Multi-Hand Effect on Royal Cycle

I was just running some numbers on VPW and noticed that when I analyzed single line 9/6 JoB, it says the Royal frequency is 1 every 40,390 hands on average. When I analyzed the same game on triple play, the program still says the frequency is 40,390. That doesn't seem right.

Now perhaps by playing three times as many hands, you are supposed to divide the total by three, making the cycle 1 in every 13,463 hands, but that doesn't seem quite right either since all the hands are correlated to some degree.

Does anyone have any input on how to correctly calculate the cycle when playing multiple hands?

Playing multiple hands does not affect the frequency of getting a certain hand, that is determined entirely by the strategy you employ.

Quote: bigfoot66

Playing multiple hands does not affect the frequency of getting a certain hand, that is determined entirely by the strategy you employ.

I don't think that's what being asked though. I think they want to know the average number of "rounds" needed to see at least one Royal. Which is a little less likely than 1 in 40,390/3. I might try to do more detailed math later if JB, Wiz, etc. don't beat me to it first...lol

Thanks Tring. I'd like to get a better idea of how the correlation affects the cycle, but dividing by the number of hands will work as an estimator for now.

If anyone has any further input, it would be appreciated.

I come up with a cycle of about 14,148 games for Triple Play and 8,625 games for Five Play, assuming 9/6 Jacks or Better played with optimal strategy.

Thanks for allowing me to allowing me to be lazy slash playing MyVegas blackjack tonight, JB. :D

Quote: JB

I come up with a cycle of about 14,148 games for Triple Play and 8,625 games for Five Play, assuming 9/6 Jacks or Better played with optimal strategy.

It's not as often as that because of the burn factors.

Quote: JIMMYFOCKER

It's not as often as that because of the burn factors.

I obtained my results as follows:

1) I counted the number of deal hands where the correct strategy play makes it possible to complete a Royal Flush on the draw. For example, if the dealt hand is a Full House, it cannot complete a Royal Flush on the draw because the correct strategy play is to hold the Full House. Only hands where the correct play can result in a royal were counted. Here are the results:

a) Hold 0 cards, 3 possible royals on the draw = 52,080
b) Hold 0 cards, 4 possible royals on the draw = 32,280
c) Hold 1 card, 1 possible royal on the draw = 390,828
d) Hold 2 cards, 1 possible royal on the draw = 204,660
e) Hold 3 cards, 1 possible royal on the draw = 27,492
f) Hold 4 cards, 1 possible royal on the draw = 936
g) Hold 5 cards, 1 possible royal on the draw = 4

2) For each "signature" listed above, I determined the probability of a single hand becoming a royal on the draw by dividing the number of possible royals by the number of cards chosen from the remaining 47. I will refer to these probabilities as "Hit" for completing a royal flush on the draw, and "Miss" for not completing a royal flush on the draw:

a) Hit = 3/1533939; Miss = 1533936/1533939
b) Hit = 4/1533939; Miss = 1533935/1533939
c) Hit = 1/178365; Miss = 178364/178365
d) Hit = 1/16215; Miss = 16214/16215
e) Hit = 1/1081; Miss = 1080/1081
f) Hit = 1/47; Miss = 46/47
g) Hit = 1/1; Miss = 0/1

3) For the 8 ways to hit and/or miss on the 3 hands in Triple-Play (HHH, HHM, HMH, MHH, HMM, MHM, MMH, MMM) I multiplied the above probabilities accordingly. For example, the probability of 3 misses (MMM) when holding 4 to a royal is 0.93751866156824595706153742427015. The probability of completing a royal on all 3 hands (HHH) when holding 3 to a royal is 0.0000000007916312286680072872916119552183.

4) For each signature, I summed the results in #3 for every outcome except MMM, since the original question asked about the cycle of seeing at least one royal in Triple Play.

5) I multiplied the sums in #4 by the probability of having been dealt a starting hand with that signature (the counts in #1 divided by 2,598,960). The final probabilities for each signature are:

a) Hold 0 cards, 3 possible royals on the draw = 0.000000117572282851336
b) Hold 0 cards, 4 possible royals on the draw = 0.0000000971641274005428
c) Hold 1 card, 1 possible royal on the draw = 0.00000252927037070693
d) Hold 2 cards, 1 possible royal on the draw = 0.0000145683676311798
e) Hold 3 cards, 1 possible royal on the draw = 0.0000293292181685062
f) Hold 4 cards, 1 possible royal on the draw = 0.0000225022827485309
g) Hold 5 cards, 1 possible royal on the draw = 0.00000153907716932927

6) I summed the probabilities in #5 to get the overall total for the game, which is 0.0000706829524985049.

7) The inverse of the result of #6 is 14,147.68. Therefore, the probability of being dealt a hand where the correct strategy play can result in a royal flush on the draw, and actually completing at least one royal flush on the draw, in Triple-Play, is approximately 1 in 14,148. Note that this is a lower probability than simply multiplying the single-hand royal flush probability by 3. The result of that is approximately 1 in 13,464. The methodology I used takes the "bottleneck" effect of the deal hand into account, and therefore reflects the reduced likelihood of seeing a royal.

I do not believe my results are incorrect but will not hesitate to admit it if they are, so please share your methodology and results so we can pinpoint where our figures disagree.

I'd personally like to know what "burn factors" even is. I have never heard of the term. Whatever it is, I'm pretty sure JB has it accounted for. I wasn't going to even examine it as close as he did and give a "close approximation" instead.

Thanks again JB!