I've heard that either the standard deviation or variance is a multiple of the square root of the number of lines times that of a single line machine, which makes intuitive sense.
----- some_bitch_already_took_sumorez <will@express-pc.com> wrote:
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>I've been trying to search online for this and I'm not coming up with anything. It's relatively easy to determine, within a standard deviation, what your probability is for single-line games such as 9/6 Jacks. What I'm not sure about, however, is for 50/100 play games, with a given bankroll.
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> For example -- a few months ago, a friend wanted me to calculate probabilities for getting Diamond in a Day on a $5 9/6 machine, without factoring in cashback/comps/etc. I came up with the following (I don't have my work here, sorry):
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> -$8,900 to -$5,200: 0.16%
> -$5,200 to -$1,500: 32%
> -$1,500 to $2,200: 59%
> $2,200 to $5,900: 4.2%
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> I'm trying to find something similar, but for, say, a $1 50-play 9/6 Jacks machine. How would I go about this? I don't know how the multi-hand aspect of it changes variance (other than to say "it increases it by some number").
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