Best videopoker bonuses

billryan
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October 27th, 2020 at 10:43:34 AM permalink
Can anyone give the frequency of getting 1)a rainbow flush-a hand with all four suits and 2)the same for all four suits and an Ace.
Would the second happen approx 1/13th of the first hand?
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unJon
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October 27th, 2020 at 11:05:30 AM permalink
Quote: billryan

Can anyone give the frequency of getting 1)a rainbow flush-a hand with all four suits and 2)the same for all four suits and an Ace.
Would the second happen approx 1/13th of the first hand?



From a five card poker hand? Dealt, no draw?
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billryan
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October 27th, 2020 at 11:29:51 AM permalink
Regular JOB game.
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gordonm888
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October 27th, 2020 at 5:20:15 PM permalink
The odds of the first five cards dealt in a VP game having all 4 suits is 26.275% (approx.)
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billryan
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October 27th, 2020 at 5:29:50 PM permalink
Thank you, and as a final hand if that can be factored in. I realize much of that depends on what is held, but even a general estimate would be helpful. It's for a silly bar bet and I thought it would be easier to find.
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October 28th, 2020 at 10:17:45 AM permalink
I spent some time thinking about the final hand and its a very complex calculation which would take me many hours. The answer is almost certainly between 15% and 35% and I would not be surprised to see it be close to the 26% number I quoted for 5 random cards from a 52 card deck.

Simulation would really be best - if one of the simulation gurus in the forum would be willing to do it (VP has so many strategy rules, it makes simulation pretty complicated.)

One approach would be to for you to play, say, 100 hands on an online video poker and keep track of how many hands were rainbow. That would provide an estimate.
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
billryan
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October 28th, 2020 at 10:19:44 AM permalink
Thank you for your efforts.
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malgorium
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October 30th, 2020 at 8:43:00 PM permalink
Quote: gordonm888

The odds of the first five cards dealt in a VP game having all 4 suits is 26.275% (approx.)



How did you calculate this? I'm getting 26.375% and I'm not sure if I'm doing something wrong...
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gordonm888
October 31st, 2020 at 7:55:06 AM permalink
Quote: malgorium

How did you calculate this? I'm getting 26.375% and I'm not sure if I'm doing something wrong...

Gordonm888 probably (pun intended) had just miss typed the 3 as 26.375% is correct.
Last edited by: miplet on Oct 31, 2020
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October 31st, 2020 at 2:54:34 PM permalink
Quote: miplet

Gordonm888 probably (pun intended) had just miss typed the 3 as 26.375% is correct.



Absolutely correct. Here is the number I calculated: 0.263745498. I got so worried about how to usefully round the digits that I mistyped it. Sorry for the confusion.
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ssho88
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November 2nd, 2020 at 4:01:49 AM permalink
Quote: billryan

Thank you for your efforts.



5 cards poker hand with 4 suits, but there are 2 cards with same suit, so total different permutations = 5!/2!.

So, probability = 13/52 * 13 /51 * 13/50 * 13/49 * 48/48 * 5!/2! = 0.2637455.
malgorium
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November 2nd, 2020 at 3:36:33 PM permalink
Quote: ssho88

5 cards poker hand with 4 suits, but there are 2 cards with same suit, so total different permutations = 5!/2!.

So, probability = 13/52 * 13 /51 * 13/50 * 13/49 * 48/48 * 5!/2! = 0.2637455.



Ah interesting. Yea, I did it like

If you want all 4 suits, then there have to be exactly 2 of 1 suit. So there are 13C2 ways to do that and 13C1 for the other suits. But the double-suit can happen for any of the 4 suits, so there are 4(13C2)(13C1)(13C1)(13C1) total ways to make a hand with all 4 suits.

And there are 52C5 total ways to make any 5 card hand, so the total probability is just the quotient of the two, which is the same value you got.

That's the great thing about math - more than 1 way to skin a cat!
marcel55
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March 20th, 2021 at 7:19:39 AM permalink
Did you do all the calculation again?
Mission146
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March 20th, 2021 at 8:48:17 AM permalink
Quote: malgorium

Ah interesting. Yea, I did it like

If you want all 4 suits, then there have to be exactly 2 of 1 suit. So there are 13C2 ways to do that and 13C1 for the other suits. But the double-suit can happen for any of the 4 suits, so there are 4(13C2)(13C1)(13C1)(13C1) total ways to make a hand with all 4 suits.

And there are 52C5 total ways to make any 5 card hand, so the total probability is just the quotient of the two, which is the same value you got.

That's the great thing about math - more than 1 way to skin a cat!



Here's another way:

nCr(13,2)*nCr(13,1)*nCr(13,1)*nCr(13,1)/nCr(52,5) = 0.0659363745498199

0.0659363745498199 * 4 (suits) = 0.2637454981992796 (Difference with SSho due to rounding)

Which is actually the same way you did it, essentially, just using the nCr function instead.
https://wizardofvegas.com/forum/off-topic/gripes/11182-pet-peeves/120/#post815219
camapl
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June 29th, 2021 at 3:29:16 AM permalink
Quote: billryan

Can anyone give the frequency of getting 1)a rainbow flush-a hand with all four suits and 2)the same for all four suits and an Ace.
Would the second happen approx 1/13th of the first hand?



For part 2) of your OP, having all four suits and an Ace… Assuming you have exactly 1 Ace, and it must match suit with one of the other cards, then the probability is (4C1)(12C1)(12C1)(12C1)(12C1) / (52C5) = 3.19%
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