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I'm interested in calculating the probability of a royal flush being dealt to ANY of the n players seated at a Texas Hold'em poker game (2 hole cards, 5 community cards). The probability of YOU being dealt a royal flush is readily available on the internet but I'm having trouble finding this particular probability, especially for variable number of players.

There are $133,784,560$ 7-hand combinations, $4,324$ of which result in a royal flush so the probability of YOU being dealt a royal flush is $\approx 0.00003232$

It seems that simply $n*0.00003232$ will give a good approximation to the probability I'm after but something seems off about this in that 1 player receiving a royal flush would negate anyone else from receiving a royal flush (except in the case of a royal being on the board).

In an attempt to come up with an exact solution I've done the following:

Number of ways to deal n-handed NLHE hand (sample space):

$$52!/(52-2n-5)! = 52!/(47-2n)!$$

Number of ways royal can be dealt using BOTH hole cards:

$$4*n*5*4*{5\choose 3}*3!*(47!/(47-2n)!)$$

(suit,player,1st hole card,2nd hole card,community placement,community order,distribute remaining cards)

Number of ways royal can be dealt using ONLY 1 hole card:

$$4*n*{2\choose 1}*5*{5\choose 4}*4!*(47!/(47-2n)!)$$

(suit,player,which hole card,hole card,community placement,community order,distribute remaining cards)

And finally, number of ways royal can be dealt using 0 hole cards (royal on board)

$$4*5!*(47!/(47-2n)!)$$

(suit,community cards,remaining cards)

Thus giving the probability of a royal being dealt to ANY player in an n-handed NLHE game being:

$$((4*n*5*4*{5\choose 3}*3!*(47!/(47-2n)!)+4*n*{2\choose 1}*5*{5\choose 4}*4!*(47!/(47-2n)!)+4*5!*(47!/(47-2n)!))/(52!/(47-2n)!))$$

For n=1 this gives the $\approx 0.00003232$ as we should expect and for n=9 this yields $181/6497400\approx0.000278573$ which is, as I'd expect, quite close to $9*0.00003232\approx0.00029088$.

So, I have 3 real questions:

1.) Are my intuitions about n*[7 card probability] being ONLY an approximation correct?

2.) Is my "exact" equation correct? Besides my n=1,n=9 substitutions are there are external sources I can check this against? It's been awhile since I've worked with probability and I'm not too confident in my solution.

3.) Is there a more succinct way of expressing the exact probability given the 7-card probability. Perhaps not simply n*[7 card probability] but a variation of this?

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Your calculation is correct, and you were actually quite close to the more elegant answer yourself.

You wrote:

"It seems that simply $n\cdot0.00003232$ will give a good approximation to the probability I'm after but something seems off about this in that $1$ player receiving a royal flush would negate anyone else from receiving a royal flush (except in the case of a royal being on the board)."

The opposite is actually true: If one player receiving a royal flush would prevent anyone else from receiving one, the events would be disjoint and the probability for any player to get a royal flush would indeed be simply $n$ times the probability for a particular player to get one. As you noticed, there is only a single case in which more than one player has a royal flush, namely if it's on the board, and we have to compensate for overcounting this case. In adding the $n$ probabilities as if they were disjoint, we're counting this case $n$ times instead of once, so we need to subtract it $n-1$ times. A royal flush being on the board is one of $\binom75=21$ ways of having a royal flush in $7$ cards. Thus the probability including the correction for overcounting is

$$ np-\frac1{21}(n-1)p=\frac{20n+1}{21}p\;, $$

where

$$ p=\frac{4\binom{47}2}{\binom{52}7}=\frac1{30940} $$

is the probability that you calculated for a $7$-card hand to have a royal flush. Thus the probability you want is

$$ \frac{20n+1}{649740}\;, $$

in agreement with your calculation.

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  • $\begingroup$ Would the 21 be the same for any hand and just plug in n and p? $\endgroup$ – paparazzo May 3 '17 at 19:00

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