chgobjpro: Calc SD with Wonging out?

[chgobjpro]

How is SD calculated when wonging out?
Example .. Today I played a bad game.
6D,H17, DAS Pen was 4.5/6.
3 other players, myself and the dealer.
Approx 100 hands per hour.
All bets were on single spot.
I was able to sit out all negs and
play only Trues =>0.
My bets per hour were as follows.
26 bets of 0 (wonged out hands)
60 bets of 10= 600
7 bets of 100= 700
4 bets of 150= 600
2 bets of 200= 400
1 bet of 300= 300
Total bet amount=2600
Is average bet for
calc SD $26 per hand(using 100 hands)
or $35 per hand(using 74 hands) per hour.

Are the wonged out hands counted in the
calculation of the sq rt of no. of
hands played?

What is one SD range of this game?

Thanks chgo

[Don Schlesinger]

> My bets per hour were as follows.
> 26 bets of 0 (wonged out hands)
> 60 bets of 10= 600
> 7 bets of 100= 700
> 4 bets of 150= 600
> 2 bets of 200= 400
> 1 bet of 300= 300

Ok, fine, but the total bets are irrelevant.

> Total bet amount=2600

Doesn't enter into the equation.

> Is average bet for
> calc SD $26 per hand(using 100 hands)
> or $35 per hand(using 74 hands) per hour.

Average bet doesn't enter into equation either. :-)

Do it this way:

1. Square each bet size (the zero bets count as zero);

2. Multiply each squared result by its corresponding frequency and by the variance of a hand of blackjack played when those bets are made. You won't necessarily know this number, but it can backed out from the Chapter 10 charts of BJA -- I explain how to do this on p. 178. If you're lazy, just take 1.33 as a ballpark estimate to be used on each line;

3. Sum vertically;

4. Take the square root of your answer.

Voila! That's the SD per 100 hands SEEN.

> Are the wonged out hands counted in the
> calculation of the sq rt of no. of
> hands played?

In a manner of speaking. They count as zero (see above).

> What is one SD range of this game?

I get 66.85 units per hour as one SD. I did it VERY quickly. Check my math.

Don

[Don Schlesinger]

It's important to understand why you can't use your average bet size to do any SD calculations, if you are employing a large spread and varying your bet sizes.

I was rushed last night, but now, I'd like to explain why this is so, wiht a simple illustration.

Suppose you made bets of, say, $10 50 % of the time and bets of $90 the other 50% of the time.

Then, of course, your average bet size is exactly $50. Now, if you use the latter value and square it, you get $2,500. But, if you use the two different values, square them, multiply by their frequencies (50% each time) and sum, you get a very different value, namely $4,100. So, the correct SD is 64% HIGHER than you would get by using the (erroneous) average-bet method.

Hope this is clear.

Don

[Don Schlesinger]

> So, the correct SD is
> 64% HIGHER than you would get by using the
> (erroneous) average-bet method.

The above, not bothering to multiply by the 1.33, for ease of illustration, would be the relative variances. You'd need to take the square root of each to get the SDs. So, SD in one case would be $50, while SD in the other case would be $64.03, which is 28% higher.

Sorry for any confusion.

Don

[chgobjpro]

I followed your formula.
All my bet units squared and mutiplied by the number of hands on each bet. Then multiplied by 1.33 and added the total, which came to 446,880. Calced the Sqrt. I came up with $668.
Is that the same as your 66.85 units and you figured $10 units?

[Don Schlesinger]

> I followed your formula.
> All my bet units squared and mutiplied by
> the number of hands on each bet. Then
> multiplied by 1.33 and added the total,
> which came to 446,880. Calced the Sqrt. I
> came up with $668.
> Is that the same as your 66.85 units and you
> figured $10 units?

Yup. Congrats. You've just passed SD 101! :-)

Don

P.S. You might also check out BJA, p. 23.

[chgobjpro]