Originally Posted by
DSchles
You went to the right place, but you asked to be a friend, which, normally, I don't accept. You needed to simply send a message (email) by putting in DSchles in the "To" box.
In any event, maybe an example will help you to understand some fundamental concepts.
Suppose you flat bet $5 a hand for 100 hands. What is the overall s.d. for that? Well, the one-hand s.d. is 1.15 x $5, which is $5.75. Since overall s.d. is a square root function, the total s.d. for 100 hands is $5.75 x 100^0.5 = $5.75 x 10 = $57.50. Simple and straightforward, right?
So, now, let's consider someone who bets $1 a hand for 50 hands and $9 a hand for another 50 hands. Clearly, that person has an average bet of the same $5 as above, and so you might be tempted to use that value and conclude that the overall s.d. for the 100 hands is the identical $57.50 as above, but you would be very wrong to do so!
As I mentioned in my previous post, if bet sizes are different (makes no difference if you are spreading as a card counter, playing progression, or whatever), you cannot use the average bet size to calculate s.d. Instead, you need to use a somewhat more complicated process that involves: using the variance (the square of s.d.), multiplying by the square of the bet size, multiplying by the frequency of that particular bet, summing all the horizontal lines vertically, and, finally, taking the square root of that sum.
Wow! All that, huh? Yup. and there are no shortcuts; that's the way to do it. Period. So, instead of getting the $57.50 that you thought might be the s.d. for the second example, what is the correct answer? Well, let's follow my instructions above:
Variance is 1.3225. First bet squared is 1. Second bet squared is 81. Frequency of each bet is 50 (hands). So, ...
[(1.3225 x 1 x 50) + (1.3225 x 81 x 50)]^0.5 = (66.125 + 5,356.125)^0.5 = 5,422.255^0.5 = 73.64. Oh!! So, $73.64 instead of $57.50, despite the same average bet of $5 each time.
Clear?
Don
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