A High HU in hold'em

[DrPhysic]

A request for help for our mathematicians, please. A bit of info that would be very helpful, that I cannot find, and absolutely do not know how to calculate.

In Holdem, playing heads up, if a statistical sample of hands are dealt to the river, what percentage of those hands would be won by an Ace High or lower?

Help: Tell me the answer. Useful info playing HU.

Better help: Tell me where to find the answer. A useful reference, no doubt.

Best help: Show me how to calculate it.


Thx in advance,

Doc

[Bozeman]

I get about 5.5% for two random hands (+~.5% ties). However, this may not be the answer you are looking for, since when a pair or better is on the board, high card will not be the winning hand, even if neither player receives help.

(I used Poker Calculator)

Craig

[DrPhysic]

Craig,
That's exactly the answer I wanted, If the board pairs, or shows a bigger hand, it's not an A high hand.

I thought that an A high won the hand a lot more often in HU than one hand in 18. I am talking about a true HU situation, not a face card enriched deck on a 9 player table after 7 others have folded.

Where is Poker Calculator?

I know the pokenum holdem hand calculator at twodimes.net, but as far as i know that will only calculate probablities for specific hands, not more general situations like this.

Thx

Doc

[Bozeman]

Poker Calculator is a nice tool ( http://koti.mbnet.fi/jraevaar/pokercalculator/ ), it can calculate for random hands, for ranges of hands, and for multiple games. It also tells what kind of hands win. Only caveat: it has at least on bug (incorrect results for omaha). Still, much easier to use than pokerstove.

For your problem, remember: one player will have a pp 11.4% of the time, only ~47% of the time will the board have no pairs, trips, or quads, and better than ~75% of the remaining (~42%) times one player at least will improve. So the answer will not be very often.

Now this is showdown poker, in actual poker there will be significantly more folding before the river.

Craig

[DrPhysic]

Craig,

Thx. D/Ling poker calculator now.

Doc

[BruceZ]

For your problem, remember: one player will have a pp 11.4% of the time, only ~47% of the time will the board have no pairs, trips, or quads, and better than ~75% of the remaining (~42%) times one player at least will improve.

The board will be unpaired just over 50% of the time, not 47%. This is 52*48*44*40*36/(52*51*50*49*48) = 50.7%. I believe you may have computed 48*44*40*36*32/(48*47*46*45*44)= 47% in an attempt to compute the probability of the board being unpaired given that the players hold no pair, but this would not be correct, and 25% of 42% or 10.5% is not the correct final answer. We can easily compute the exact probability of neither player making a pair and the board being unpaired as follows.

The probability of one player having no pocket pair is 16/17. The probability of the other player holding no pocket pair when the first player holds no pocket pair is 1 - 72/1225 since there are 72 pairs left out of 50*49/2 = 1225 possible hands. So the probability of neither player holding a pocket pair is 16/17*(1 - 72/1225) = 88.6% as you also found (1 - 11.4%). The probability of the board being unpaired and not pairing either of the player's hands is 36*32*28*24*20/(48*47*46*45*44) = 7.5%. So putting it all together, the probability of neither player ending up with so much as a pair is

16/17*(1 - 72/1225)*36*32*28*24*20/(48*47*46*45*44) = 6.7%.

The only hard part about the original question is the removal of straights and flushes from these hands without a pair. So 6.7% is an upper bound.


I get about 5.5% for two random hands (+~.5% ties). However, this may not be the answer you are looking for, since when a pair or better is on the board, high card will not be the winning hand, even if neither player receives help.

(I used Poker Calculator)

I read this to mean that poker calculator is including hands with pairs on the board in this 5.5%, and we want to not include them. Is that right? Does it remove flushes and straights?

[Bozeman]

Bruce, I think I clearly stated that my numbers were merely approximations. I wanted to suggest to Dr. why his intuition was incorrect. ~47% is quite close for 48 unknown cards. The 5.5% is good, it doesn't count pair on board.

Craig

[BruceZ]

~47% is quite close for 48 unknown cards.

No, if you remove 4 cards of different ranks from a 52 card deck, the probability of the board being unpaired is essentially the same as it is for the 52 card deck. Instead of 50.7% it becomes 50.9%. Here is the detail:

[ 36*32*28*24*20 +
5*12*36*32*28*24 +
C(5,2)*12*9*36*32*28 +
C(5,3)*12*9*6*36*32 +
C(5,4)*12*9*6*3*36 ] / (48*47*46*45*44) = 50.9%.

So a complete deck is a good approximation for the 48 card case. 47% corresponds to removing 4 cards of the same denomination. That isn't an approximation for anything relevant to this problem.


I wanted to suggest to Dr. why his intuition was incorrect.

His intuition was that the answer is greater than 5.5%. Your analysis would confirm that intuition since ~25% * ~42% = ~10.5% > 5.5%. The problem is your statement that ~75% of the time one player gets help. This should be evaluated under the condition that the board is not paired, and then this would be 85%, and the result would follow as 88.6% * 50.9% * 15% = 6.7%.

In any case, it is most direct and exact to simply say that 88.6% of the time neither player would hold a pocket pair, and of that 88.6%, 7.53% of the time the board will not pair and the board will not pair either hand, as shown in the above post. 88.6% * 7.53% = 6.7%.