# Puzzling Probability Question?

A friend flips one penny and one nickel and says that at least one is Heads. What is the probability that both are Heads?

Suppose he said that the penny is Heads. Then what is the probability that both are Heads?

What is the difference between these problems? Why should knowing which coin is Heads change the probability that both are Heads?

### 3 Answers

- PuzzlingLv 71 month agoFavorite Answer
This involves conditional probability.

In the first case, there are *3* ways the friend could make that statement about there being at least one head.

penny heads, nickel heads --> HH

penny heads, nickel tails --> HT

penny tails, nickel heads --> TH

Only 1 out of these 3 possible cases has both of them being heads. So the probability is 1/3.

In the second case, there are only *2* ways the friend could make the statement that the penny is heads.

penny heads, nickel heads --> HH

penny heads, nickel tails --> HT

Here 1 out of the 2 options has both being heads, so the probability is 1/2.

UPDATE:

The key point in case 1 is that we aren't told that a *specific* coin is heads or tails. We are instead told that "at least one" is heads. The assumption is the person looked at both coins before making that statement and there are 3 ways that statement could be made truthfully (HH, HT, TH).

If instead the coins are covered and the person looks at one coin and discovers it is heads, they can say "at least one coin is heads". In this case, it *is* equivalent to case 2 because they have looked at a specific coin to make the statement, not both coins. In looking at just one coin, they are saying "I know one coin is heads, but I don't know about the other". In that case it is 1/2 whether the other is heads.

But what about if the coin they looked at is tails? They can't accurately answer the question about whether at least one coin is heads without looking at the second coin. And if they do and they give away the fact that not both coins are heads.

That's why I think we have to assume they just look at both coins to accurately answer the question of whether there is at least one head or not. In that case, the probability is 1/3 of them being both heads.

UPDATE #2:

This is also known as the "Boy or Girl" paradox. I've included a link to the Wikipedia article below. The cases are in the opposite order to here but it's the same thing. In their first case, they say the older child is a boy. In the second case, they say there is at least one boy in this family (of two children). The respective probabilities are 1/2 and 1/3.

- atsuoLv 61 month ago
If the friend said nothing then we can think 4 cases.

case1. (penny,nickel) = (H,H)

case2. (penny,nickel) = (H,T)

case3. (penny,nickel) = (T,H)

case4. (penny,nickel) = (T,T)

When the friend said "At least one is Heads", case4 is rejected. So the probabilities of case1, case2 and case3 becomes 1/3, 1/3 and 1/3. Therefore, the probability that both are Heads is 1/3.

When the friend said "The penny is Heads", case3 and case4 are rejected. So the probabilities of case1 and case2 becomes 1/2 and 1/2. Therefore, the probability that both are Heads is 1/2.

In these two problems, the number of possible cases is changed so the probability is changed, too.

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- Mr. SmartypantsLv 71 month ago
It doesn't really matter. In either case you have ONE coin that could be heads or tails, and the probabilities are 50/50.

Your logic seems right. But the logic saying prob= 1/3 in first case seems right too. But both can't be right.

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It seems counterintuitive, but if you try it yourself, you'll see. Toss a penny and a nickel. If at least one is heads, then count that case. Then check if they are both heads. If you toss it multiple times, about 1/3 of the cases they'll both be heads.