Question

Your friend tosses three coins and you roll a single die. If the number on the die you roll is less than or equal to the number of heads that your friend tosses, you win $X. If not, you lose $1. How large should X be in order for this to be a fair game?

Answer 1

I'm not sure how this is calculus... it sounds more like plain probability. So you can figure that there are 2^3*6=48 total possibilities for the outcomes of your tosses (6 for the die and 2 for each coin which are then multiplied together). If your friend only tosses 1 or no heads at all, you lose no matter what you roll. If your friend gets 2 heads, there are 3 ways in which he/she can get it (HHT, HTH, THH) and you have to roll a one. Thus there are 3 ways here in which you can win. If your friends flips all three as heads, there are 2 ways you can win (rolling a 1 or a 2) and thus there is a total of 5 ways for you to win. Now, there are also 48-5=43 ways for you to lose, so the ideal ratio for wins loses (because of the odds) is 43:5 = 8.6:1. Therefore X should be 8.6.