Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Design a game where three coins are tossed and the player "wins" if he or she tosses either all three heads or all three tails. The question is if the "wager" is $1, then what is a fair winning payout for the player such that he or she would break about even playing this game over time?

share|cite|improve this question

HINT: Only two of the possible outcomes, TTT and HHH, are wins. The other $n$ are losses, where I leave it to you to determine just what $n$ is. The $n+2$ outcomes are equally likely. If the winning payout is $p$, therefore, on average the player pays $n+2$ dollars and wins $2p$ dollars per $n+2$ games and has net winnings of $2p-(n+2)$ dollars. In a fair game the net winnings are $0$.

share|cite|improve this answer

The probability of $3$ heads in a row is $\left(\frac{1}{2}\right)^3$, as is the probability of $3$ tails in a row. So the probability of winning is $\frac{1}{4}$.

It is not precisely clear what "payout" means. We assume that you get your dollar back plus an amount $a$ that we will call the payout,

Then the random variable $X$, which is the amount you win, is $-1$ with probability $\frac{3}{4}$ and $a$ with probability $\frac{1}{4}$.

It follows that $E(X)=(-1)\left(\frac{3}{4}\right)+(a)\left(\frac{1}{4}\right)$. For a fair game, we need $E(X)=0$. Solve the equation $(-1)\left(\frac{3}{4}\right)+(a)\left(\frac{1}{4}\right) =0$ for $a$. We get $a=3$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.