I can't think of a title.

[1800GAMBLER]

I was wondering how the figures for recommended bankroll were made, so someone please answer me this.

I make a game up with i plan to bust my gamble addict rich friend for everything he owns then give it to his divorced wife who cheated on him with 3 men in 1 night, filmed it, and sent it to him. I think poker has made me numb, anyhow!

In this game - it's pretty basic but he'll play - every hand i have QQ and he has AKo. No blinds, max bet of $1 only preflop and then we deal 5 cards out.

So really it's just a $1 bet against my $1 call then deal cards.

Because i get bored easier than him i just set this up on a computer and let him click away all day long, my question is..

How would i calculate the bankroll for this machine? Wanting the bear min. never wanting to go bust.

Thanks.

[BruceZ]

This is the classical gambler's ruin problem.

Let p = probability QQ wins, and q = probability that QQ loses. According to www.twodimes.net/poker, p = 56.99%, and q = 42.66%. This assumes 4 different suits.

Assume initially that your opponent has enough bank to play forever. Then if r is the risk of ruin, we can write:

r = q + pr^2

That is, there are 2 ways we can go bust. We can go bust by losing the first hand, and that happens with probability q, or we can win the first hand with probability p, and then go broke after that. When we win the first hand, we double our bankroll to $2, so at that point our probability of ruin becomes whatever it was squared since now we must lose our bankroll twice. When we have $1 the risk of ruin is r, so when we have $2 the risk of ruin is r^2. Rearranging the above gives:

pr^2 - r + q = 0

Factoring and using p + q = 1:

(pr - q)(r - 1) = 0

r = q / p.

This says that the risk of losing a $1 bankroll is just the ratio of the probability of losing a hand to the probability of winning a hand. The other possible solution r = 1 can be ruled out, but that requires a more advanced derivation than what I'm giving here.

The probability of losing a bankroll of size B is

r = (q/p)^B

since we must lose $1 B times. This is the risk of ruin formula for this game assuming your opponent can't go broke. You can never make your risk of ruin exactly 0, but you can make it as small as you like by increasing your bankroll. and it gets smaller exponentially.

Solving for B gives the bankroll requirement:

B = ln(r) / ln(q/p)


Now if there is a significant chance that your opponent can go broke, the risk of ruin formula changes. Say your opponent's bankroll is A. Then the risk of ruin formula becomes:

r = [ 1 - (p/q)^A ] / [ 1 - (p/q)^(B+A) ]

This is also a classic result. It comes from solving the difference equation:

q[B] = p*q[B+1] + q*q[B-1]

with initial conditions:

q[0] = 1, q[A+B] = 0.

Where q[B] is the risk of losing a bankroll of size B.


The bankroll requirements and risk of ruin for poker should be computed based on your EV and standard deviation using the formulas I posted in this thread. This is the most accurate way to derive bankroll guidelines. These formulas are derived in a similar manner as the above by creating a coin flip game which has a given EV and sigma for N flips. For a derivation of these formulas, look here.

[BruceZ]

For a 1% risk of going broke, you would only need a bankroll of about ln(.01) / ln(43/57) = 16 bucks. Less than you thought, eh?

[1800GAMBLER]

Wow, yeah, it is.

I was just wondering about roulette wheels and such how much they hold to make sure they don't go broke and wondered about the maths behind it.

I start my degree course in maths next year; my 4th year includes a major in gambling math. should be interesting.

Thanks for all the effort.