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26th June 2008, 04:33 PM | #1 |
Did you spill my pint?
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Location: Scotland
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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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26th June 2008, 04:37 PM | #2 |
I would save the receptionist.
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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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26th June 2008, 04:54 PM | #3 |
Muse
Join Date: Aug 2005
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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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26th June 2008, 04:54 PM | #4 |
Penultimate Amazing
Join Date: May 2008
Posts: 10,696
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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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"The method of science is tried and true. It is not perfect, it's just the best we have. And to abandon it with its skeptical protocols is the pathway to a dark age." -Carl Sagan "They say a little knowledge is a dangerous thing, but it's not one half so bad as a lot of ignorance."-Terry Pratchett |
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26th June 2008, 07:25 PM | #5 |
Critical Thinker
Join Date: Jun 2008
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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%. |
26th June 2008, 08:00 PM | #6 |
Illuminator
Join Date: Aug 2007
Posts: 4,438
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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%. |
26th June 2008, 08:43 PM | #7 |
Illuminator
Join Date: Aug 2007
Posts: 4,438
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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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28th June 2008, 09:07 PM | #8 |
Muse
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Location: Oxford
Posts: 999
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