> Am I doing this right? If I have a wager with a 30%
> winrate and a 3 to 1 payback, and I bet $1000 here 10
> times, then:

> Average squared result = [(0.30 * $3000^2) + (0.70 *
> (-$1000)^2)] = $3,400,000

Right. Easier to call $1,000 your "unit," do all the calculations with the smaller numbers, then convert at the end. But, then, I didn't do it that way in my example, above, so can't find any fault with your approach.

> Variance for 1 bet = $3,400,000 - $200^2 = $3,360,000
> S.d. for 1 bet = $3,360,000^0.5 = $1833.03

Right.

> Variance for 10 bets = $3,360,000 * 10 = $33,600,000
> S.d. for 10 bets = $33,600,000^0.5 = $5796.55 (or just
> $1833.03 * 10^0.5)

Right.

> Also, just to confirm, in the chapter 2 s.d. tables,
> the variances listed in that one column assume a 1
> unit bet per hand played and we then multiply by the
> bet squared to get the variance for that particular
> number of units, correct?

Yes. But, there are some two-hand wagers, where the variance is increased by the covariance (roughly 0.50).

> And I think I have this right, but just to confirm:
> 1) Once all the variances in the table have been
> summed, we divide by the sum of the frequencies
> because we aren't playing all 100 hands; we just take
> out the 26.77% divisor for each TC frequency as
> common, correct?

Right.

> 2) If we didn't divide by the sum of the frequencies,
> the resulting s.d. would be for 100 hands played &
> observed, correct?

Not sure about this one. If you didn't divide by the sum of the frequencies, in the above examples, then you'd get the wrong answer. :-) The resulting s.d. wouldn't be for 100 hands played, because the frequencies don't add to 100%. I think what you meant to write was, "If this were for 100 hands played (the frequencies add to 100%), then we wouldn't have to bother to divide by the sum of the frequencies."

> (Sorry, I keep saying last question and I always have
> a few more; feel free to cut me off anytime.)

Be my guest. You're doing great.

Don