# texas holdem probability question

**URL:** <https://boards.straightdope.com/t/texas-holdem-probability-question/611723>\
**Category:** The Game Room\
**Created:** [February 4, 2012, 3:07am UTC](https://boards.straightdope.com/t/texas-holdem-probability-question/611723 "2012-02-04T03:07:26Z")\
**Posts on this page:** 11\
**Page:** 1

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**Author:** ![copperwindow](https://avatars.discourse-cdn.com/v4/letter/c/4af34b/32.png) [@copperwindow](https://boards.straightdope.com/u/copperwindow)\
**Post date:** [February 4, 2012, 3:07am UTC](https://boards.straightdope.com/t/texas-holdem-probability-question/611723/1 "2012-02-04T03:07:26Z")

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I don’t under stand why this question is so hard. Me and my opponent go all in preflop. I have the ace of hearts and the king of spades. My opponent has the ace of spades and the king of hearts. What is the probability that we **won’t** split the pot?

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**Author:** ![notfrommensa](https://avatars.discourse-cdn.com/v4/letter/n/f14d63/32.png) [@notfrommensa](https://boards.straightdope.com/u/notfrommensa)\
**Post date:** [February 4, 2012, 3:32am UTC](https://boards.straightdope.com/t/texas-holdem-probability-question/611723/2 "2012-02-04T03:32:15Z")

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According to this Texas Holdem Calculator, and assuming 48 live cards, Each player has 0.74% or winning the hand and a 98.52% chance of splitting the pot.

reported for a forum change to game room.

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**Author:** ![ZenBeam](https://avatars.discourse-cdn.com/v4/letter/z/3ab097/32.png) [@ZenBeam](https://boards.straightdope.com/u/ZenBeam)\
**Post date:** [February 4, 2012, 3:37am UTC](https://boards.straightdope.com/t/texas-holdem-probability-question/611723/3 "2012-02-04T03:37:00Z")

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Wouldn’t the only way to not split the pot be if four or five of the flop, turn, and river cards are spades, or four or five are hearts?

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**Author:** ![copperwindow](https://avatars.discourse-cdn.com/v4/letter/c/4af34b/32.png) [@copperwindow](https://boards.straightdope.com/u/copperwindow)\
**Post date:** [February 4, 2012, 3:52am UTC](https://boards.straightdope.com/t/texas-holdem-probability-question/611723/4 "2012-02-04T03:52:49Z")

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> [@notfrommensa](#):
>
> According to this Texas Holdem Calculator, and assuming 48 live cards, Each player has 0.74% or winning the hand and a 98.52% chance of splitting the pot.
> 
> reported for a forum change to game room.

I don’t understand. 98.52% + 0.74% does not equal 100%. What are the other outcomes?

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**Author:** ![keeper0](https://avatars.discourse-cdn.com/v4/letter/k/b19c9b/32.png) [@keeper0](https://boards.straightdope.com/u/keeper0)\
**Post date:** [February 4, 2012, 3:56am UTC](https://boards.straightdope.com/t/texas-holdem-probability-question/611723/5 "2012-02-04T03:56:19Z")

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> [@copperwindow](#):
>
> I don’t understand. 98.52% + 0.74% does not equal 100%. What are the other outcomes?

But 98.52% + 0.74% + 0.74% = 100%

EACH player has a 0.74% chance of winning.

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**Author:** ![The\_Controvert](https://avatars.discourse-cdn.com/v4/letter/t/dec6dc/32.png) [@The\_Controvert](https://boards.straightdope.com/u/The_Controvert)\
**Post date:** [February 4, 2012, 3:56am UTC](https://boards.straightdope.com/t/texas-holdem-probability-question/611723/6 "2012-02-04T03:56:38Z")

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98.52% split pot  
.74% you win  
.74% opponent wins

This totals 100%.

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**Author:** ![x-ray\_vision](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/x-ray_vision/32/351_2.png) [@x-ray\_vision](https://boards.straightdope.com/u/x-ray_vision)\
**Post date:** [February 4, 2012, 4:39am UTC](https://boards.straightdope.com/t/texas-holdem-probability-question/611723/7 "2012-02-04T04:39:09Z")

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In other words, the probability is 0.0148 (1.48%).

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**Author:** ![Colibri](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/colibri/32/1841_2.png) [@Colibri](https://boards.straightdope.com/u/Colibri)\
**Post date:** [February 4, 2012, 4:53am UTC](https://boards.straightdope.com/t/texas-holdem-probability-question/611723/8 "2012-02-04T04:53:45Z")

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Moving to the Game Room.

Colibri  
General Questions Moderator

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**Author:** ![copperwindow](https://avatars.discourse-cdn.com/v4/letter/c/4af34b/32.png) [@copperwindow](https://boards.straightdope.com/u/copperwindow)\
**Post date:** [February 4, 2012, 5:29am UTC](https://boards.straightdope.com/t/texas-holdem-probability-question/611723/9 "2012-02-04T05:29:48Z")

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> [@Colibri](#):
>
> Moving to the Game Room.
> 
> Colibri  
> General Questions Moderator

No need. The question has been answer. May as well just close it.

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**Author:** ![Gray\_Ghost](https://avatars.discourse-cdn.com/v4/letter/g/c0e974/32.png) [@Gray\_Ghost](https://boards.straightdope.com/u/Gray_Ghost)\
**Post date:** [February 4, 2012, 5:37am UTC](https://boards.straightdope.com/t/texas-holdem-probability-question/611723/10 "2012-02-04T05:37:37Z")

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Not sure which probability calculator Mensa used—there are several—but I just wanted to chime in with one of my favorite pieces of freeware: [Pokerstove](http://www.pokerstove.com/).

There are more powerful equity calculators out there, but I think you have to pay for them. With Stove, it’s easy to figure out ranges and odds of winning/splitting for a variety of ranges; both your own and as many as 9 different opponents. It can’t do Bayesian weights for elements within the range, but that’s really a small quibble.

Now if I could just find a version of it for Omaha…

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**Author:** ![notfrommensa](https://avatars.discourse-cdn.com/v4/letter/n/f14d63/32.png) [@notfrommensa](https://boards.straightdope.com/u/notfrommensa)\
**Post date:** [February 4, 2012, 3:52pm UTC](https://boards.straightdope.com/t/texas-holdem-probability-question/611723/11 "2012-02-04T15:52:36Z")

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I used the one at [cardplayer.com](http://www.cardplayer.com/poker-tools/odds-calculator/texas-holdem). I thought I inserted a hyperlink, but apparently not.

> [@](#):
>
> Wouldn’t the only way to not split the pot be if four or five of the flop, turn, and river cards are spades, or four or five are hearts?

Assuming no other player is in the pot, yes.
