a good rule of thumb to calculate standard deviation for any session is
Average Bet x (SQRT of Hands played) x 1.1
Assume a bet spread of 10-100 over a time span of 2 hours (and play 50 hands an hour)
lets just assume that most of your play will be between TC 1 and TC 2 which is something like a 25 dollar bet and house edge (due to Advantage play) of about 0%
your standard deviation for the session will be
25x10x1.1 = 275 (or 27.5 units)
so if you think about bankroll and long-run ROR, all you have to do is approximate the "infinite session". lets assume that you will play 500 hours of blackjack in your lifetime, at 50 hands per hour.
you will need an approximate bankroll of
25x 158. x 1.1 = $4350 (or 435 units) IN ORDER TO WITHSTAND ONE STANDARD DEVIATION over that entire "session"
the normal distribution graph shows that 68% of the time (in a completely even game), your profits or losses will fall within one standard deviation from the mean (the mean is zero, because we are playing a fully 0% ev game).
This means that 16% of the time, you will bust out if you only have a bankroll of 435 units. This would equate to a 16% Risk of ruin.
A safer BR size would be 1000 units of your minimum bet in order to withstand 2+ standard deviations of play across this lifetime of 500 hands. why? because 2+/- standard deviations in the normal distribution graph account for 95% of all scenarios. This equates to a a 2.5% risk of ruin.Why 2.5% and not 5%? because standard deviation can be really bad or really good! in calculating risk of ruin, we only worry about the really bad variance that can hurt us. Therefore if 100 people play exactly the same way as described above, about 2.5 of them on average will go bust with a 1000 unit bankroll.
Lastly please note that this is assuming an even 1-1 game with no house edge.
This math really puts into perspective the value of wonging out during negative expectation shoes and only playing favorable counts which give the player long-term advantage.
If we adjust the EV of this game to say, +1% (the normal edge than a poor advantage player can squeak out)
then the MEAN of that normal distribution graph is nonzero. Specifically it will be (%Edge x dollars wagered)
in our 500-hour lifetime scenario, average bet=25 and hands played = 25,000 so lifetime dollars wagered would be 625,000. A 1% player edge would mean that we can expect a profit of $6,250 over this lifetime, however due to normal variance across the 500 hours, (i.e. $4350 that we calculated earlier), 68% of the time we could end up anywhere between a profit of $1900 or a profit of $10,600.