1. ## pm: SD question

Say I run a sim and I get an SD of 30 units per 100 hands played, 1-12 spread.

1) If the count called for a bet of 4 units and I decided to bet 8 units instead, and I did this exactly twice out of every 100 hands, how would I calculate the effect these two plays would have on my SD (assuming all other bets are optimal)?

2) If the count called for a bet of 8 units and I decided to bet 4 units instead, and I did this exactly twice out of every 100 hands, how would I calculate the effect these two plays would have on my SD (assuming all other bets are optimal)?

3) If I decided to make a bet of 3 units at a count of -1 exactly twice out of every 100 hands, how would I calculate the effect these two plays would have on my SD (assuming all other bets are optimal)?

I was wondering if there's a general way to make these calculations assuming that I've obtained all the pertinent info from CVData (i.e. SD per hand at specific counts, e.v. at specific counts, or whatever other info is necessary).

Hopefully these questions make some sort of sense.

pm

2. ## Don Schlesinger: Re: SD question

> Say I run a sim and I get an SD of 30 units per 100

OK. But, the spread is irrelevant, once you have the final number.

> 1) If the count called for a bet of 4 units and I
> decided to bet 8 units instead, and I did this exactly
> twice out of every 100 hands, how would I calculate
> the effect these two plays would have on my SD
> (assuming all other bets are optimal)?

Optimal doesn't matter. What we assume is that all the other bets are exactly the same as when you produced the hourly s.d. of 30 units, and that now you're changing two bets. See p. 20 of BJA3 for the methodology (only here, we're not back-counting, so you're playing 100% of the seen hands per hour).

Basically, to the hourly variance of 900 (the square of s.d.), you're now adding 2 x 4^2 x 1.3 (two hands of four extra units, assuming average variance of about 1.3). That raises the variance by 32 x 1.3 = 41.6, to 941.6, and, therefore, the s.d. to sqrt(941.6), or 30.69. So, no big deal.

> 2) If the count called for a bet of 8 units and I
> decided to bet 4 units instead, and I did this exactly
> twice out of every 100 hands, how would I calculate
> the effect these two plays would have on my SD
> (assuming all other bets are optimal)?

Same idea as above. Variance decreases by 41.6, to 858.4; therefore s.d. becomes 29.30.

> 3) If I decided to make a bet of 3 units at a count of
> -1 exactly twice out of every 100 hands, how would I
> calculate the effect these two plays would have on my
> SD (assuming all other bets are optimal)?

Assuming that those plays would normally be for one unit, you've increased variance by 2 x 2^2 1.3 = 10.4, to 910.4. New s.d. is 30.17.

Bottom line: None of these plays makes any meaningful difference.

> I was wondering if there's a general way to make these
> calculations assuming that I've obtained all the
> pertinent info from CVData (i.e. SD per hand at
> specific counts, e.v. at specific counts, or whatever
> other info is necessary).

See above.

> Hopefully these questions make some sort of sense.

Perfect sense.

Don

3. ## pm: Thanks much

I appreciate the help as always. The chapter 2 s.d. table also seems to make more sense now.

Just to confirm, the additional 1.3 variance that we have to multiply by is due to splits and doubles, correct?

Also, in CVData, there are two sets of s.d.'s and variances given, real and normalized. What are the "normalized" values? I read the explanation in the manual but I still didn't understand.

Thanks.

pm

4. ## Don Schlesinger: Re: Thanks much

> I appreciate the help as always. The chapter 2 s.d.
> table also seems to make more sense now.

> Just to confirm, the additional 1.3 variance that we
> have to multiply by is due to splits and doubles,
> correct?

And 3-to-2 payoffs for naturals.

> Also, in CVData, there are two sets of s.d.'s and
> variances given, real and normalized. What are the
> "normalized" values? I read the explanation
> in the manual but I still didn't understand.

I'm sorry, but I'm not familiar with that feature. I'll ask Norm to respond.

Don

5. ## Norm Wattenberger: Re: Thanks much

The "normalized" numbers ignore the bets. They assume flat betting.

norm

> I appreciate the help as always. The chapter 2 s.d.
> table also seems to make more sense now.

> Just to confirm, the additional 1.3 variance that we
> have to multiply by is due to splits and doubles,
> correct?

> Also, in CVData, there are two sets of s.d.'s and
> variances given, real and normalized. What are the
> "normalized" values? I read the explanation
> in the manual but I still didn't understand.

> Thanks.

> pm

7. ## pm: Re: Thanks much

Don, just one last general question about s.d., just out of curiosity. Isn't s.d. also a function of the win percentage? Say you took 10 \$10 bets that only paid 1 to 10, but at a 95% winrate. Wouldn't this \$100 of action have a far lower s.d. than \$100 wagered at the same edge but with a lower win percentage (say a 1.09 to 1 payout with a 50% winrate, for the same 4.5% edge)? How would the s.d.'s be calculated in these situations?

pm

8. ## Don Schlesinger: Amended response

In my first answer to the above, I made a mistake, expressing the per-hand e.v. as \$4.50, instead of \$0.45 (4.5% of \$10). So, I deleted that response and have made the correction, below.

> Don, just one last general question about s.d., just
> out of curiosity. Isn't s.d. also a function of the
> win percentage?

In general, yes. But, for BJ, the average squared result of a hand (roughly 1.3) dwarfs the average expectation for a hand (which is virtually zero), so the e.v., or edge, is negligible in the s.d. calculation. When we subtract the latter from the former (to get per-hand variance), and then take the square root, to get s.d., all that really matters is the average squared result, not the edge.

However, from your example below, I see that, by "win percentage," you don't mean "edge," but rather the frequency with which we win or lose our wager. So, yes, this matters a great deal.

> Say you took 10 \$10 bets that only
> paid 1 to 10, but at a 95% win rate. Wouldn't this \$100
> of action have a far lower s.d. than \$100 wagered at
> the same edge but with a lower win percentage (say a
> 1.09 to 1 payout with a 50% win rate, for the same 4.5%
> edge)?

Yes, absolutely.

> How would the s.d.'s be calculated in these situations?

The squares of the two (identical) edges (both 4.5% of \$10, or \$0.45) would be subtracted from what I'm about to do, to get variances. Then, sqrt(var.) = s.d.

For the first example, the average squared result is:
[(0.95 x \$1^2) + (0.05 x -\$10^2)] = \$5.95, while for the second example, the average squared result is a much larger:
[(0.50 x \$10.90^2) + (0.50 x -\$10^2)] = \$109.405.

So, the two respective s.d.s are:

1. (\$5.95 - \$0.45^2)^0.5 = \$2.40, and
2. (\$109.405 - \$0.45^2)^0.5 = \$10.45

Note, as mentioned above, how the squares of the edges are virtually inconsequential, when compared to the average squared results.

Don

9. ## pm: Re: Amended response

Thanks a bunch for the info; the concepts are starting to make a lot more sense to me now.

pm

10. ## pm: Confirmation

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

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

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)

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?

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

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

pm

11. ## Don Schlesinger: Perfect!

> 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

12. ## pm: Re: Perfect!

I'm elated; I actually understand some of the fundamentals now. The reason I was asking was because I'm concocting some cover plays that can't be simmed, so the only way I can figure out what effect they'll have on my risk of ruin and e.v. is by doing the calcutions by hand. Hopefully I won't screw anything up too badly now.

Thanks; I'm sure I'll have some more questions along the way.

pm

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