Buy in requirements

[jay_shark]

For this post , i'll come up with different buy ins according to your win rate and the chance that you'll go bust .

If you win y % of your games , then the chance you lose is (1-y)% . The probability with N buy ins that you'll go broke at some point is [(1-y)%/y%]^N . So here is the breakdown .

Here is the probability you'll go bust with 5 buy ins
as a
a)55 % player .
b)60% player
c) 65% player

sol.
a)[.45/.55]^5= 36% which is too high .
b)[0.4/0.6]^5=13 % which is still too high
c) [0.35/0.65]^5 = 4.5% which is not bad but still high

The probability you go bust with 10 buy ins as a
a) 55% player
b) 60% player
c) 65% player .

sol.
a) [0.45/0.55]^10 = 13%
b) [0.4/0.6]^10 = 1.7% which is fairly low
c) [0.35/0.65]^10 = 0.2% This is extremely conservative .

The probability you go bust with 15 buy ins
a) 55% player
b) 60 % player
c) 65% player

sol.
a) [0.45/0.55]^15=4.9%
b) [0.4/0.6]^15 = 0.228% very conservative
c) [0.35/0.65]^15=0.00927% You'll never go broke

So if you're only working with 5 buy ins , then there is a significant risk of going broke even if you win 65 % of your games .On the other hand , with 10 buy ins you're pretty much playing comfortably if you can win 65 % of your games .If you're working with 15 buy ins as a 55 % player , then there is still a significant risk at 4.9% of going broke .

Make sure you think along these lines when you determine how many buy ins you need to play comfortably according to your desired ror .

[jay_shark]

To be even more specific , your RoR depends on your s.d since it is possible that two players may have the same win rate at 60% but one player plays with more volatility .This same player would need more buy ins to account for this .

[Goldmund]

Interesting post...This calculation is a "lifetime" calculation I suppose no?

[jay_shark]

Oh yes , this is the probability that if you play the game for ever that you'll go broke at some point . It uses the fact that there will always be a game waiting for you no matter what . Your opponents' bankroll is infinite .

[solinar]

Your missing the variability of variability though [img]/images/graemlins/smile.gif[/img]

You are assuming you always play your average opponent. In order to have an average opponent, you must play vs some opponents where your winrate is higher than normal, and vs some opponents where it is lower than normal. Sometimes, in a given sample, you will end up playing more tough opponents, and your sample winrate has a good chance at being lower than normal. During this sample, your chance at going bust due to variance is much higher than your normal chance. Since you only need to go bust once, this is what you have to plan for.

[omgwtfnoway]

good post jay, nice work.

[jay_shark]

Well to be precise the correct formula to use is the following :

r0r= e^(-2ub/sigma^2)

ror= risk of ruin or the probability you go broke
u= hourly rate which is $/h over time
B=bankroll
sigma=standard deviation


Every player has a different standard deviation and you should be able to work it out yourself easily . Using 30 sessions , you should be able to determine your standard deviation which gives you all the information you need to determine your ror given various bankrolls .

[Goldmund]

Is the rake included in your calculation?

[cwar]

Cool calculations but I think kind of unnecesarry, I have a great bankroll strategy that has 0% ROR.

[jay_shark]

It is very necessary as it doesn't get talked about in this kind of detail .

No matter what bankroll strategy you use , there is always a risk in going broke . Also ,it is perfectly acceptable to accept a 1 % risk in busting .