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Question: Four fair coins are flipped. You get $\$1$ each time head appears. The person tells you that there are at least two heads. What is the expected payoff?

I cannot seem to grasp conditional expectation and PMFs. I understand how to arrive at the expected value of getting heads in the four tosses if you are given $\$1$ each time head appears $E[X] = \$2$, but when the conditional is placed stating that the person flipping the coin tells you that there are at least $2$ heads, I do not understand how to construct the conditional expectation.

How would I extend this to the person telling me there are at least $3$ heads? Or $1$ head?

Thank you.

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4 Answers 4

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The idea is to examine all the possible situations given that you got two heads. There are ways to shorten the following, but for now I will just do the straight-forward thing.

Without any other information, there are $16$ possible toss permutations (for convenience I shall call getting tails $0$ and getting heads $1$):

1) $0000$

2) $0001$

3) $0010$

4) $0011$

5) $0100$

6) $0101$

7) $0110$

8) $0111$

9) $1000$

10) $1001$

11) $1010$

12) $1011$

13) $1100$

14) $1101$

15) $1110$

16) $1111$

Without further information each of these occurs with equal probability.

Now, the condition that we have at least $2$ heads puts some restrictions on what the tosses could be. From the tosses we can then count that $11$ have at least two heads (or equivalently that $5$ have one or fewer heads). Hence we want to calculate the expected value of heads given that our toss is equally likely to be any one of these $11$, and cannot be one of the remaining $5$. Of the $11$ tosses, $6$ have two heads, $4$ have three heads and $1$ has four heads, giving us $$\frac{6\times \$2+4\times \$3+1\times\$4}{11}=\$\frac{28}{11}\approx \$2.55$$ as the expected outcome.

If instead our condition was $3$ heads or $1$ head we would do the same thing except that we have a different number of choices ($5$ choices in the $3$ heads case and $15$ choices in the $1$ heads case).

Now, I said that I gave the straight-forward way first, but there is another way to do it with binomial coefficients. Notice that the number of tosses with exactly $2$ heads is $\binom{4}{2}=6$, and the number with exactly $3$ heads is $\binom{4}{3}$, etc. So instead of counting out the possibilities we can directly compute the solution by taking $$\frac{\binom{4}{2}\times \$2+\binom{4}{3}\times \$3+\binom{4}{4}\times\$4}{\binom{4}{2}+\binom{4}{3}+\binom{4}{4}}=\$\frac{28}{11}\approx \$2.55$$

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  1. Number the coins $1, 2, 3, 4$.
  2. Enumerate all of the possible elementary outcomes of the tosses of the four coins, in order: so for example, you might see an outcome of $(H, T, T, H)$. There should be $16$ such outcomes, without regard to the number of heads or tails.
  3. Among these outcomes, which ones have at least two heads?
  4. For each of the outcomes with at least two heads, count the actual number of heads. This gives you the payoff for each outcome.
  5. Since each of these outcomes from step (3) is equally likely, the expected payoff is...
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Count the number of combinations:

  • Exactly $2$ heads: $\binom{4}{2}=6$
  • Exactly $3$ heads: $\binom{4}{3}=4$
  • Exactly $4$ heads: $\binom{4}{4}=1$

Calculate the probabilities:

  • Exactly $2$ heads: $\frac{6}{6+4+1}=\frac{6}{11}$
  • Exactly $3$ heads: $\frac{4}{6+4+1}=\frac{4}{11}$
  • Exactly $4$ heads: $\frac{1}{6+4+1}=\frac{1}{11}$

Calculate the expected number of heads: $2\cdot\frac{6}{11}+3\cdot\frac{4}{11}+4\cdot\frac{1}{11}=\frac{28}{11}=2.\overline{54}$

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Simply add up how much money you would get for every outcome, and divide by the number of outcomes. The unconditional case is:

(T,T,T,T) -> 0
(H,T,T,T) -> 1
(T,T,T,H) -> 1
(H,T,T,H) -> 2
(T,T,H,T) -> 1
(H,T,H,T) -> 2
(T,T,H,H) -> 2
(H,T,H,H) -> 3
(T,H,T,T) -> 1
(H,H,T,T) -> 2
(T,H,T,H) -> 2
(H,H,T,H) -> 3
(T,H,H,T) -> 2
(H,H,H,T) -> 3
(T,H,H,H) -> 3
(H,H,H,H) -> 4

Add the amounts in each case, which is 32, then divide by the number of cases, 16. That is all there is to finding an expected value. Now, just do it again with the new rule. So, list the cases again, removing the ones that cannot appear.

(H,T,T,H) -> 2
(H,T,H,T) -> 2
(T,T,H,H) -> 2
(H,T,H,H) -> 3
(H,H,T,T) -> 2
(T,H,T,H) -> 2
(H,H,T,H) -> 3
(T,H,H,T) -> 2
(H,H,H,T) -> 3
(T,H,H,H) -> 3
(H,H,H,H) -> 4

Now the sum is 28, and there are 11 cases, so the expected payoff is 28/11, which rounds to $2.55.

It seems to me that people often over complicate problems like this. Just add up the values for each case and divide by the number of cases. That is really all there is to expected value.

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