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#1
12-12-2004, 09:31 AM
 AA suited Junior Member Join Date: Apr 2004 Posts: 14
Heads up: % of the time high card wins?

Final 2 in SnGs.

What % of the time does the person w/the high card win played to showdown? (ie: board helps neither person)

thx
#2
12-12-2004, 01:00 PM
 Paul2432 Senior Member Join Date: Jun 2003 Location: Bryn Mawr, PA USA Posts: 374
Re: Heads up: % of the time high card wins?

This question is impossible to answer. For each matchup the answer will be different (between 50-75%)

Paul
#3
12-12-2004, 05:38 PM
 gaming_mouse Senior Member Join Date: Oct 2004 Location: my hero is sfer Posts: 2,480
Re: Heads up: % of the time high card wins?

Well, not impossible to answer, but a pointless and uninformative exercise for sure. I mean you could write a program to enumerate over all possible matchups, which would count the number of times the person with the high card wins.

But this is surely not what the OP really wants to know. I'd guess there are certain situations (hand types) that he's interested in. But who knows.

AA Suited, could you clarify?

gm
#4
12-13-2004, 01:10 AM
 AA suited Junior Member Join Date: Apr 2004 Posts: 14
Re: Heads up: % of the time high card wins?

when heads-up, what % of the time does the board (flop/turn/river) not help either player?
#5
12-13-2004, 01:18 AM
 joeboss Junior Member Join Date: Jan 2004 Posts: 8
Re: Heads up: % of the time high card wins?

ill take a stab at it...
the chances of pairing are 6/50 + 6/49... ect...
thats about 62.55% of course this is not including flushes ect...
#6
12-13-2004, 02:15 AM
 gaming_mouse Senior Member Join Date: Oct 2004 Location: my hero is sfer Posts: 2,480
Re: Heads up: % of the time high card wins?

Without some more assumptions about the 2 players hands it becomes a pain to calculate, and not too useful either. So let's assume that neither player has a pocket pair, and that players share no cards. That is, between the two players there are a total of 4 different card ranks. In that case, the chance that neither player pairs by the river is:

(36 choose 5) / (48 choose 5) = 0.220166512

gm
#7
12-13-2004, 08:26 AM
 yoshi_yoshi Member Join Date: Dec 2004 Location: Cambridge, MA Posts: 54
Re: Heads up: % of the time high card wins?

As a rough estimate, I ran this through the twodimes poker analyzer...

Hands:
As 7c
Jh 3s
2c 2s

Dead Cards: 2d 2h

Results:
cards EV
As 7c 0.489
3s Jh 0.369
2s 2c 0.141

This has the effect of a twelve-card deck (since all the 2s are gone) and I've chosen two hands that minimize straight/flush possibilities.

So based on a twelve-card deck, the odds of one of the two players hitting something is approx. 85%. For a regular 13-card deck, you could estimate it at 75-85% I think.

What about hole cards that work together more? I also ran...

cards EV
6s 5s 0.347
Jc Tc 0.567
2s 2c 0.086 (dead cards: 2d 2h)

So with two suited connectors, one of them will hit something in a 12-rank deck a good 93% of the time, meaning 85-90% in a regular deck.
#8
12-13-2004, 06:23 PM
 AA suited Junior Member Join Date: Apr 2004 Posts: 14
Re: Heads up: % of the time high card wins?

[ QUOTE ]
Without some more assumptions about the 2 players hands it becomes a pain to calculate, and not too useful either. So let's assume that neither player has a pocket pair, and that players share no cards. That is, between the two players there are a total of 4 different card ranks. In that case, the chance that neither player pairs by the river is:

(36 choose 5) / (48 choose 5) = 0.220166512

gm

[/ QUOTE ]

exactly what i was looking for.

22% seems like a low # though? That means 39% of the time one person will pair up. Thus the person with the high card has a 22+39 = 61% chance of winning?
#9
12-14-2004, 01:52 AM
 gaming_mouse Senior Member Join Date: Oct 2004 Location: my hero is sfer Posts: 2,480
Re: Heads up: % of the time high card wins?

[ QUOTE ]
22% seems like a low # though? That means 39% of the time one person will pair up. Thus the person with the high card has a 22+39 = 61% chance of winning?

[/ QUOTE ]

No. You can say that 78% of the time at least one of them will pair up. It does not follow that 39% of the time exactly one will pair up. You have to do a different calculation to solve that.

But it seems the question you are interested in is: How often does the player with the high card win, when that player has not improved to a pair?

So imagine yourself on the river with your A-high hand, say. You now want to know the chance your oppo has paired.

There are 15 unseen cards which your oppo could hold to make a pair with the board, out of a total of 45 unseen cards. We also assumed that your oppo's cards are different ranks than yours, which means that there are really only 39 unseen cards from which his hand can be composed.

Thus your oppo could hold any one of (39 choose 2) = 741 possible hands. However, we also assumed that your oppo does not hold a pocket pair. There are 5*3 + 6*4 = 39 pocket pairs possible. So there are really 702 possible hands for your oppo.

How many don't contain a card that pairs the board? (24 choose 2) = 276, not counting possible pocket pairs. There are 24 possible pocket pairs, giving our oppo 252 possible hands that don't pair that board and that agree with our assumptions. Thus the chance that your high card wins is:

252 / 702 = 0.358, about 36%

Note that this does not take into account the information you glean by your oppo's betting patterns. For example, if he has checked raised the turn, you should probably give him credit for a pair.

gm

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