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  #1  
Old 03-11-2005, 04:36 AM
Elvis Elvis is offline
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Location: Stockholm, Sweden
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Default Probability question

In an all-in situationen against AA, I always thought that the best possible hand to hold was either 87s, 76s, or 65s. I also thought that these three hands had exactly the same chance of beating AA. However, look at these results from twodimes.net:

http://twodimes.net/h/?z=3426
pokenum -h ah ad - 8s 7s
Holdem Hi: 1712304 enumerated boards
cards win %win lose %lose tie %tie EV
Ad Ah 1315602 76.83 391672 22.87 5030 0.29 0.770
8s 7s 391672 22.87 1315602 76.83 5030 0.29 0.230


http://twodimes.net/h/?z=52865
pokenum -h ah ad - 7s 6s
Holdem Hi: 1712304 enumerated boards
cards win %win lose %lose tie %tie EV
Ad Ah 1315168 76.81 391637 22.87 5499 0.32 0.770
7s 6s 391637 22.87 1315168 76.81 5499 0.32 0.230


http://twodimes.net/h/?z=17228
pokenum -h ah ad - 6s 5s
Holdem Hi: 1712304 enumerated boards
cards win %win lose %lose tie %tie EV
Ad Ah 1314307 76.76 391582 22.87 6415 0.37 0.769
6s 5s 391582 22.87 1314307 76.76 6415 0.37 0.231

Could someone please explain to me why 87s wins more often than 65s, and why 65s ties more often than 87s?
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  #2  
Old 03-11-2005, 06:29 AM
3N1GM4 3N1GM4 is offline
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Default Re: Probability question

erm, dunno whether I've misread your post (the columns on your figures don't line up), but it seems that they all beat AA the same amount: 22.87% of the time.
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  #3  
Old 03-11-2005, 06:36 AM
jdl22 jdl22 is offline
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Location: Pittsburgh, PA
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Default Re: Probability question

Without too much thought here is what I've come up with:

AA vs suited connectors tie with:
- quads and ace on board (equally likely in all cases)
- flush on board not helped by either (slightly more likely to occur with 65s than 87s because there are more higher spades available)
- straight on board not helped by either

in this last case they are different. Let's look at all possible straights that cause a chopped pot:

QJT98
JT987
T9876
98765
87654

Notice that the first pairs one card from the 87 and none from the other two. That makes this straight more likely to occur if you have 76 or 65 than if you have 87. Similarly the J high straight pairs both cards from the 87 hand, one from the 76 hand and none from the 65 hand. This makes it more likely to occur if you're holding 65, followed by 76 and 87. So if you look at the total number of paired cards from these straights you get:
87 - 9
76 - 7
65 - 3

That makes 65 most likely to chop, followed by 76 and 87.

I think they all lose equally often, if not I don't know why.
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  #4  
Old 03-11-2005, 07:10 AM
Elvis Elvis is offline
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Default Re: Probability question

[ QUOTE ]
I think they all lose equally often, if not I don't know why.

[/ QUOTE ]

I agree completely. Could it be that twodimes.net is slightly off, or could someone else enlighten us?
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  #5  
Old 03-11-2005, 10:48 AM
Kurn, son of Mogh Kurn, son of Mogh is offline
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Location: Cranston, RI
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Default Re: Probability question

Those results are exactly equal. The variation you see is statistically insignificant.

Twodimes ran 1712304 boards. There are actually over 205,000,000 possible boards. Thus, there will always be some slight variance in the numbers.
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  #6  
Old 03-11-2005, 02:27 PM
Spekkio Spekkio is offline
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Default Re: Probability question

Could someone please explain to me why 87s wins more often than 65s, and why 65s ties more often than 87s?

A. Because if you hold 65s, there is a higher chance that you'll play the board. With 87s, there is a higher chance that at least the 8 will be part of the 5 cards you show down with.
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  #7  
Old 03-11-2005, 03:17 PM
pzhon pzhon is offline
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Default Re: Probability question

Two Dimes appears to be exactly correct. It tabulated all 48 choose 4 possible boards.

Most of the differences come from straights, but 87s also wins more frequently when both players make a full house or both players make a flush/SF.

When the board has a full house, the suited connector can still win with an overboat.

With 65s, this can happen when AA makes Xs full of aces, while 65 makes sixes full of Xs or fives full of Xs, and there are 3 choices for X: 2, 3, and 4. That totals 3*4*2*3=72 ways 65s can win with an overboat.

With 87s, there are 5 choices for X: 2, 3, 4, 5, and 6. That gives 87s 120 ways to win with an overboat.

Suppose the board is all spades. How many ways can the suited connector's flush/SF be counterfeited? With 65s, there are 8 cards above a 6, so there are more times both players play the board's flush. There are 4 SF ties, and another 43 flush ties if I counted correctly. With 87s, if the board comes all spades, there are only 3 SF ties (A-5, 9-K, T-A), and 2 flush ties.
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  #8  
Old 03-11-2005, 05:46 PM
Kurn, son of Mogh Kurn, son of Mogh is offline
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Default Never Mind

I realized my error when reading the later posts. Duh. [img]/images/graemlins/blush.gif[/img]
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