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

My brother and I made a bet which I took. He had two coffee cups which were roll up the rim to win. The odds of a cup winning were 1 in 6. He was feeling very confident and said "if I don't win with both of these cups I pay your a dollar, but if I do win with both of these cups you pay me 10 dollars".

I understand that the probability of both cups winning is 1/6 * 1/6 = 1/36, which is the basis of why I took the bet. But how does the payout ratio affect this bet?

share|cite|improve this question
up vote 0 down vote accepted

Your expected return from the bet is $\frac {35}{36}\cdot 1 + \frac 1{36} \cdot (-10)=+\frac {25}{36}$ so the bet is in your favor. If you won $1$ but had to pay $40$ the return would be $\frac {35}{36}\cdot 1 + \frac 1{36} \cdot (-40)=-\frac {5}{36}$ and the bet would be unfavorable. You would still win most of the time, but if you made the bet a large number of times you would lose in total.

share|cite|improve this answer
Thanks for your response that was just what I was looking for. – Matt Feb 21 '13 at 22:09

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.