Standard Deviation
As an example, what is the difference in playing one Blackjack hand at $60,000 versus playing six Blackjack hands at $10,000? How much more risky is it to play the one hand at $60,000 versus spreading the same $60,000 over six hands?
Thank you for your help.
Quote: markdcasinoI am wondering how to calculate the standard deviation on a Blackjack game for one hand versus multiple hands. If the standard deviation in Blackjack is 1.14149, how does that differ if you play multiple hands?
As an example, what is the difference in playing one Blackjack hand at $60,000 versus playing six Blackjack hands at $10,000? How much more risky is it to play the one hand at $60,000 versus spreading the same $60,000 over six hands?
Thank you for your help.
The following reply applies to playing six hands of blackjack in one round.
Here's the Wizard's page on Variance in Blackjack: https://wizardofodds.com/games/blackjack/variance/
He gives this formula for the total variance for playing n hands each of bet 1 unit: nv + n(n-1)c, where v is the variance for each hand and c is the covariance between each hand.
In your example, the standard deviation is 1.14149, which makes the variance 1.141492 = 1.303, which is the variance for a hand of 6-deck, DAS, surrender, re-split aces blackjack. And the Wizard gives 0.479 as the covariance for those rules.
Using the Wizard's formula, the variance for betting one unit on each of six hands would be: 6(1.303)+6(5)(0.479) = 22.188.
If we call $60,000 one unit, then if we divided $60,000 equally over six hands, then each hand has a bet of one-sixth of a unit. Then since the variance of betting one unit on each of six hands is 22.188, the variance for betting one-sixth of a unit on each of six hands would be 22.188 / 62 = 0.61633. And the standard deviation is 0.616331/2 = 0.78507.
The standard deviation for betting $60,000 on one hand of blackjack is 60,000(1.14149) = $68,489. And the standard deviation for betting $10,000 on each of six hands of blackjack is 60,000(0.78507) = $47,104.
The above is for playing six hands in one round, but if you want to know the answer for the standard deviation for playing $10,000 per round for six rounds, you can use the same formula, nv + n(n-1)c, but with setting the covariance to zero.
Surely, Standard deviation only really has meaning if the distribution of the results fits into a smooth(ish) curve, not for a small number of results like 1, 2, or 6 wagers.Quote: markdcasinoI am wondering how to calculate the standard deviation on a Blackjack game for one hand versus multiple hands. If the standard deviation in Blackjack is 1.14149, how does that differ if you play multiple hands?
As an example, what is the difference in playing one Blackjack hand at $60,000 versus playing six Blackjack hands at $10,000? How much more risky is it to play the one hand at $60,000 versus spreading the same $60,000 over six hands?
Thank you for your help.
Playing 1 hand of Blackjack at $60,000 can only have one of four outcomes of winning $90,000, $60,000, $0, or losing $60,000
Whereas, playing six hands at $10,000 gives the same min and max outcomes, but with lot's of potential intermediate results.
Maximum/minimum loss is more like $480,000 depending on split/double rules. There are many more outcomes than those listed. Which is why the game is so difficult to quantify without simulationQuote: OnceDear
Playing 1 hand of Blackjack at $60,000 can only have one of four outcomes of winning $90,000, $60,000, $0, or losing $60,000.
?Quote: Ace2Maximum/minimum loss is more like $480,000 depending on split/double rules. There are many more outcomes than those listed. Which is why the game is so difficult to quantify without simulation
I'd assumed that his $60k was total bankroll, thus making splitting or doubling (unfortunately) impossible
Sitting down without the bankroll to split/double is probably worse EV than playing at a 6:5 tableQuote: OnceDear?
I'd assumed that his $60k was total bankroll, thus making splitting or doubling (unfortunately) impossible
