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08-04-2002, 03:52 AM
hi,

could someone please help me with a bank roll query (not related to poker). If someone is playing a game with an overall payout of 100.2% and a 16.7% chance of each bet winning, what bankroll would be needed in terms of units to have a 95% chance of not going bankrupt?

if possible please show me the formulae you have used (kelly or other wise).

thank you, insights are much appreciated.

Arrash

08-04-2002, 05:49 PM
From Blackjack Attack by Don Schlesinger p. 140:

ruin = [(1-w/sd)/(1+w/sd)]^(B/sd)

ruin = probability of ruin

w = average win

sd = standard deviation

B = bankroll

Solving for B gives:

B = ln(ruin^sd)/ln[(1-w/sd)/(1+w/sd)]

He says this can be more simply approximated under certain conditions to the following which gives exactly the same result in your case:

B = -[sd^2/(2w)]ln(ruin)

In your case I will assume that when a bet wins it always pays the same which is w = .002 bets. Otherwise I need more information to get the sd. To get sd, use the fact that 5/6 of the time you will lose 1 bet, and 1/6 of the time you will win 5.012 bets, and use:

var = E(x^2) - [E(x)]^2

var = (5/6)(-1)^2 + (1/6)(5.012)^2 - (.002)^2

var = 5.02 bets

std = sqrt(var) = 2.24 bets

pluging into either of the above equations give a bank of 3758 bets.

This is what you need to bet in whatever manner gives you the win rate and standard deviation we assumed. Kelly betting is different in that you have a theroretical risk of ruin of 0 assuming you can continue to shrink your bets indefinitely when you lose.