A thoery question?

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stevel96a1
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A thoery question?

Post by stevel96a1 »

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?

billryan
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Post by 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.

onemoretry
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Post by onemoretry »

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.

Stan_7777
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Post by Stan_7777 »

Would not the factor that the odds would pertain to each of the 40000 machines, each having played only 1 of 40,000 plays?

billryan
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Post by billryan »

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.

BobDancer
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Post by BobDancer »


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.

Vman96
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Post by Vman96 »

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

billryan
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Post by billryan »

[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.

BobDancer
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Post by BobDancer »





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.


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