My slot machine has 3 slots, and each time the user pulls the lever I want them to be given a different set of symbols in a random order every time including amount of each symbol.

For example, if hitting a row of apples has a chance of winning x100, what would be the math that determines how many apples should be in a row if I wanted the winning amount (in this case x100) to equal to probability of winning (1 in 100). So if there was 10 symbols which would be 10*10*10 to produce 1000 possible combinations (forgetting the fact that an apple can take up more than 1 of those 10 symbols). What would be the math used to determine how much of those 10 symbols needs to be apples to make it exactly a 1 in 100 chance of hitting it?

Would putting only 1 apple in each slot be 1 in 1000? Does that make 2 apples in each slot 1 in 500?

Just did some basic math but don't know if this is right:

A) If I have 10 symbols there is 1000 possibilities.

B) If 2 of those symbols where apples in each slot, there would be 8 out of 1000 possible Apple wins.

C) So does that mean there is an average 125 spins to get all apples? The closest to 1 in 100 chance? I am confused.


Well, I believe you have the same probability for each slot. $\left(\frac{a^3}{10}\right)^3=\frac1{100}$ so $a=\sqrt[3]{10}$ so it must occupy $\sqrt[3]{10}$ slots, whatever that means.

If you put $1$ apple in one slot, you would get $\frac1{1000},$ but if you put $2$ apples, then there are $2\cdot2\cdot2=8$ combinations for a total of $\frac1{125}.$

Note: If you allow different (assuming integral) numbers of slots, it can be shown it must be a permutation if $(1,5,2)$ or $(1,1,10)$ slots.

EDIT: There are a few more questions asked.

A) Yes. There are $10$ combinations for each slot for a total of $10\cdot10\cdot10=1000$ combinations.

B) Yes. The probability for one slot is $\frac2{10}=\frac15.$ The probability of three is $\left(\frac15\right)^3=\frac1{125}=\frac8{1000}.$

C) Yes, the expected value of spins required is $125.$ It is the closest unless you allow one slot to have more apple slots than another.

  • $\begingroup$ Do you mean $\frac1{1000}$ if there was just 1 apple? $\endgroup$
    – Joe
    Sep 7 '18 at 4:40
  • $\begingroup$ Yes... edited :) $\endgroup$
    – Jason Kim
    Sep 7 '18 at 4:50

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