# Casino/House bankroll for Blackjack

Given a maximum bet size of "x" in blackjack, how many maximum bets should the house keep in its bankroll/account to prevent it ever realistically losing all its money against players playing perfectly? (not including card counting/other factors)

For arguments sake, assume the house probability of winning is 50.5% per game.

• Are you planning to take into account that on a single hand in a casino the results among different players can be positively correlated (e.g. if the house reaches $21$ or goes bust)? – Henry Nov 1 '18 at 14:16
• @Henry For simplicity, let's assume this isn't necessary (provided it doesn't change things by an order of magnitude or more) – user4779 Nov 1 '18 at 15:54

If you treat this as $$\pm1$$ random walk where the independent probability of $$+1$$ steps is $$p$$ and of $$-1$$ steps is $$1-p$$ then it is a known result that with $$p \lt \frac12$$ the probability of the sum of the steps ever reaching some positive integer $$n$$ is $$\left(\dfrac {p}{1-p}\right)^n$$

Here $$p =0.495$$. You need to quantify realistically: if you want the probability or risk of ever losing a net $$n$$ or more to be less than some $$r$$ then you need $$r \lt \left(\dfrac {p}{1-p}\right)^n$$ and so want $$n > \dfrac{\log(r)}{\log(p) - \log(1-p)}$$

For some illustrative values of the risk $$r$$ (the first shown is $$1\%$$ and the fifth $$0.0001\%$$) you get suggested values of $$n$$. You can round down if you want the risk of strictly exceeding these positions to be $$r$$ or less.

  r           n
0.01        230.25
0.001       345.38
0.0001      460.50
0.00001     575.63
0.000001    690.75

• Thank you! Very interesting to see that with only 230n the risk of the casino ever busting (in any practical timeframe) is only 1%. – user4779 Nov 2 '18 at 2:58