There is a game where you are asked to roll two fair six-sided dice. If the sum of the values equals 7, then win £21. However, must pay £5 to play each time both dice are rolled. Do you play this game?
One way to think about this is that getting a 7 comes with 1/6 chance, and to make money we need to get 7 at a rate of 1/4, so the answer is not to play.
Another way to think about it is: what is my chance of throwing a 7 at least once in every 4 throws? In which case I would calculate a probability of not throwing a 7 4 throws in a row (5/6)^4, and then subtract this from 1 to get a probability of throwing at least one 7. Which is 1 – (5/6)^4 = 0.52. By this logic I would play the game.
Both of these answers cannot be correct. Could someone explain to me which one is incorrect and why? Thanks!
EDIT: wow, this is the first time I asked a question on StackOverflow, did not expect to get so many responses. Thank you all, I am very grateful!
Best Answer
Your first approach is correct, at least approximately. (If you won £20 on a 7, then you'd need to roll 7's at a rate of $\frac14$ to break even. Since you win £21, you can afford a slightly lower rate, though not as low as $\frac16$.)
Your second approach is wrong. It's true that more than half the time, you win at least once over a sequence of four throws. However, all that tells you is: more than half the time, you don't lose money. This doesn't mean that you break even on average! In fact, it tells you that:
This doesn't look as good anymore. In general, just because a random variable is positive more than half the time, doesn't mean its expected value is positive.