The Cincinnatti Kid

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August 1st, 2013 at 8:37:26 PM permalink
paisiello
Member since: Oct 30, 2011
Threads: 20
Posts: 546
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.
August 2nd, 2013 at 7:18:22 PM permalink
paisiello
Member since: Oct 30, 2011
Threads: 20
Posts: 546
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.
August 2nd, 2013 at 7:24:04 PM permalink
Ibeatyouraces
Member since: Jan 12, 2010
Threads: 43
Posts: 5710
Four of those straight flushes in your calculations are royals. Would that make a difference?
"Shut up Meg." Peter Griffin, Family Guy
August 2nd, 2013 at 9:20:21 PM permalink
tringlomane
Member since: Aug 25, 2012
Threads: 2
Posts: 3737
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.
August 3rd, 2013 at 10:19:06 AM permalink
paisiello
Member since: Oct 30, 2011
Threads: 20
Posts: 546
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.
August 8th, 2013 at 6:57:04 PM permalink
paisiello
Member since: Oct 30, 2011
Threads: 20
Posts: 546
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.
August 8th, 2013 at 8:12:35 PM permalink
miplet
Member since: Dec 1, 2009
Threads: 5
Posts: 1117
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.
August 8th, 2013 at 8:23:07 PM permalink
tringlomane
Member since: Aug 25, 2012
Threads: 2
Posts: 3737
I'm sure it is. BruceZ (poker probability God) even had to correct Wiki on probability of dominated pairs.
August 9th, 2013 at 1:24:18 PM permalink
paisiello
Member since: Oct 30, 2011
Threads: 20
Posts: 546
Should someone correct the wiki article then?
August 9th, 2013 at 1:29:30 PM permalink
tringlomane
Member since: Aug 25, 2012
Threads: 2
Posts: 3737
Quote: paisiello
Should someone correct the wiki article then?


Probably, but I have never messed with any articles before.
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