Caribbean Stud Poker : probability for the player winning with a royal flush
May 11th, 2011 at 12:50:41 PM
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In the Caribbean Stud Poker table, for the player winning with a royal flush, I can see there are 16759740 combinations. As the total number of combinations is 19933230517200, the probability is 8.40794E-07.
I should have calculated this probability this way :
The odds of a royal flush are 4 in 2,598,960. The probability is 1.53908E-06
If the player gets a royal flush then the odds of the dealer getting a royal flush is reduced to 3 in 1,533,939. The probability is 1.95575E-06
The probability of this happening in the same time (for a push) is 3.01E-12 (The product of the two prior values).
In conclusion, I should have said that the probability for player winning with royal flush is : 1.53908E-06 - 3.01E-12 = 1.53907E-06 far from the initial 8.40794E-07 value.
Please, could you tell me where I am making the mistake ?
Thanks
I should have calculated this probability this way :
The odds of a royal flush are 4 in 2,598,960. The probability is 1.53908E-06
If the player gets a royal flush then the odds of the dealer getting a royal flush is reduced to 3 in 1,533,939. The probability is 1.95575E-06
The probability of this happening in the same time (for a push) is 3.01E-12 (The product of the two prior values).
In conclusion, I should have said that the probability for player winning with royal flush is : 1.53908E-06 - 3.01E-12 = 1.53907E-06 far from the initial 8.40794E-07 value.
Please, could you tell me where I am making the mistake ?
Thanks
May 11th, 2011 at 1:10:58 PM
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The probability of a player getting and winning with a royal flush are (odds of player royal)-(odds of coincidental dealer royal). The odds of the player winning the hand once the player's royal is already known are 1-(odds of coincidental dealer royal). I'm not sure what your subtraction is all about at the end there...
"So as the clock ticked and the day passed, opportunity met preparation, and luck happened." - Maurice Clarett
May 11th, 2011 at 1:35:01 PM
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Thanks for the reply.
That's exactly what I meant:
The odds of player royal are 4 in 2,598,960, which is 1 in 649,740.
The odds of dealer royal is reduced to 3 in 1,533,939, or 1 in 511,313.
The odds of coincidental dealer royal is 1 in 332,220,508,620 (The product of the two prior values).
The probability of the push hand is 1/332,220,508,620 =3.01E-12 (very weak)
So the probability for player winning with royal flush should be near 4/ 2,598,960 because the coincidence is weak, isn't it ?
Quote: rdw4potusThe probability of a player getting and winning with a royal flush are (odds of player royal)-(odds of coincidental dealer royal).
That's exactly what I meant:
The odds of player royal are 4 in 2,598,960, which is 1 in 649,740.
The odds of dealer royal is reduced to 3 in 1,533,939, or 1 in 511,313.
The odds of coincidental dealer royal is 1 in 332,220,508,620 (The product of the two prior values).
The probability of the push hand is 1/332,220,508,620 =3.01E-12 (very weak)
So the probability for player winning with royal flush should be near 4/ 2,598,960 because the coincidence is weak, isn't it ?
