What does the variance in blackjack mean?

According to the Wizard of Odds, the variance for six decks, dealer stands on soft 17, no double after split, no re-splitting aces, and no surrender is 1.295. Does this mean that if the expected return is 99.57%, the probability that the return will be between 96.156% (99.57-(3*√1.295)) to 102.984% (99.57+(3*√1.295)) is about 99.7%? Also, how do I use the covariance?

Quote: BlackjackLover

According to the Wizard of Odds, the variance for six decks, dealer stands on soft 17, no double after split, no re-splitting aces, and no surrender is 1.295. Does this mean that if the expected return is 99.57%, the probability that the return will be between 96.156% (99.57-(3*�ã1.295)) to 102.984% (99.57+(3*�ã1.295)) is about 99.7%? Also, how do I use the covariance?

On this Wizard page, the expected value, variance, and covariance for one hand of blackjack with the rules dealer stands on soft 17, no double after split, no re-splitting aces, and no surrender are -0.00573, 1.295, and 0.478, respectively.

The return is 1 - 0.00573 = 0.99427. Suppose you are flat betting one unit on one hand of blackjack per round. The standard deviation of your return depends on how many rounds you play. If you play 10,000 rounds, for example, the standard deviation of your average return is the square root of this quantity: 1.295/10,000, which is 0.01138. So your average return would have a 99.73% chance of falling in the range 0.99427 - 3*(0.01138) to 0.99427 + 3*(0.01138), which is 96.01% to 102.84%.

Now, here is where the covariance, 0.478, is used: suppose instead of betting one unit on one hand of blackjack per round, you bet 1/2 unit on each of two hands of blackjack per round. The variance per round would then be 2 * 0.5^2 * 1.295 + 2 * 0.5 * 0.5 * 0.478 = 0.8865. And if you play 10,000 rounds, the standard deviation of your average return is the square root of 0.8865/10,000 = 0.009415. Then your average return would have a 99.73% chance of falling in the range 0.99427 - 3*(0.009415) to 0.99427 + 3*(0.009415), which is 96.60% to 102.25%.

Thank you very much for the detailed explanation. Does this mean that if I bet 1/5 unit on each of five hands, the variance per round will be 5*0.2^5*1.295 + 5*0.2^5*0.478 = 0.0028368?

Quote: BlackjackLover

Thank you very much for the detailed explanation. Does this mean that if I bet 1/5 unit on each of five hands, the variance per round will be 5*0.2^5*1.295 + 5*0.2^5*0.478 = 0.0028368?

Good question.

The variance in this case is 5*0.2^2*1.295 + 20*0.2^2*0.478 = 0.6414.

An expression for the variance per round betting 1/n on each of n hands would be:
n * (1/n)^2 * 1.295 + n*(n-1) * (1/n)^2 * 0.478