Harry Porter wrote:
aaquad250 wrote:
> if i do a coin-toss version of multistrike:
>
> outcome prob win ev=pw var=p(w-1)^2
> L 0.5 0 0 0.5
> W,L 0.25 0.5 0.125 0.0625
> W,W,L 0.125 1.5 0.1875 0.03125
> W,W,W,L 0.0625 3.5 0.21875 0.390625
> W,W,W,W 0.0625 7.5 0.46875 2.640625
> sum 1 1 3.625
>
> may indicate that multistrike variance is 3.625 times greater than
> the base game since regular coin toss has a variance of 1.
That's an analogy that was introduced in a parallel discussion over on
acvpp (AC vp group) by Effen Dolts.
Are you saying that the above analogy was introduced by Effen, or that the discussion itself was introduced by Effen and that someone else introduced the above analogy in that discussion? ... because I'm a bit confused (not hard to do, I guess). I recall that Effen had actually figured the variance to be 58 in this identical coin-flip scenario; I recall that because I had replied to his post after doing the math myself to confirm that "58" was correct. Now we have this "3.625" figure, and I'm trying to figure out if Effen has decided that the "58" was wrong for some reason. Furthermore, in looking at the above, which I commented on awhile ago (maybe yesterday), I can't quite follow it, but at the same time I have great regard for you and your abilities, Harry, as well as for Effen, so I'm respectfully asking you both to be a little bit patient with me and help me along here.
I've always understood Variance to be equal to [sum(x - mean)^2] / n ... where x is each sample, mean is the average value of all samples, and n is the number of samples. I probably haven't expressed that very well, but I'm sure you know the formula and what I mean by the above. [Of course, sometimes the denominator is (n-1), but in large number like "long-term" play, (n-1) approaches n, so I don't bother with that in calcs such as these.] Now, in light of what I know is supposed to be the correct 'formula' for variance, I have studied the above 'analogy' to see if it is the same thing, but just expressed in another way, and I'm having trouble seeing that it is -- not to mention the fact that if you do the arithmetic with the above formula and plug in 16 values representing exact 50/50 values at each of the four levels, you will come out with "58" -- so it is impossible that the above procedure is the same as the conventional formula for variance. But since you haven't spotted that, or commented on it, and not only "feel that it has some validity" but actually suggest that it "is a bit inflated", I am getting really confused about whether we're discussing the same thing or not.
It would help alot if I understood what the formula is this analogy ... var=p(w-1)^2 ... means and where it comes from. From the rest of the 'table', it looks like it means ...
... variance = probability x [(winnings - 1)squared]. For starters, I can't see where the -1 comes from; is it supposed to be a subtraction of the mean, with the mean being equal to 1? "1" is the average value in a 50/50 game where a lose pays 0 and a win pays 2, but in this particular scenario, we are betting 4 at a time (paying for all four levels up front), and the average is certainly not 1 if we are supposed to use the average for the "game" (from start to finish).
In case you missed my earlier post, here is the math portion using the conventional formula, but altered to follow the above-format:
L 8 out of 16 -- pays (0) = 0 x 8 = 0
W,L 4 out of 16 -- pays (2) = 2 x 4 = 8
W,W,L 2 out of 16 -- pays (2 + 4) = 6 x 2 = 12
W,W,W,L 1 out of 16 -- pays (2 + 4 + 8) = 14 x 1 = 14
W,W,W,W 1 out of 16 -- pays (2 + 4 + 8 + 16) = 30 x 1 = 30
Totals 16 Total = 64 / 16 = 4
Average Winnings per 'game' is "4"
We then take the difference between each sample and the average:
When we win 0 the difference is -4, squared is 16 ... x 8 = 128
when we win 2 the difference is -2, squared is 4 ... x 4 = 16
when we win 6 the difference is +2, squared is 4 ... x 2 = 8
when we win 14 the difference is 10, squared is 100 ... x 1 = 100
when we win 30 the difference is 26, squared is 676 ... x 1 = 676
Total = 928
which we then divide by the total number of games (16) to get 58.
Have I made a mistake in here somewhere, or am I applying the wrong formula?
Thanks.
Bill Velek
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I feel that it has some validity, but my gut feeling is that the
resulting indicator of MS variance is a bit inflated based upon my
limited experience with the game and various anecdotal player reports.
Keep in mind that the mechanism here is far simpler than that of video
poker. Effen describes the analogy as "only an approximation".
However, he does provide some strong reasoning as to why the
approximation should be in the ballpark.
Again, bottom line, I think the question of MS variance is only going
to be resolved when someone decides to crank through the tedious
arithmetic required.
- Harry
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