# 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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Of course, part of the problem is semantics. Let me make the intuition clearer for you by adding a 3rd neutral party. Neutral party tosses two coins. And covers up each coin with a piece of cloth. Situation (A) your truthful observer picks one piece of cloth, the neutral party reveals the answer under that piece of cloth. Truthful observer reports back to you. Situation (B) your truthful observer looks under both pieces of cloth, and reports back to you. By adding the piece of cloth I am making it clear that in (A), the observer makes his report with no knowledge whatsoever of the other coin –  Willie Wong Dec 24 '10 at 9:57
whereas in (B), the observer has knowledge of both coins. So this emphasizes how when the truthful observer decides one coin to observe ad to ignore the other coin, he makes whatever result he will report independent of the outcome in the other coin. –  Willie Wong Dec 24 '10 at 10:00
@Willie It seems absurd to me that the probabilities could be different due to the different knowledge the observer has of the outcomes: knowing about only one of the coins versus knowing about both the coins. What he tells me is identical (and truthful) in both situations: "At least one of the coins is a head". –  NotSuper Dec 24 '10 at 10:22
@NotSuper: But what information that statement gives you is different in the two situations. Think of it this way, perhaps. In situation A, there was a priori only a 50-50 chance that the observer would be able to state that one of the coins is heads. So when he does make that claim, it gives you some more information about the state of the coins. In situation B, the observer has it easier: in 3 out of 4 cases, he'll be able to state that at least one of two coins is heads. When he does, it's not as surprising, and so less informative; you know less about the state of the coins than in A. –  Rahul Dec 24 '10 at 18:58
@NotSuper: This applies only to your original question, not to your edit. –  Rahul Dec 24 '10 at 18:59

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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@Rahul Sure I want what you have to say. In my last comment both of us were dealing only with situation A. You aren't disagreeing with my argument there, are you? As for the observer's strategy, I don't believe I've left room for him to have one. So please clarify. As for probability spaces, I was familiar with them before I asked the question -- I just don't agree that they are of much use here, or have been used in a way that doesn't reflect the situations I've posed. For example, in the one-or-two girls problem, some were allowing for boy-boy to be an original possibility -- –  NotSuper Dec 25 '10 at 10:55
whereas I was presenting situations where to begin with there already is one girl. So I visit a family who have two kids. One comes in the living room to meet me. She's a girl. So I'm wondering to myself if the other child is a girl -- and thinking that the probability of the other child being a girl is 1/2. There is no boy-boy. Only girl or boy. Some have set the situation up as examining a random sample of 2-child families. But there is no study. This is a one-off. I just happened to visit this family one day. Examining the gender or number of their kids was not on my mind. –  NotSuper Dec 25 '10 at 10:56
However, I meet one of their children and wonder about the other one. Similarly with the coins. The situations are given, as I've stated. There IS one heads (this is a given), no matter whether the observer looks at only one (it may be misleading to call this one coin "the first coin") or both. In both A and B he supplies me with information which together with some logical deduction on my part, informs me that there is at least one heads. The observer has no strategy. He simply reports. It doesn't matter whether he looked at both coins or only one before reporting. Your turn. Thanks, Rahul. –  NotSuper Dec 25 '10 at 10:58
@NotSuper: Well, no, in my last comment, I was comparing situations A and B, responding to your proposition that "I receive exactly the same information from the observer in both cases". I was trying to show that your knowledge of the coins is different in situations A and B. Do you disagree with this? –  Rahul Dec 25 '10 at 22:03
@NotSuper: As for the rest of your comments, it seems you are running up against a deeper issue here, namely that when you speak of probabilities, you are implicitly considering a random sample in a space of possibilities! It makes no sense to ask of the probability of a single one-off event in a vacuum. For example, what is the probability is that I have a brother? The question can only make sense if you consider me a random person from a large sample space, and ask instead the probability that a random person from that set has a brother. –  Rahul Dec 25 '10 at 22:10