What are the odds of getting a quad aces + royal flush in a Texas Hold'em game?

[Warp]

There's this youtube video about a Texas Hold'em match where one player gets four aces and the other a royal flush. I was wondering what are the statistical chances of this happening (the answer preferably in the form "1 in x".)

 

[billy_joule]

There are probabilities for poker hands here:

https://en.wikipedia.org/wiki/List_of_poker_hands

To get to "1 in x" form, convert from % to a decimal then take the reciprocal.

 

[phinds]

I'd say the odds are zero since a royal flush requires an ace and there are only four in the deck and you are describing a scenario where there are 5 of them.

 

[Mentallic]

I'd say the odds are zero since a royal flush requires an ace and there are only four in the deck and you are describing a scenario where there are 5 of them.

Not quite. In texas holdem you only hold 2 cards in your hand anyway, so the other 2 for the quads have to be on the table, which are shared amongst both players.

If I have time over the next few days, I'll write a program to simulate the scenario and find the probability that way.

 

[Orodruin]

In a randomly dealt hand (heads-up) which goes to the river, the probability of this happening would be 1/877961175 (assuming I got all the factors right, which is never assured). With a more people at the table, the chances would go up. This is a rather trivial exercise in combinatorics.

 

[Warp]

I'd say the odds are zero since a royal flush requires an ace and there are only four in the deck and you are describing a scenario where there are 5 of them.

I find it curious that I always get this same answer from somebody when I ask this question.

(The reason I'm asking here is that I have asked in two other forums, and got two completely different answers. So I wanted to double-check also here.)

In a randomly dealt hand (heads-up) which goes to the river, the probability of this happening would be 1/877961175 (assuming I got all the factors right, which is never assured). With a more people at the table, the chances would go up. This is a rather trivial exercise in combinatorics.

Could you show me the individual steps you used to arrive at that number?

(Incidentally, that answer is also completely different than the ones I got from the other forums. Go figure. This seems to be a much harder problem than it might sound.)

 

[Orodruin]

Could you show me the individual steps you used to arrive at that number?

(Incidentally, that answer is also completely different than the ones I got from the other forums. Go figure. This seems to be a much harder problem than it might sound.)

There are a number of things to consider:

• The board has to look as: AAYYX, where YY are among the KQJ10 of the suite that the royal is in.
• One of the aces on the board has to be of the suite that the royal is in.
• The hand of the royal has to contain the remaining YY.
• The other hand has to be the remaining AA.
• X can be any card not already accounted for.

You need to find:

• The number of possible suites for the royal.
• The number of possible boards given that suite (the board completely fixes the cards of each player).

Disclaimer: My calculation was for giving a particular player the royal and the other player the AAAA. If you do not care who gets what, the probability is doubled.

Actually - scratch my number. There is an additional probability coming from being able to obtain AAAA with only one ace on hand. This probability should be easily computable according to the same steps, i.e., board is AAAYY and the AAAA player's hand is AX. To get the total, add the probabilities.

Edit: And in addition, there is the AA vs YX on a AAYYY board. So three scenarios that need to be solved for, but they are all distinct so the probabilities add.

 

[willem2]

There are 52!/(43! * 5! * 2! * 2!) = 2781381002400 ways to chose the two two-card hands and 5 cards on the table.

There are 3 possibilities.
1. Two aces on the table and 4 cards of the flush
hand1: A A
hand2: F x
table: A Af F F F

there are 6 ways to choose which aces go in hand1.
there are 2 ways to choose the colour of the flush
there are 4 ways tho choose which one of the flush cards goes into hand 2.
there are 44 ways to choose the extra card in hand 2.
total 6*2*4*44 = 2112

2. Two aces on the table and 3 cards of the flush
hand1: A A
hand2: F F
table: A Af F F x

there are 6 ways to choose which aces go in hand1.
there are 2 ways to choose the colour of the flush
there are 6 ways tho choose which one of the flush cards goes into hand 2.
there are 44 ways to choose the extra card on the table.
total 6*2*6*44 = 3168

3. Three aces on the table and 3 cards of the flush
hand1: A x
hand2: F F
table: A A Af F F

there are 4 ways to choose which ace goes in hand1.
there are 3 ways to choose the colour of the flush
there are 6 ways tho choose which one of the flush cards goes into hand 2.
there are 44 ways to choose the extra card in hand 1
total 4*3*6*44 = 3168

total: 3168*2+2112 = 8448. Probability 8448/2781381002400 = 3.03734 * 10^(-9).
For two players, multiply this by 2, because it doesn't matter who has the aces.
end result 6,07468 * 10^(-9) or approximately 1 in 165 million.

 

[Orodruin]

total: 3168*2+2112 = 8448. Probability 8448/2781381002400 = 3.03734 * 10^(-9).

I agree with this, although I chose to first look at the board cards, e.g., for the AAXFF board:
4 ways of chosing the suite
3 ways of chosing the extra A
44 ways of chosing the X
6 ways of chosing the flush cards on the board
1 way of assigning the hands (2 if you do not care about who has aces)
Naturally, the result is the same.

because it doesn't matter who has the aces.

The guy with the aces will disagree ;)

 

[Warp]

total: 3168*2+2112 = 8448. Probability 8448/2781381002400 = 3.03734 * 10^(-9).

That's actually the same result as one of those I got in another forum (the person didn't seem to take into account that either player could get the aces), which gives me confidence that it's the correct answer (well, at least if we ignore that the hands could go to either player).

 

[Orodruin]

That's actually the same result as one of those I got in another forum (the person didn't seem to take into account that either player could get the aces), which gives me confidence that it's the correct answer (well, at least if we ignore that the hands could go to either player).

The difference (as noted both in posts #7 and #8) is a simple factor of 2. As long as you know what you are quoting, you can give either probability.

 

[PeroK]

I find it curious that I always get this same answer from somebody when I ask this question.

(The reason I'm asking here is that I have asked in two other forums, and got two completely different answers. So I wanted to double-check also here.)

Could you show me the individual steps you used to arrive at that number?

(Incidentally, that answer is also completely different than the ones I got from the other forums. Go figure. This seems to be a much harder problem than it might sound.)

It's easy to go wrong with these sorts of problems. There's nothing theoretically difficult, but you have to be very careful to do the right calculations. @willem2 is correct.

I did it a slightly different way. First, I looked for a Royal Flush in Hearts and four Aces (then multiplied this answer by four). And, assumed the first player has the flush and the second the four Aces. The three cases are then:

$A_HF_HF_HAX \ \ \ F_HF_H \ \ \ AA$ with $6 \times 3 \times 44 \times 1 \times 1 = 18 \times 44$ possibilities

$A_HF_HF_HAA \ \ \ F_HF_H \ \ \ AX$ with $6 \times 3 \times 1 \times 1 \times 44 = 18 \times 44$ possibilities

$A_HF_HF_HF_HA \ \ \ F_HX \ \ \ AA$ with $4 \times 3 \times 44 \times 1 = 12 \times 44$ possibilities

So, the total number of possibilities is the sum of these: $48 \times 44 = 2112$

Then, multiply this by $4$ to take account of flushes in the other suits: $8448$

Now, you calculate the total number of all hands of the form 5-cards, 2-cards 2-cards, which is:

$N = \binom{52}{5} \binom{47}{2} \binom {45}{2} = 2.78 \times 10^{12}$

The odds are then 1 chance in $N/8448 = 3.29 \times 10^{8}$

And, as above, for two players either could have the Flush and the other the Aces, so we can halve the odds to get 1 chance in $1.65 \times 10^{8}$

Finally, if you have 5 players, say, then there are 20 different possibilities for who has the Flush and who has the Aces, so the odds in a game of 5 players are 1 chance in $1.65 \times 10^{7}$

 

[Orodruin]

Finally, if you have 5 players, say, then there are 20 different possibilities for who has the Flush and who has the Aces, so the odds in a game of 5 players are 1 chance in 1.65×107

... or more generally, in a game of $n$ players, there are
$$
\begin{pmatrix}n \\ 1,1,n-2\end{pmatrix} = \frac{n!}{1! 1! (n-2)!} = n(n-1)
$$
different ways of selecting who has the AAAA and who has the royal, leading to $p_n = n(n-1)p_0$, where $p_0 \simeq 3\cdot 10^{-9}$ is the probability of any given distribution of the hands to the players. For a full table with $n=10$, you would therefore have $p_{10} \simeq 2.7\cdot 10^{-7}$ - about once every 4 million hands. Playing live at a full table with ca 30 hands per hour, it would take you 14 years of continuous playing (no breaks, no sleeping) to experience this at your table (70 years if you want to be a part of the action, 140 if you want to be winning) - and this assumes you will see the hand to the river, either hand may fold at an earlier street.