Odds of the same Royal Flush 2x within 10 hands

I have a question for the Wizard...

I had some unbelievably good luck recently on Video Poker. At a casino near my home in Colorado I drew a Royal Flush in Spades holding 3 of 5 cards and drawing the remaining 2. Within 10 hands of getting the first Royal, the exact same thing happened again on the same machine. Second Royal was also in Spades and also holding 3 of 5 and drawing 2.

What are the odds of that occurring??

Hi, and welcome to WoV,

Not the Wiz, but here is my attempt at an answer for "JoB 9/6" Video Poker:

1/561,105,327 for both of them happening in spades (for any starting hand)

1/3,516,764,163 for the "Royal Flush Spades holding 3 of 5 cards and drawing the remaining 2", starting hands.

Note 1: The math will possibly be a bit different for other kinds of video poker, which is why I put "JoB 9/6" above (since i don't think you said what it was).

Note 2: If someone else provides a different answer, then I would guess that mine is likely to be wrong.

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Edit:
My second figure (the one I think you are trying to find the answer to) is just an "educated guess", so it could be out by quite a bit.

Thanks!!

Coincidentally, it occurred on two different games, but on the same machine. The first one was on 9/5 Double Double Bonus. After I got paid, I switched to 8/5 Bonus Poker Deluxe which is where the second one occurred.

Quote: ksdjdj

(snip)
1/3,516,764,163 for the "Royal Flush Spades holding 3 of 5 cards and drawing the remaining 2", starting hands.
(snip)

I would guess that the probability was as low a 1/3.8 billion, playing the games you mention ( when using basic strategy).

Also, below are the links that I used to help me work this stuff out.

link_1 , link_2 , link_3

Also, the odds of this happening to you are likely not as rare as what the above odds would make you think it is, because you have probably played more than 10 hands of VP in your life, right? (see below)

Say that you have played 19 total hands in your life, then there are 10 ways to get the ""Royal Flush Spades holding 3 of 5 cards..." twice, within any of the "10 hand blocks of hands".
You roughly have a 1/380 million chance of seeing this pattern, in 19 hands.

Let me rewrite the question as follows:

What is the probability that in ten hands I get a royal flush in the same suit, twice, both times holding three to the royal?

I show the answer to that question is 1 in 375,301,378.

Let p = probability of a royal flush, holding 3 = 4*combin(5,3)*combin(47,2)/combin(52,5) = 0.00092507

Let q = All other events = 1-p = 0.99998461

The answer is combin(10,2)*p^2*q^8*(1/4)

The reason for the 1/4 is the two royals must be in the same suit.

If both royals must be in spades, then divide again by 4.

Quote: Wizard

Let me rewrite the question as follows:

What is the probability that in ten hands I get a royal flush in the same suit, twice, both times holding three to the royal?

I show the answer to that question is 1 in 375,301,378.

Let p = probability of a royal flush, holding 3 = 4*combin(5,3)*combin(47,2)/combin(52,5) = 0.00092507

Let q = All other events = 1-p = 0.99998461

The answer is combin(10,2)*p^2*q^8*(1/4)

The reason for the 1/4 is the two royals must be in the same suit.

If both royals must be in spades, then divide again by 4.

Well, I would agree with this but:

Occasionally you will be dealt 3 to a Royal but have a made straight flush, flush or straight and thus not re-draw. So, technically, the odds are slightly longer than those posted above. Aren't they?

You worded my question much better than I did myself!

Thanks!!!

Quote: Wizard

I show the answer to that question is 1 in 375,301,378.

Let p = probability of a royal flush, holding 3 when your high royal card is:

= 4*combin(5,3)*combin(47,2)/combin(52,5) = 0.00092507

Let q = All other events = 1-p = 0.99998461

The answer is combin(10,2)*p^2*q^8*(1/4)

The reason for the 1/4 is the two royals must be in the same suit.

If both royals must be in spades, then divide again by 4.

I am having another problem, namely, the math and the equations. First

4*combin(5,3)*combin(47,2)/combin(52,5) =0.016637424 NOT 0.00092507 as stated above.

In fact, the Wizard's 0.00092507 = 1/combin(47,2)


I think the probability equation for a Royal Flush with 3 initial Royal cards is:

When 3 royal cards are A High, p= 4*6*(c(47,2)-c(8,2) -1*3*3)/c(52,5)/c(47,2) = 0.00000891839
When 3 royal cards are K High, p= 4*3*(c(47,2)-c(8,2) -2*3*3) /c(52,5)/c(47,2) =0.00000442075
When 3 royal cards are Q High, p= 4*1*(c(47,2)-c(8,2) -3*3*3)/c(52,5)/c(47,2) =0.00000146077

The sum of the 3 probabilities above p(tot) = 0.00001479991
= 1/67567.9596

And q = 0.99998520

Edit: So, I get that the answer to the question is 1 in 405,863,977.

Edit: The terms in bold type in my equations account for the fact that you don't draw to 3 Royal Cards when you have a made Flush,SF or Straight.

Did I make a mistake?

Quote: Wizard

Let me rewrite the question as follows:

What is the probability that in ten hands I get a royal flush in the same suit, twice, both times holding three to the royal?

I show the answer to that question is 1 in 375,301,378.

Let p = probability of a royal flush, holding 3 = 4*combin(5,3)*combin(47,2)/combin(52,5) = 0.00092507

Let q = All other events = 1-p = 0.99998461

The answer is combin(10,2)*p^2*q^8*(1/4)

The reason for the 1/4 is the two royals must be in the same suit.

If both royals must be in spades, then divide again by 4.

I got the final formula the same as you, but why was this wrong? (see below):

Here is what I did:

1) Find p by doing (1/92 x 1/1,081) = 1/99,452***
*** : I multiplied the figures in the two tables in this link, see "RF3"
2) Find p (in spades) by dividing the figure in 1) by 4 = 1/397,808
3) Plug the figure above into a spreadsheet using the formula below:

Formula used (in the spreadsheet): (COMBIN(C7;2))*(B7^2)*(1-B7)^(C7-2)

Note: 10= "the number of hands played" = "C7", and p = "B7"

Therefore, my attempt at an answer = ~1/3,516,764,163

Even if I do it an alternative way,

1) Find p by doing (1/92 x 1/1,081) = 1/99,452***
*** : I multiplied the figures in the two tables in this link, see "RF3"
2) Plug the figure above into a spreadsheet using the formula here >>> (COMBIN(C7;2))*(B7^2)*(1-B7)^(C7-2) = approx. 1/219,811,021
3) Divide the above answer by 16 (4 x 4) to get the probability that both are in spades

Therefore, my attempt at an alternative answer would = ~1/3,516,976,339

Note: I know that you don't always keep "3 to a Royal", but I didn't really know how to account for that exactly.

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Again, these other people that have answered are a lot better at this type of math (and probably math in general too) so they (or others) may be able to explain what I did wrong.

Thanks in advance

Quote: gordonm888

Occasionally you will be dealt 3 to a Royal but have a made straight flush, flush or straight and thus not re-draw. So, technically, the odds are slightly longer than those posted above. Aren't they?

I agree. However, I'm not going to muddy the waters with those exceptions.

Quote: Wizard

What is the probability that in ten hands I get a royal flush in the same suit, twice, both times holding three to the royal?
...
Let p = probability of a royal flush, holding 3 = 4*combin(5,3)*combin(47,2)/combin(52,5) = 0.00092507

If I'm reading you right, I think your answer assumes that both times, the player starts with 3 to a royal. But there's a 1/92 chance to be dealt 3 to a royal to begin with. Maybe the OP wants to know the chances of getting two royal flushes in 10 hands, where each royal started with 3 to a royal? Is that the question you answered? If not, the odds are gonna be longer....

Quote: Wizard

I agree. However, I'm not going to muddy the waters with those exceptions.

Hey Wizard, scroll up. Note that I did indeed calculate the odds while factoring in those exceptions.

And my question is: why quote your answers to 9 significant digits when you are ignoring aspects of the problem that affect the 5th or 6th digit?

Here's my stab at it:

What are the chances of getting dealt 3 to a royal in spades twice over 10 hands, and converting those to royal flushes both times?

The chances of 3 to a royal on the deal are 1/92.

For the 3RFs to be spades, ÷ 4 = 1/368.

Using the binomial distribution formula, the chances of that happening twice over 10 hands is 1/3076.

The chances of converting 3 to a royal, to a royal flush, is 1/1081. So, ÷1081 for each of the two hands, = 1 in 3,594,493,636.

Howzat?

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