paisiello
paisiello
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August 1st, 2013 at 8:37:26 PM permalink
The Cincinnatti Kid final hand
If you read the wikipedia entry on this movie:
wiki
It talks about the probability of getting a straight flush vs. a full house. The number given is truly a lottery-winning probability event:
45,102,781 to 1
However, if you try and figure out how this number is derived then their number seems to come up a little high.
paisiello
paisiello
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August 2nd, 2013 at 7:18:22 PM permalink
Here is my calculation to show that the wiki page is wrong:

Five card draw
Total number of possible hands 2,598,960

Number of possible full house hands 3744
Number of possible straight flush hands 40

Probability of these hands in two separate deals = (3744 / 2598960) (40 / 2598960)
= 0.000002217% or 1 in 45,102,785

Since this number is the same as the wiki number then clearly the referenced source has underestimated the probability of the occurence of this event. The actual probability should be much lower since some cards have been removed from the deck.
Ibeatyouraces
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August 2nd, 2013 at 7:24:04 PM permalink
deleted
DUHHIIIIIIIII HEARD THAT!
tringlomane
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August 2nd, 2013 at 9:20:21 PM permalink
Quote: Ibeatyouraces

Four of those straight flushes in your calculations are royals. Would that make a difference?



No. It's about being able to arrange 10 cards into one straight flush (including a Royal) and the other into a full house. The lack of pairs in the straight flush will make it a little more difficult for the other person to make a full house.

Lets arrange the straight flush first because that's easiest:

4 suits x 10 high ranks (5 high to Ace high) = 40

Total combinations: 2,598,960

Full houses that can be made from the remaining cards:

Full House from 2 ranks of 3 cards each:
5*C(3,3)*4*C(3,2) = 5*1*4*3 = 60

Full House from 2 ranks of 4 cards each:
8*C(4,3)*7*C(4,2) = 8*4*7*6 = 1344

Full House from a trips rank of 3 cards and a pair rank of 4 cards:
5*C(3,3)*8*C(4,2) = 5*1*8*6 = 240

Full House from a pair rank of 3 cards and a trips rank of 4 cards:
5*C(3,2)*8*C(4,3) = 5*3*8*4 = 480

Total ways: 60 + 1344 + 240 + 480 = 2124

Total remaining combinations for the 47-card stub: C(47,5) = 47!/42!/5! = 1533939

So the probability of both being dealt between them is:

(40/2,598,960)*(2124/1,533,939) = 1 in ~46,923,800.65

This also assumes these hands go to these specific players (it's more likely the hand would show up if it wasn't heads up, obviously) and assumes the second hand is precisely a full house, not a full house or better.
paisiello
paisiello
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August 3rd, 2013 at 10:19:06 AM permalink
We are looking for the chances that this situation arises.

You calculated the probability that the first hand dealt was a straight flush and the second hand was a full house. Shouldn't you add to this the probability that the reverse order event occurs also? i.e. the first hand dealt is a full house and the second hand is a straight flush?

So I think your number should be doubled:

2*(40/2,598,960)*(2124/1,533,939) = 1 in ~23,461,900

Which would agree with my intuition that the number should be much less than the number given in the wiki article.
paisiello
paisiello
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August 8th, 2013 at 6:57:04 PM permalink
I am now trying to determine where the wiki page's second number comes from: 332,220,508,619 to 1. It states the number is for when both hands occur in the same deal. However, I think it was shown conclusively above that the actual number is much smaller than this. So then, where does this 332,220,508,619 number come from?

One guess is that it is the probability of getting a full house with the pair only contained in the straight flush. Using the numbers from above, this can be calculated as:
(40/2,598,960)*(480/1,533,939) * 2 = 1 in ~103,818,909

This is a bigger number but still far off from 332,220,508,619.

So now try and get an upper bound on the total number of all possible hands. I think this can be estimated simply as:
(1/2,598,960)*(1/1,533,939) = 1 in 3,986,646,103,440

Well, this now is a larger number than the the wiki number but only 13 times or so. Something tells me that as rare as the situation is it is not so rare as to be only 1/13 times smaller than all possible hands.

So now I think the referenced source in the wiki article is suspect.
miplet
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August 8th, 2013 at 8:12:35 PM permalink
Quote: paisiello

I am now trying to determine where the wiki page's second number comes from: 332,220,508,619 to 1. It states the number is for when both hands occur in the same deal. However, I think it was shown conclusively above that the actual number is much smaller than this. So then, where does this 332,220,508,619 number come from?

One guess is that it is the probability of getting a full house with the pair only contained in the straight flush. Using the numbers from above, this can be calculated as:
(40/2,598,960)*(480/1,533,939) * 2 = 1 in ~103,818,909

This is a bigger number but still far off from 332,220,508,619.

So now try and get an upper bound on the total number of all possible hands. I think this can be estimated simply as:
(1/2,598,960)*(1/1,533,939) = 1 in 3,986,646,103,440

Well, this now is a larger number than the the wiki number but only 13 times or so. Something tells me that as rare as the situation is it is not so rare as to be only 1/13 times smaller than all possible hands.

So now I think the referenced source in the wiki article is suspect.


Yes, I agree the source needs to use Excel or something. Even getting exactly a Q high straight flush vs Aces full of 10's is 1 in 83,055,127,155; half of that if you don't care who loses.
“Man Babes” #AxelFabulous
tringlomane
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August 8th, 2013 at 8:23:07 PM permalink
I'm sure it is. BruceZ (poker probability God) even had to correct Wiki on probability of dominated pairs.
paisiello
paisiello
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August 9th, 2013 at 1:24:18 PM permalink
Should someone correct the wiki article then?
tringlomane
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August 9th, 2013 at 1:29:30 PM permalink
Quote: paisiello

Should someone correct the wiki article then?



Probably, but I have never messed with any articles before.
mickeycrimm
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August 9th, 2013 at 4:12:03 PM permalink
Lancey's pulling $5000 out of his pocket on the end and raising is a no no in a table stakes games. Lancey bragging about being the best, after the hand played out, is B.S. I'm sure the Cincinnati Kid had figured out at that point that Lancey was a card chaser. He just didn't have the money to keep playing. But I'm sure the Kid would be prepared the next time he bankrolled up and ran into Lancey. The best quote in the movie came early on when someone said the Kid might be cheating because he ran over the game. "I don't have to cheat to win" said the Kid.
"Quit trying your luck and start trying your skill." Mickey Crimm
paisiello
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August 9th, 2013 at 9:41:55 PM permalink
OK, I updated the page:
wiki
I made a direct reference to this thread.
paisiello
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August 9th, 2013 at 9:59:05 PM permalink
Quote: mickeycrimm

Lancey's pulling $5000 out of his pocket on the end and raising is a no no in a table stakes games.

The game was open stakes which apparently was more common in the past such as the 1930's when this movie takes place.

Quote: mickeycrimm

I'm sure the Cincinnati Kid had figured out at that point that Lancey was a card chaser.

Not sure because of one hand that you could say he was a card chaser. I think Lancey had become desperate and made a bad call.

Quote: mickeycrimm

He just didn't have the money to keep playing.

Not sure that was relevant since the Kid could have easily been staked if he wanted to continue.
paisiello
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August 10th, 2013 at 12:33:20 PM permalink
Or maybe not. I think he actually had pot odds to make the call.

No antes or blinds in a heads up game of no limit 5 card stud.
Lancey "The Man" Howard is dealt the 8♦ and the Cincinnati Kid the 10 of clubs. The Kid bets $500 and Howard calls. Was this a good call? Probably not a bad one since his holding J beats a 10.

Pot size = $1,000.
Howard gets the Q♦ and The Kid the 10 of spades. The Kid bets $1,000 and Howard raises $1,000. The Kid calls. Was this move by Lancey the "wrong move at the right time"? Maybe he thought representing a pair of Q's would get The Kid to fold or at least set him up for a big bluff later? Once he makes this raise he will become pot committed with the next hand.

Pot size = $5000.
Howard is dealt the 10♦ and The Kid gets the A of clubs. The Kid bets $3,000 . At this point the pot odds would have been $3,000 to win back $11,000 total or 27.3%. The Man has 9 outs for a flush draw plus another 3 outs for a gut shot straight draw = 12 outs total. 12/44 = 27.3% chance to win. Since this equals the pot odds then it makes sense to call. Plus the implied odds give even more of an incentive to call. The Man calls.

Pot size = $11,000.
The Man's final card is the 9♦; The Kid gets the A of spades. The Kid checks. The Man bets $1,000. The Kid raises $3,500 and is all in. Howard reaches into his wallet and raises another $5,000. The Man agrees to take his marker and the Kid calls the bet.

Pot size =$30,000.
Howard turns over the J♦ for a queen-high straight flush. The Kid turns over the A of hearts to show his full house. Maybe not a bad beat after all.
paisiello
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August 13th, 2013 at 3:05:20 PM permalink
Anybody have access to this site or this specific article?
card player

I would sign up myself but not too keen on giving my credit card info.
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