A thoery question?
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A thoery question?
Every 40k turns should produce s royal? So the same should happen if i had 40000 vp machines lined up and 40000 people all hit once at same time 1 should produce a royal?
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I think you would have a 50-50 chance of someone hitting one. You'd also have a 50-50 chance of there being two.
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Every 40k turns should produce s royal? ? That's not really correct.
The fact that the statistical average is around one royal per 40000 hands does not mean you "should" hit one in that number of hands.
The fact that the statistical average is around one royal per 40000 hands does not mean you "should" hit one in that number of hands.
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Would not the factor that the odds would pertain to each of the 40000 machines, each having played only 1 of 40,000 plays?
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I would think 40,000 people having one spin on 40,000 machines would be no different than 40,000 people each taking one spin on one machine, or one person taking one spin on 40,000 machines, or one person taking 40,000 spins on one machine. I think the only factor that matters is the 40,000 spins.
Could be wrong. Certainly wouldn't be the first time. Or the last.
Could be wrong. Certainly wouldn't be the first time. Or the last.
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I think you would have a 50-50 chance of someone hitting one. You'd also have a 50-50 chance of there being two.It's actually closer to:1/3 of the time nobody would hit a royal: 1/3 of the time there would be exactly one royal; 1/3 of the time two or more would be hit.if you go to the www.gamblingwithanedge.com website, my weekly column from 10 days ago gave a discussion of this type of situation.
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I think you would have a 50-50 chance of someone hitting one. You'd also have a 50-50 chance of there being two.
It's closer to a 1 - 1/e chance of someone hitting at least one.
e = 2.718281828459045
1 - 1/e = 0.6321205588
More accurate answer for a hypothetical game with a royal cycle of exactly 40,000.
1 - (39999/40000)^40000 = 0.6321251574
It's closer to a 1 - 1/e chance of someone hitting at least one.
e = 2.718281828459045
1 - 1/e = 0.6321205588
More accurate answer for a hypothetical game with a royal cycle of exactly 40,000.
1 - (39999/40000)^40000 = 0.6321251574
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[QUOTE=billryan] I think you would have a 50-50 chance of someone hitting one. You'd also have a 50-50 chance of there being two.
It's closer to a 1 - 1/e chance of someone hitting at least one.
e = 2.718281828459045
1 - 1/e = 0.6321205588
More accurate answer for a hypothetical game with a royal cycle of exactly 40,000.
1 - (39999/40000)^40000 = 0.6321251574[/QUOTE]
While I appreciate your response, I really don't understand it. Could you dumb it down a half dozen notches or so? Thanks.
It's closer to a 1 - 1/e chance of someone hitting at least one.
e = 2.718281828459045
1 - 1/e = 0.6321205588
More accurate answer for a hypothetical game with a royal cycle of exactly 40,000.
1 - (39999/40000)^40000 = 0.6321251574[/QUOTE]
While I appreciate your response, I really don't understand it. Could you dumb it down a half dozen notches or so? Thanks.
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While I appreciate your response, I really don't understand it. Could you dumb it down a half dozen notches or so? Thanks.I already did.While VMAN correctly said the chance for at least one royal was 63.1%, my "simplified" answer was 66.7% --- which I believe is close enough for most people trying to get a general grasp of the situation.If you're doing exact calculations, his number is much better than mine.