0
$\begingroup$

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.

Thanks!

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.