You flip a coin. If you get heads you win \$2 if you get tails you lose \$1. What is the expected value if you flip the coin 1000 times?

I know that the expected value of flipping the coin once is $\frac{1}{2}(2) - \frac{1}{2}(1) =0.50$

Would the expected value be 500?

  • $\begingroup$ yes, this is a binomial distribution whose mean is $np = 500$ to success and of course $n(1-p) = 500$ to lose. Hence, $500*2-500*1=500$ $\endgroup$ – user164550 Mar 16 '16 at 16:14
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    $\begingroup$ $\frac12 (2) - \frac12 (1) = .5$. But your reasoning is correct, and the answer is $500$. $\endgroup$ – Michael Harrison Mar 16 '16 at 16:14
  • $\begingroup$ Do you have any reason to believe the coin is fair? $\endgroup$ – Eric Towers Mar 16 '16 at 16:16
  • $\begingroup$ Yes, it is a fair coin $\endgroup$ – johnbowen Mar 16 '16 at 16:22

Let $X$ be the random number of heads obtained in $n = 1000$ flips of a fair $(p = 1/2)$ coin. Then $$X \sim \operatorname{Binomial}(n = 1000, p = 1/2),$$ and the expected value of the number of heads is $$\operatorname{E}[X] = np.$$ Now, the net winnings (or losses) is equal to 2 dollars for each head minus 1 dollar for each tail observed. So if we observed $X$ heads, then presumably we also observed $1000-X$ tails, and the net winnings/losses is therefore $$W = 2X + (-1)(1000-X) = 3X - 1000.$$ Because $X$ is a random variable, $W$ is also a random variable. The expectation of $W$ is simply $$\operatorname{E}[W] = \operatorname{E}[3X - 1000] = 3 \operatorname{E}[X] - 1000 = 3 np - 1000 = 3(1000)(1/2)-1000 = 500.$$

Notice that this approach lets us calculate the expected winnings for any general payoff structure where we win $s$ dollars for each head and lose $t$ dollars for each tail. Then $$W = sX - t(n-x) = (s+t)X - nt,$$ and $$\operatorname{E}[W] = (s+t)(np) - nt = n(sp - t(1-p)).$$ What this tells us is that the expected winnings/losses for a fixed $s$ and $t$ will in general become more extreme as the number of trials increases (if $sp \ne t(1-p)$).

We can also use the above to calculate the variance; e.g., $$\operatorname{Var}[W] = (s+t)^2 \operatorname{Var}[X] = (s+t)^2 np(1-p).$$


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