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4 coins are in a bucket: 1 is gold on both sides, 1 is silver on both sides, and 2 are gold on one side and silver on the other side.

I randomly grab a coin from the bucket and see that the side facing me is gold. What is the probability that the other side of the coin is gold?

I had thought that the probability is $\frac{1}{3}$ because there are 3 coins with at least one side of gold, and only 1 of these 3 coins can be gold on the other side. However, I suspect that the sides might be unique, which derails my previous logic.

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This question is very similar to this earlier one, where cards were used instead of coins. –  Sasha Sep 11 '12 at 0:22
    
near-duplicate of Probability problem –  MJD Sep 11 '12 at 2:09
    
almost-duplicate of Statistics: Bertrand's Box Paradox –  MJD Sep 11 '12 at 2:10
    
And this time with cards: Card probability problem –  MJD Sep 11 '12 at 2:10
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5 Answers

up vote 7 down vote accepted

Here's a more intuitive answer.

You have two random things going on: your choice of coin is random, and the side facing up is random. So in fact you're picking one of 8 coin-sides out of the bag, and each one occurs with equal weighting.

So let's label the sides: G1 and G2, S1 and S2, G3 and S3, and G4 and S4 (where G = gold, S = silver, and they happen to be fused together as I have written them). You choose a random coin-side. It's gold, so, it must be G1, G2, G3 or G4 - all equally probable. The reverses of those coins are, in order, G2, G1, S3 or S4 - all still equally probable, of course. So the probability is 1/2.

This simply corresponds to the fact that, if you pick a random coin-side out of the bag and it's gold, it's twice as likely to be the G1-G2 coin than the G3-S3 coin (because that coin has twice as many gold sides).

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50%. GIVEN that the first side you see is gold, what is the chance that you have the double-gold coin?

Assume you do this experiment a hundred times. In 50% of the cases you pull out a coin and see a gold side; the other 50% you see a silver side. In the latter case we have to discard the experiment and only count the cases where we see gold.

There is initially a 25% chance of double-gold, 25% chance double-silver, and 50% chance half-and-half. We discard the 25 cases where you draw the double-silver, and the 25 cases where you draw a half-and-half silver-side up. So of the 50 cases remaining, half are double-gold and half are gold-up silver-down.

Hence given that you have drawn a coin and see gold on top, there is a 50% there is gold on the bottom.

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If one is careful, one can find an informal but correct argument that gives the right answer. However, it is useful to know how to do a formal conditional probability calculation. One reason is that the intuition can be treacherous.

Let $S$ be the event that the side we see is gold, and let $D$ be the event we have drawn the double-gold coin. We want $\Pr(D|S)$. By a standard formula, we have $$\Pr(D|S)\Pr(S)=\Pr(D\cap S).\tag{$1$}$$

We first find $\Pr(S)$. The event $S$ can happen in two ways: (i) We drew the double-gold, and the side we see is gold or (ii) We drew a "mixed" coin, and the side we see is gold.

To find the probability of (i), note that with probability $1/4$ we draw the double-gold. If this is the case, then the side we see is gold with probability $1$. So the probability of (i) is $(1/4)(1)$.

To find the probability of (ii), note that with probability $2/4$ we draw a mixed coin. given that we do, the probability of seeing the gold side is $1/2$. So the probability of (ii) is $(2/4)(1/2)$.

Thus $\Pr(S)=(1/4)(1)+(2/4)(1/2)=1/2$.

Note that $\Pr(D\cap S)$ is just the probability of (i), which is $(1/4)(1)$.

Now from Equation $(1)$ we find that $\Pr(D|S)=\frac{1/4}{1/2}=1/2$.

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How many coins in the bucket are the same on both sides? 2

How many coins in the bucket are different on the two sides? 2

So probability that you selected a coin that is the same on both sides ( gold in your example) is 1/2.

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That doesn't answer the question because the questions asks GIVEN the observed side is gold what is the probsility that the other side is gold. The coin with two sides silver is out of the question. There are three possibilities all equally likely given the observed side is gold and hence the probability is 1/3 as the OP originally thought. –  Michael Chernick Sep 11 '12 at 1:22
    
Sorry it is not 1/3 because the three possibilities are not equally likely. But a formal argument like Andre's is neede to show that it is 1/2. –  Michael Chernick Sep 11 '12 at 1:27
    
He says he randomly grabs a coin and sees that the side facing him is X and asks for the probability of the other side being X. Given the symmetry between silver and gold, it is irrelevant what X is. He happened to pick gold. I think my simple analysis is correct, not just the answer. –  Robert Miller Sep 11 '12 at 4:23
    
If you condition on a gold coin being picked the set of possible ouycomes does not include the case of two silver. So including the two silver case in the argument is wrong @RobertMiller. –  Michael Chernick Sep 11 '12 at 11:05
    
Robert, if you remove the coin with silver on both sides, the answer to the original question is still 1/2. But the probability that the selected coin will be the same on both sides goes down to 1/3. So your analysis would give 1/3 as the answer, which is incorrect. –  Deedlit Sep 13 '12 at 2:23
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I randomly grab a coin from the bucket and see that the side facing me is gold.

As any good brain-teaser, it distracts you from the fact that, for all purposes, you randomly selected a coin side that happened to be golden (4 total), not a coin with at least one goden side (3 total). There is 2/4 chance of having selected a coin with both sides golden, and 1/4 chance each for the G/S coins - meaning that there's a total 1/2 chance flipping the coin would reveal the other side is golden as well.

This is a variation of Bertrand's box paradox.

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