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  #1  
Old 12-12-2005, 04:14 AM
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Default Odds of winning vs. a random hand

I'm trying to determine the probabilty of winning against a random. Given the following scenario how would I go about caculating this. Any help would be much appreciated.

Hero Ks Js
Villan ??

Board 4h 9c 6c 3h
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  #2  
Old 12-12-2005, 04:29 AM
ThinkQuick ThinkQuick is offline
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Default Re: Odds of winning vs. a random hand

Get Pokerstove.
<font class="small">Code:</font><hr /><pre>
Text results appended to pokerstove.txt

45,540 games 0.016 secs 2,846,250 games/sec

Board: 4h 9c 6c 3h
Dead:

equity (%) win (%) / tie (%)

Hand 1: 32.3715 % [ 00.32 00.00 ] { KsJs }
Hand 2: 67.6285 % [ 00.67 00.00 ] { random }


---</pre><hr />
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Old 12-12-2005, 04:35 AM
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Default Re: Odds of winning vs. a random hand

Thanks but I am a little more concerned about how I would go about doing the actual math behind it rather then just the pure percent result from pokerstove.
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Old 12-12-2005, 01:25 PM
mostsmooth mostsmooth is offline
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Default Re: Odds of winning vs. a random hand

can a hand really be classified as random if it gets to the turn?
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  #5  
Old 12-12-2005, 02:30 PM
LetYouDown LetYouDown is offline
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Default Re: Odds of winning vs. a random hand

[ QUOTE ]
Thanks but I am a little more concerned about how I would go about doing the actual math behind it rather then just the pure percent result from pokerstove.

[/ QUOTE ]

Well, there are 1035 hands the other player could have. Just run through each one and see how many rivers result in a win for K-J vs. the random hand and do a sum of both, then divide. Or get PokerStove.
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Old 12-12-2005, 02:45 PM
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Default Re: Odds of winning vs. a random hand

Thanks. Thats the way that I was kinda thinking about doing the calculations was just wondering if there might be a simpler way of doing it without having muscle through each hand.
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  #7  
Old 12-12-2005, 10:37 PM
ThinkQuick ThinkQuick is offline
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Default Re: Odds of winning vs. a random hand

[ QUOTE ]
Thanks. Thats the way that I was kinda thinking about doing the calculations was just wondering if there might be a simpler way of doing it without having muscle through each hand.

[/ QUOTE ]

wow.
um, why again?


there are many subcategories like unpaired suited, unpaired gutshot, etc.; but you'd need to be careful not to double - count.
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Old 12-12-2005, 10:54 PM
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Default Re: Odds of winning vs. a random hand

[ QUOTE ]

wow.
um, why again?


[/ QUOTE ]

I need this for a project for a probability class I'm taking and the only way that comes to mind on how to solve this problem is to determine every possible hand combination and figure out the probabiliy of winning vs. those hands. I was just trying to see if there was another way that I could go about figuring this out other than the way that I was thinking. Unfortunatly is seems like im just gonna have to sit down and grind out some numbers.
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  #9  
Old 12-12-2005, 11:48 PM
AaronBrown AaronBrown is offline
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Default Re: Odds of winning vs. a random hand

You can make it a little easier on yourself. You have seen six cards, so there are 46 unknown ones. They can be arranged 46*45*44/2 = 45,550 ways.

But for most purposes, only the ranks matter, not the suits. There are only 1,183 different rank combinations. Only if the KJ hand wins do you have to ask if all three of the new cards (the other player's hand plus the river card) are clubs or hearts. It's easy to do that as a side calculation.

Within the ranks, you can simplify further. There are two main possibilities for KJ, it pairs one of those cards or it doesn't. There are four possibilities for the unknown hand: no pair, one pair, two pair and three of a kind. Only in the no pair/no pair and one pair/one pair cases do you have to analyze further to see who wins.

Finally, you have to figure out the various straight possibilities for the other player's hand. There aren't many of these.

You could do this with pencil and paper in half an hour or so, by enumerating cases as above. Or you could spend about the same amount of time writing a computer program to do it in a few milliseconds.
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