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Tags gambling , probability , slot machines

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Old 26th June 2008, 04:33 PM   #1
mummymonkey
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Odds

Given 3 reels on a slot machine and 14 symbols on each reel.
What are the odds of:

1. Any pair
2. Any triple?

Ta.
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Old 26th June 2008, 04:37 PM   #2
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Originally Posted by mummymonkey View Post
Given 3 reels on a slot machine and 14 symbols on each reel.
What are the odds of:

1. Any pair
2. Any triple?

Ta.

1. 1 in 14

2. 1 in (14x14); or 1 in 196

Think of it this way: The first wheel spins. It doesn't matter what symbol comes up on it. The second wheel spins. The chance that it will match the first is 1 in 14. Now, if it matches the first, we don't care what the third wheel does. But, if it doesn't match the first, the third wheel has a 1 in 14 chance of matching.

The more I think about it, the less certain I am that the odds of a double are 1 in 14. It may be 27 in 196. Oh lord, I hope someone comes along who knows.
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Old 26th June 2008, 04:54 PM   #3
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The probability of a double that is not a triple is the number of ways you can select reels to make that happen (3: AAB, ABA, BAA) times the number of ways the symbols can make it work (14 choices for A, 13 for B, and since they're not indistinguishable no dividing by 2), or 3*14*13=546 successful ways, all divided by the total possible results (14^3), for about 19.9%.

The triple is as LL said, 0.5%.

And neither, for sanity, is (14*13*12) / 14^3 = 79.6%
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Old 26th June 2008, 04:54 PM   #4
paximperium
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Actually, in a real modern slot machine, the odds are preprogramed. Higher cost slots are programed with better odds compared to penny slots.
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Old 26th June 2008, 07:25 PM   #5
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Because the wheels are independent from one another we don't have to worry about the order they turn up in. We're not repeatedly pulling samples from the same set like we would for a deck of cards.

Let's pick any random pair, say cherries, and calculate the odds for it. One wheel will have a 1/14 chance of turning up cherries, the sencond will also have 1/14 odds for a combined total of 1/(14*14). The third wheel will have to be something that prevents a triple turning up. The third whee's odds will then be 13/14. So, the combined odds of getting a pair of cherries will be 13/(14^3) = 0.4738%. If we don't care which pair turns up(cherries lemons, whatever), then it could be any of 14 pairs for a total of 14*0.4738% = 6.633%.

Calculating the odds of triple cherries is easier. It's simply (1/14)*(1/14)*(1/14). For ANY triple it would be 14*(1/14)*(1/14)*(1/14) = 0.1052%.
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Old 26th June 2008, 08:00 PM   #6
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I calculate the chance of a pair as follows:

Reel 1 has some value. Reel 2 has a 1 in 14 chance of having the same value. Reel 3 also has a 1 in 14 chance of matching reel 1's value. Reel 3 has an independant 1 in 14 chance of matching reel 2. Three chances in 14 of a pair, assuming all 14 values on all 3 reels are equally likely, which Google tells me is 21.4%.
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Old 26th June 2008, 08:43 PM   #7
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Just wrote a program, 560 pairs out of 2744 possible combinations = 20.4%, 14 triples = .51%, 546 pairs which are not triples = 19.9%.
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Old 28th June 2008, 09:07 PM   #8
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Originally Posted by mummymonkey View Post
Given 3 reels on a slot machine and 14 symbols on each reel.
What are the odds of:

1. Any pair
2. Any triple?

Ta.
can we assume that the reels are independent, and each of the symbols is equally likely?

what odds would you bet on if you do not make these assumptions?
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