You are offered a game of 12 coin tosses, wherein you'll win 150 USD on heads, and lose 100 USD on tails. What is the probability of ending up with a net loss after this game? Enter answer as a decimal between 0 and 1, accurate to 5 decimal places.

I know that you'd have to lose 8 games at least to have a net loss, but I'm stuck after that.

Need this done asap

isn't that just a list of all combinations where you lose 8 games? eg (12 choose 8) / 2^12

then combine that idea with losing 9 games, 10 games, 11 games, 12 games

This post was edited by carteblanche on Oct 10 2013 07:02pm

I forgot to add the 1 chance to lose all 12 games. It changed the percentage by 2 hundredths of a point, kind of dumb.

In order for you to lose money, you must have 4 or fewer heads.

Therefore the probability of losing money is having 0, 1, 2, 3, or 4 heads.

So your formula becomes:

Do this for 0, 1, 2, 3, and 4 and sum them together.



This post was edited by Dontrunaway on Oct 10 2013 07:15pm

Yeah I got it, just forgot to add the one chance of getting 12 losses.

Thanks for the quick response though

Np, needed to reinforce it for myself either way, lol.