AceHound
AceHound
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June 26th, 2012 at 11:56:09 AM permalink
In texas hold'em what are the odds of making a royal flush and the player must be holding the ace as one of their hole cards?

To follow up, what are the odds of making a royal flush in a particular suit with the player having the ace as one of their hole cards?

1 in ?????????? hands
Thanks in advance
EdCollins
EdCollins
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June 26th, 2012 at 3:13:06 PM permalink
Approximately 1 in 108,695 hands.

I say "approximately" because the above answer is the result of a computer simulation.

I simulated the event 100,000,000 times, for three different runs.

On the three runs there were 914, 920, and 962 times a royal flush was made WITH the player holding the ace.

920/100,000,000 = 1 in 108,695 hands. (Ball park figure)

I'm sure someone else will give you a more exact figure. If not, maybe this figure is good enough for your purposes.
CrystalMath
CrystalMath
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June 26th, 2012 at 4:11:37 PM permalink
I calculate 1 in 46,060 hands. I don't know if I'm missing something though.

I assumed that you have the Ace and the other card is not part of the royal. In the next five cards, you have to get the remaining 4 cards.

There are combin(50,5) ways to deal the remaining cards.

There are 46 ways to deal the royal because the last card can be any of the 46 remaining cards that aren't part of the royal.

So the frequency of a royal, in this scenario, is combin(50,5)/46.

Also, I don't know the strategy for this game, so this assumes that you will always play through the river.
I heart Crystal Math.
EdCollins
EdCollins
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June 26th, 2012 at 5:20:12 PM permalink
When I set up my simulation, I didn't assume the player already had the Ace.

Also, my simulation assumed the other player's card might be part of the royal. (The question didn't say that it couldn't)

I'm sure that accounts for our big difference in our answers.
CrystalMath
CrystalMath
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June 26th, 2012 at 7:36:22 PM permalink
Quote: EdCollins


When I set up my simulation, I didn't assume the player already had the Ace.

Also, my simulation assumed the other player's card might be part of the royal. (The question didn't say that it couldn't)

I'm sure that accounts for our big difference in our answers.



Yes, that's it.
I calculated the probability of getting one ace and a non-royal, two aces, and an ace with another card to the royal. The odds are 1 in 106033.9583.

If the player is dealt an Ace and a non-royal card, the conditional odds are 1 in 46060.
If the player is dealt two Aces, the conditional odds are 1 in 23030.
If the player is dealt an Ace and another card to the royal, the conditional odds are 1 in 1960.
I heart Crystal Math.
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