A man will win $100 if he throws either a 7 or 12 in the roll of a pair of dice.?

Question

He will lose $25 if he does not. What is the man's mathematical expectation? Is this game fair?

Answer 1

The game is not fair. The chances of rolling a 7 OR a 12 is 7/36(6/36 for 7 + 1/36 for 12), or about 19.444%

If you lose $25 when you don't roll 7 or 12, and GAIN $100 if you do, then you want at least 1/5, or 20% of your rolls to be 7 or 12(e.g. four consecutive rolls not yielding 7 or 12, then the fifth one being 7 or 12 breaks even.) So, in the long run, you lose.

Answer 2

The possible outcomes of a pair of dice are:

6*6=36 possible outcomes

of these, 7 can be made as:

2+5
3+4
6+1
5+2
3+4
1+6

and 12 can be made as

6+6
6+6

So there are 8 successes in 36 possibilities, for an 8/36 probability, or 2/9

betting 1/4 odds on a 2/9 success is not a good bet.

Answer 3

E(game)=p(success)*v(success). p(7 or 12)=1/6+1/36=7/36; v(success)=$100. $100*(7/36)=(25*7)/9, 175/9, 19.444...

The game is weighted towards the maker, of course.