#### DSchles

##### Well-Known Member
Enough. Variance is a precise mathematical term equal to the average squared result E(X^2) minus the square of the average result E(X)^2. Since, in blackjack, the mean, or average result, E(X), is very close to zero, we don't lose much accuracy by describing variance as, simply, the average squared result. And standard deviation is the square root of variance.

And, again, standard deviation and variance are measures of dispersion around the mean. They permit you to make probabilistic statements about how likely any result is to be any particular distance to either side of the mean. That last part is important. Variance does NOT signify just one-sided movement (for example, to the left, or losses). It applies equally to both sides. Analogously, in the stock market, volatility refers to movement in either direction, up or down, and not just to downside movement. Of course, you wouldn't know that by watching TV.

It's fine to say that you don't understand a term. That's why we're here. But when we're talking about precise mathematical experssions, it's a little silly to make up your own definitions or offer "opinions" of what you think they may mean. We don't ask, "What is your opinion of 2 + 2?" This isn't social studies.

Don

#### aceside

##### Active Member
DSchles said:
Enough. Variance is a precise mathematical term equal to the average squared result E(X^2) minus the square of the average result E(X)^2. Since, in blackjack, the mean, or average result, E(X), is very close to zero, we don't lose much accuracy by describing variance as, simply, the average squared result.

Don
Hi Don. Can you help find what is wrong with my calculation of the Kelly wager? I studied a little more about the variance in Blackjack on the Wizard of Odds. His simulation shows that for the 6-deck game with No DAS and No Surrender rules, the EV is -0.573% and the variance is 1.295. With the same set of rules, the variance increases (but converges) with the number of decks. This is for a basic strategy player flat betting one unit.

For an AP player varying bet amount with the true count, let me do some exercises using real numbers to better understand the Kelly wager. Firstly, we set the EV of -0.573%=0%. Consider these three situations:

When TC<1, the player’s advantage is 0.0% and the variance is 1.295. The player must bet 1 unit.
When 1=<TC<2, the player’s advantage is 0.5% and the variance becomes 2*2*1.295=5.180 because of betting 2 units. The player should bet k*0.5%/(2*2*1.295), so k=2*[2*2*1.295/(0.5%)].
When 2=<TC<3, the player’s advantage is 1.0% and the variance becomes 4*4*1.295=20.720 because of betting 4 units. The player should bet k*1.0%/(4*4*1.295), so k=4*[4*4*1.295/(1.0%)].

The last two situations give different k values, and this inconstancy has bothered me for some time. This is the part I do not understand Kelly. Thank you in advance.

#### DSchles

##### Well-Known Member
You're confusing two issues. If you look at Table 2.1 on page 20 of BJA3, you'll see very clearly the methodology for calculating global standard deviation (or variance) for three different scenarios. Everything should be very clear. That calculation involves squaring the bet size and also, of course, the frequencies of each true count.

Calculation of optimal bet sizes is a much more complicated process when the bets are constrained on the low end by table minimums and on the high end by maximums or a desire to limit one's bet spread. But, apart from those min and max bets, all intermediate Kelly bets in the ramp are simply (bankroll x edge)/variance. And, of course, both the edge and the variance are a function of the true count. There is no squaring of the bet in the Kelly calculation.

Again, all the Kelly optimal bets are part of the ramps in each of the hundreds of charts in chapter 10 of BJA3.

Don