I've picked up a project from someone else who has created a spreadsheet that is meant to represent some slot machine mathematics and i'm struggling to understand the math that is used to get certain figures.

The game is built on 3 sets of balls that are draw. 10 balls, 15 balls and 20 balls. Each set has 3 balls drawn from them giving you 3/10, 3/15 and 3/20. These ball sets are unique and individual and have no correlation to each other. From there we're building a set of paytables on those.

The first tier that was created is: Minimum 1 matched from each set which is calculated as:

ball set of 10 = 0.708333333

ball set of 15 = 0.516483516

ball set of 20 = 0.403508772

Then he multiplied all of those together to get a probability to calculate the rest of that line in the table.

I need to understand how he got those 3 figures that make up 1 probability figure.



In the first case, you are drawing 3 balls from a group of 10, and you want to calculate the chance that at least one of them is one of the 3 'winning' balls.

This is (1-P(none of them 'winning')) = $1-(\frac{7}{10})(\frac{6}{9})(\frac{5}{8}) = 0.708333333$

The other numbers are calculated similarly.

  • $\begingroup$ Hey, that's absolutely perfect. Thanks so much. Could you assist with the payout of "Exactly 1 matched from each set" - How one would generate: 10 ball set = 0.525, 15 ball set = 0.435164835, 20 ball set = 0.357894737. Thanks again! $\endgroup$ – Quinn Olive Nov 10 '16 at 11:33
  • $\begingroup$ For each set, you've got the probability of at least one match in that set. The event of getting a match in any given set is independent of the event of getting a match in another set. For indepdenent events, the probability of all of them occuring is the product of their individual probabilities. $\endgroup$ – Iadams Nov 10 '16 at 11:48
  • $\begingroup$ Sorry, I msread your supplementary. In the 10-ball case, you get exactly one winner, by drawng one of the three winners plus two of the seven losers. There are 3 ways to pick a sinlge winner and $frac{7 x 6}{2}$ ways to pick two losers - making 3 x 21 = 63 ways to have exactly one winning ball. There are ${3 \choose 10}$ = 120 ways to pick any three, so the chance of exactly one winner is $frac{63}{120} = 0.525$ $\endgroup$ – Iadams Nov 10 '16 at 15:42

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