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.

  • $\begingroup$ 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)? $\endgroup$ – Henry Nov 1 '18 at 14:16
  • $\begingroup$ @Henry For simplicity, let's assume this isn't necessary (provided it doesn't change things by an order of magnitude or more) $\endgroup$ – 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
  • $\begingroup$ Thank you! Very interesting to see that with only 230n the risk of the casino ever busting (in any practical timeframe) is only 1%. $\endgroup$ – user4779 Nov 2 '18 at 2:58

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