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I play a game. Suppose I toss 4 fair coins onto a table. For each head I see on the table, I earn a dollar unless I chose to retoss: I get to re-toss just once if I want to, after which I would earn a dollar for each head I see after the retoss. What are my expected earnings if I retoss only if my first toss has 0, 1, or 2 heads?

Here's what I've tried. Let $X$ be the number of heads on the first toss and $Y$ be my final earnings in dollars. By conditioning on $X$, I get

$$ E(Y) = E(Y | X = 0, 1, 2)P(X = 0, 1, 2) + E(Y | X = 3)P(X = 3) + E(Y | X = 4)P(X = 4) $$

However, how do I find $E(Y | X = 0, 1, 2)$, $E(Y | X = 3)$, and $E(Y | X = 4)$?

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Maybe the symbols got in the way, it it a very concrete situation. –  André Nicolas Dec 3 '12 at 7:28

1 Answer 1

up vote 2 down vote accepted

$E(Y | X = 0, 1, 2)$ is just the expected number of heads you would get if you were not allowed to re-toss. Hence we have $E(Y | X = 0, 1, 2)=4\cdot\frac{1}{2}=2$.

$E(Y | X =3)=3$ since if you get $3$ heads on your first toss, you stop and collect your $3$ dollars in winnings.

Likewise, $E(Y | X =4)=4$.

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Thanks, the latter 2 values make sense. For $E(Y | X = 0, 1, 2)$, I could retoss though, right? Why is the expected value just 2 in that case? –  David Faux Dec 3 '12 at 7:31
@DavidFaux If you get $0$,$1$ or $2$ heads on your first toss, you can (and will) re-toss just once. So the amount of money you get is exactly equal to the number of heads you will get on your re-toss. –  Sarastro Dec 3 '12 at 7:34
Thank you, that makes sense! –  David Faux Dec 3 '12 at 16:36

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