Question

The probability that the house wins in p. The probability that the gambler wins is (1-p). The probability that the house wins-loses-wins is p(1-p)p. The house wins if it is up by 2, and the gambler wins if he is up by 1.
I THINK that the
sum (1,inf) of (1-p)^(n-1) * p describes the house winning a round. But how do I relate the house being up by 2 to win and the series I just wrote above? and How do I find a break even point?

Answer 1

Not sure why it didn't type in, but it's a series of rounds of a game of odds at a casino.

Answer 2

you can sum a geometric series if you like or you can precede by another method

Answer 3

\[\sum_{0}^{\infty}(1-p)^n\] ?

Answer 4

we can compute the probability the gambler wins as follows
either he wins on the first game, with probability \(1-p\) or he loses, the house loses, and then we start the game all over
so if we put \(x\) as the probability the gambler wins, we can wrote the equation as
\[x=1-p+px\] and solve for \(x\) in term of \(p\)

Answer 5

hope the logic is clear
if he loses twice, the house wins and the game is over, so we can ignore that possibility if we are computing the probability the gambler wins.
otherwise you sum a geometric series, but you will get the same answer for the probability the gambler wins

Answer 6

hmm maybe i made a mistake, let me check

Answer 7

yea i made a mistake
gambler loses with probability \(p\)
house loses with probability \(1-p\) and so "gambler loses, house loses, start over" should look like \((1-p)px\) and the equation should look like
\[x=1-p+

Answer 8

damn
should look like
\[x=1-p+p(1-p)x\]