# 2 slightly different situations in which 2 coins are tossed. Does the knowledge of an observer effect the probabilities of the outcomes?

Situation A: Once only, I toss 2 identical fair coins and don't look at the outcomes. A truthful observer looks at one of the coins and tells me that at least one of the coins is a head.

Situation B: Once only, I toss 2 identical fair coins and don't look at the outcomes. A truthful observer looks at both of the coins and tells me that at least one of the coins is a head.

In A, what is the probability that there are 2 heads?

In B, what is the probability that there are 2 heads?

Aren't both probabilities 1/2?

EDIT

Let me refine the question. The agreement I have with the observer is this: 1) I will toss 2 identical coins.
2) In situation A a third party will cover the coins with a cloth. The observer will look only at the outcome of the coin that comes to rest nearest him and report it to me. What is the probability that the second outcome will be the same as the first?
3) In situation B the observer will look at both outcomes and report only the state (heads or tails) of the coin that came to rest nearest him. What is the probability that the second outcome will be the same as the first?

2nd EDIT

Changing 3) above to: 3) In situation B the observer will look at both outcomes, choose one of them, and truthfully report, "There is at least one heads", or "There is at least one tails", whichever is the case. What is the probability that the second outcome (that of the other coin) will be the same as the first?

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You've just confused the issue with your edit, NotSuper. In situation B, why should the observer look at both coins, when the state of the farther coin is irrelevant? It might as well be covered by your cloth. This is really beginning to look like a Monty Hall debate. – TonyK Dec 24 '10 at 16:19
@TonyK So what are the probabilities for A and B as of my first edit? I'll now edit again and have the observer look at both coins in B and pick one, about which he will tell me, if it's heads, "There is at least one heads", or if tails, "There is at least one tails". What is the probability of the other outcome being heads or tails, respectively? – NotSuper Dec 24 '10 at 18:51
I give up...... – TonyK Dec 24 '10 at 19:04

No. There are four possibilities with two coins. For the time being identify one coin as X and one coin as Y. You have either

X = 1, Y = 1

X = 0, Y = 1

X = 1, Y = 0

X = 0, Y = 0

(where 1 is heads and 0 is tails). A truthful observer who only looks at coin X and ignores coin Y will say: "there is at least one head" in cases 1 and 3. A truthful observer who only looks at coin Y and ignores coin X will say: "there is at least one head" in cases 1 and 2. In either of those situations, only one of the two cases have the other coin as a head. So the probability of two heads is 1/2.

On the other hand, a truthful observer who looks at both coins will say: "there is at least one head" in cases 1, 2, and 3. Out of those only 1 case has two heads. So the probability of that is 1/3.

By choosing only one of the coins to look at, you are looking at the probability

$$\mathbb{P}\{ X = 1 | \mbox{ Observer chooses }Y\mbox{ and }Y = 1\}$$

plus the same with $X$ and $Y$ swapped. Since $X$ and $Y$ are independent, it is equal to

$$\mathbb{P}\{X = 1\} \cdot \mathbb{P}\{\mbox{ Observer chooses } Y\} + \mathbb{P}\{Y = 1\} \cdot \mathbb{P}\{\mbox{ Observer chooses } X\} = 1/2$$

In the case where the observer sees both coins, you are looking at the conditional probability

$$\mathbb{P}\{ Y + X = 2 | Y + X \geq 1 \}$$

that the sum of the variables $Y$ and $X$ is 2 when we know their sum is at least 1. Clearly the conditioning is not independent of what you are testing! You can establish that the above is 1/3 by counting the total number of ways $A+B$ can be at least 1 as was done above.

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A is easy.
In B, if you know in advance that the observer is just going to let you know whether the number of heads is zero or non-zero, then the probablity is 1/3.
If you don't know in advance what the observer is going to reveal, then the probability calculation involves the observer's strategy (what information the observer intends to reveal depending on the number of heads). So you can't calculate the probability. This is at the heart of the Monty Hall problem.

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 With my edit I believe I've removed any possible analogy with the Monty Hall problem. How about it? – NotSuper Dec 24 '10 at 13:20

I'm posting this as a separate answer so as to stop bothering Willie with incessant comment notifications. Of course, the other answers that have been posted are completely correct, and I'm only trying to help you reconcile your intuitive reasoning with their results.

In situation A, the information you obtain about the state of the coins is, "The first coin is heads."

In situation B, the information you obtain is, "At least one of the coins is heads."

These are not equivalent; the former is not implied by the latter, because in B you cannot conclude that the first coin is heads. In other words, the observer's statement in conjunction with your knowledge of the observer's strategy enables you to make a stronger conclusion about the coins in situation A.

If you wish, I can clarify (a) how the observer's strategy affects things, as TonyK has also mentioned, and/or (b) why in situation B the probability is 1/3 and not 1/2 as you expect.

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