2E, the odds of the

second being a 1E is 2/3, but it is

obviously not. Best regards.

first coin is retrieved. After that

point, it is 1/2. Without considereation

of the fist coin, the logic behind the

2/3 answer would say that if the first

coin drawn was a 2E, the odds of the

second being a 1E is 2/3, but it is

obviously not. Best regards.

I really annoyed myself with this one. I had agreed 100% with all the arguments saying that the answer was 50%, and I sat down to proved it by writing down all the possible outcomes. Call the coins 1a (the known 1 euro), 1b (the unknown coin if it is a 1 euro), and 2 (the unknown coin if it is a 2 euro). Now if the first coin drawn is 1a, then the second coin could be either 1b or 2. The other option is that the first coin drawn is 1b. If this is the case, then the other coin can only be 1a. You can now see that there are actually three possible outcomes, two of which are that the coin will be a 1 euro. This really annoyed me because as I said I was 100% sure that the answer was 50%. The reason it is not 50% is that it is not a simple black and white answer because you don't KNOW that the 2 even exists, so therefore there is more chance that it is a 1 because you KNOW that at least one coin is a 1. I felt I had to put this in writing in simple enough terms because all of the complex mathematical formulas listed above did nothing to convince me the answer was 2/3. Hope this has helped!

My god ! i can't believe so many people can have a problem with this teaser. It is just a matter of dependence. Poor cath...so much trouble to justify an obvious answer..2/3

You sure know how to generate a lot of comments with your conditional probability problems cath. But you do get the answers right. Rathree above has explained the theoretical basis to your correct answer. Let me see if I can simplify it. A person whom you don't know is walking along and in their pocket they have a 1 euro painted blue, a 1 euro painted red and a 2 euro in their pocket. They drop 2 of these coins into the river. You can tell from the size of ONE of them that it was a 1 euro but you didn't see the colour, nor the size of the other coin. What could be in the water? #1 (1 red euro+1 Blue Euro) #2 (1 Red euro + a 2 euro) #3 (1 Blue euro + a 2 euro). The sample space is 3. You reach in and pull out a 1 euro coin. By the way, the other person has walked off so you don't know what the third coin left in their pocket is. Probability is defined as the number of outcomes described (ie you have a 1 euro and there is still a 1 euro in the river) divided by the sample space which we have already found to be 3. Either you pull out a blue euro and there's a red one still in the water or you pull out a red euro and theres a blue one still in the water. That's 2 ways the event can happen out of a sample space of 3 = 2/3. I loved the teaser! Keep it up!

There is just one extra thing that is necessary before you go pulling the coins out of the water, the (1+1) euro event and the (1+2) euro event must be equally likely events. So when the two coins A and B are dropped initially (before you saw one) the may have been one of 4 combinations. (A=1,B=1) , (A=1,B=2),(A=2,B=1), (A=2,B=2). But since you saw that one was a one euro, it cant be the 4th case! That leaves ((A=1,B=1),(A=1,B=2) , (A=2,B=1). Now it MAY appear at this stage that a 1 + 2 euro in the water is twice as likely as a 1+1 case. (If this is so then the answer is 1/2 because all of cath's and my reasoning too is based on equally likely outcomes). But on reading the wording of the teaser carefully it says "you saw one falling in". In other words you know that coin A is a 1 euro. Coin B could be either. Then the outcomes (1,1) and (1,2) are equally likely, the reasoning above is correct and the answer is 2/3.{Lastly, note that this is different from someone saying I dropped 2 coins into the river and one of them was a 1 euro. In this case, either coin A or coin B could have been the one referred to as the 1 euro when dropped in and the situation of the coins would NOT be equally likely events}. I have not been able to convince all of my professional colleagues on this point but I remain convinced myself! Answer = 2/3.

You drop two coins into the river. You know for a fact that one of them is a 1 Euro. There are two possible scenarios. The first scenario is that both coins are 1 Euro. Regardless of which of the two coins you retrieve first, the other has to also be a 1 Euro. So far, 2 chances in 2. The second scenario is that a 1 Euro and a 2 Euro are dropped. Since the coin that you retrieved is a 1 Euro, it is obvious that the other is the 2 Euro. 0 chance in 1. Combine the two scenarios, for 2 chances in 3.

When pullng the money out of the water there are three possible combinations. You could first pull out a 1 euro then a 2 euro. You could also pull out a 1 euro then another 1 euro. Lastly you could pull out a 2 euro then a 1 euro. By telling us what the first coin was you eliminated the last combination thus leaving us with only 2 possible combinations to choose from. If you hadn't told us what the first coin was then the answer to your problem would be correct. Since you told us the first coin was a 1 euro the probability of the of the second coin being a 1 euro is 50/50.

Well done Cathal. Another nice probability teaser that you have to think about.

Nice, but wrong. (I'll assume that the probability of the 2nd coin being a $2 versus a $1 at the beginning of the problem is 50/50; the problem fails to mention that).

You have 2 coins in the river: 1 is a $1 coin, the other one is either a $1 coin or a $2 coin. If you had come back with the $2 coin, then you would know that the other one is a $1 with 100% probability. But you did not. It does not matter what your initial probability of pulling a $1 out of the river was; that's not the question asked here. Now that you have a $1 in your hand, what is the probability that the other coin is a $1. Well, it's 50/50. Because your question is not about the probability of pulling first a $1 coin, and then going back and pulling another $1 coin. The problem is this: you have 1 coin in the river, and it could be a $1 or a $2. And at this point probability is simply equal that it be one or the other. Probability is exactly equal to the initial probability that it was a $1 or a $2 that you dropped in the river. What the other coin is is behind the point. Think of it this way: assume you know that you dropped either a $2 or a $3 with your $1 in the river. You pulled the $1 out. Now what is the prob that you have a $3 in the river? it's 50%. That you call the $3 a $1 is irrelevant. It's 50% should you call it a $1, a $3 or anything.

You have 2 coins in the river: 1 is a $1 coin, the other one is either a $1 coin or a $2 coin. If you had come back with the $2 coin, then you would know that the other one is a $1 with 100% probability. But you did not. It does not matter what your initial probability of pulling a $1 out of the river was; that's not the question asked here. Now that you have a $1 in your hand, what is the probability that the other coin is a $1. Well, it's 50/50. Because your question is not about the probability of pulling first a $1 coin, and then going back and pulling another $1 coin. The problem is this: you have 1 coin in the river, and it could be a $1 or a $2. And at this point probability is simply equal that it be one or the other. Probability is exactly equal to the initial probability that it was a $1 or a $2 that you dropped in the river. What the other coin is is behind the point. Think of it this way: assume you know that you dropped either a $2 or a $3 with your $1 in the river. You pulled the $1 out. Now what is the prob that you have a $3 in the river? it's 50%. That you call the $3 a $1 is irrelevant. It's 50% should you call it a $1, a $3 or anything.

Cath is right. If you list them as known1$/unknown, then this is what you get:

If you pick up the one that is known first, then there are 2 combinations:

1/1 and 1/2

If you pick up the unknown one first, then there is only one combination (because you know that the known is 1$, and you picked up the unknown, which is also $1)

1/1

This leaves 3 combinations:

1/2

1/1

1/1

you cannot have a 2/1 or a 2/2 because you know that the coin before the/ is a 1.

If you pick up the one that is known first, then there are 2 combinations:

1/1 and 1/2

If you pick up the unknown one first, then there is only one combination (because you know that the known is 1$, and you picked up the unknown, which is also $1)

1/1

This leaves 3 combinations:

1/2

1/1

1/1

you cannot have a 2/1 or a 2/2 because you know that the coin before the/ is a 1.

2/3 is clearly the answer. 'Nuff said on that.

But perhaps a more important question: why are you digging in the mud, for Euros?

But perhaps a more important question: why are you digging in the mud, for Euros?

The answer is 1/2 ..... or 2/3 !!!!

The issue I think is that the 2/3 crowd are distinguishing between the two 1E coins. They are in effect saying there are three outcomes:

1,2

1,1

1,1

This to me is really only two outcomes - we are not asked specifically which coin is pulled, just that it be a 1E.

On the other hand , the 1/2 crowd (myself included) do not distinguish between the two 1E coins. Hence, the final coin pull is simply a 1 in 2 chance.

Ambiguity is the great destroyer of such problems....

The issue I think is that the 2/3 crowd are distinguishing between the two 1E coins. They are in effect saying there are three outcomes:

1,2

1,1

1,1

This to me is really only two outcomes - we are not asked specifically which coin is pulled, just that it be a 1E.

On the other hand , the 1/2 crowd (myself included) do not distinguish between the two 1E coins. Hence, the final coin pull is simply a 1 in 2 chance.

Ambiguity is the great destroyer of such problems....

Bonus, you read all of these comments and still get the wrong answer?

Think of it this way: you also know that one of the two coins has red paint on the back of the coin. You don't see the back of the coins when they fall in the water. Now you have the following possibilities for the coins you started with, plus the coin you saw the front of when you dropped them, plus the one you retrieve:

1R + 1 : 1R : 1R

1R + 1 : 1R : 1

1R + 1 : 1 : 1R

1R + 1 : 1 : 1

1R + 2 : 1R : 1R

1 + 2R : 1 : 1

There are six equally likely possibilities. In four of these cases the unretrieved coin is a 1, so there is a 4/6 = 2/3 probability that the unretrieved coin is a 1.

Now just to complete this, the river washes the paint off the coin. This doesn't change the probabilities, but now we can't tell the difference between two 1 Euro coins. We won't know if we pulled the coin that started with paint or not, but it is still a different coin than the one that started without paint.

Think of it this way: you also know that one of the two coins has red paint on the back of the coin. You don't see the back of the coins when they fall in the water. Now you have the following possibilities for the coins you started with, plus the coin you saw the front of when you dropped them, plus the one you retrieve:

1R + 1 : 1R : 1R

1R + 1 : 1R : 1

1R + 1 : 1 : 1R

1R + 1 : 1 : 1

1R + 2 : 1R : 1R

1 + 2R : 1 : 1

There are six equally likely possibilities. In four of these cases the unretrieved coin is a 1, so there is a 4/6 = 2/3 probability that the unretrieved coin is a 1.

Now just to complete this, the river washes the paint off the coin. This doesn't change the probabilities, but now we can't tell the difference between two 1 Euro coins. We won't know if we pulled the coin that started with paint or not, but it is still a different coin than the one that started without paint.

This type of problem is everywhere.

I flip 2 coins, I know one of them is heads. I pick one and see heads. What is the probability the other is heads.

My neighbor has 2 kids. I know at least one of them is a girl. If I ring their doorbell and see a girl, what is the probability the other kid is also a girl.

etc. etc.

It's just 2/3.

I flip 2 coins, I know one of them is heads. I pick one and see heads. What is the probability the other is heads.

My neighbor has 2 kids. I know at least one of them is a girl. If I ring their doorbell and see a girl, what is the probability the other kid is also a girl.

etc. etc.

It's just 2/3.

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