The probability of an unmade hold’ em hand at showdown is 17.4% (Wikipedia). That applies for 1 player. What is the probability that in a heads-up match with both players not having a pair and going to showdown, both fail to hit.


Since they share the 5 card board, there is obviously a strong correlation.

This would be relevant to deciding whether a high card is good enough to stay in the hand.


I think it is likely such an evaluation would best be done using simulation.

I simulated one million heads-up hold-em deals and obtained 59,416 deals where both players had high-card hands.

Of course, that is 5.9416%.

Using the formula for two standard errors (in each direction), this percentage should be within 0.05% of the true pct around 94% of the time.

I tried to modify existing code that did an exhaustive equity calculation and got 214450/3792217 =~ 5.65%. I hacked the existing code pretty hard so I probably made a mistake, though.

I enumerated all starting hands and boards, skipping those where villain or hero started with a pair. I was hoping to closely match your result. Maybe I'll double-check tomorrow.

I don't know if this is at all relevant, but my simulation did not take into account if the hand was chopped or not. That is, my simulation merely tallied how often both players have a high-card hand.

In some of those hands the pot is chopped (e.g., both players play the board, both players have the exact same hand, both players have the same high card, etc.).

Maybe that affects your equity calculation in some way?


Edit: I could re-run my simulation only tallying non-chopped pots if that makes sense.

For what it is worth, I ran a new simulation of one million deals of heads-up hold-em and removed all chopped pots. The new result is that there were 54,841 deals won by a player with a high-card hand. Of course, this is 5.4841%.

It shouldn't. I have code that performs a game tree search for a special form of fixed limit hold'em. I deleted the transposition table and call/fold decision logic, and modified the part that counts wins to count "neither player has a pair or better," and deleted the push count. I might have accidentally deleted something important.

I'm not counting wins/pushes, just the number of times neither player ends up with a pair or better divided by the total number of showdowns. I skipped hands where either player started with a pocket pair, also, per the OP.

Right now I can switch the code back and forth from calculating wins/high cards by changing a couple lines. Just to test, the code gives(* means all hands, !RR means exclude pocket pairs):

3c2c vs. *
694128063 wins for 3c2c

*!RR vs. *
1222690930388 wins for *!RR
propokertools.com matches these results exactly.

By the way, I actually didn't calculate the total number of showdowns in my program, I used propokertools. For the high cards simulation I used the notation !RR vs. !RR in propokertools to get a denominator of 2,463,909,566,976, perhaps this is where I went wrong?

I tried to simplify the code, deleting everything that was unrelated to this problem, and for some reason now I get a totally different answer for the high cards calculation, still not the same as yours:

164934908760/2463909566976 =~ 6.7%

It's more likely that my code has a problem than yours. I think I'll just give up.

Edit: I believe I figured it out. You are counting hands where either player started with a pocket pair in the total. If I use the total that counts hands where either player started with a pocket pair I get:

164934908760/2781381002400 =~5.93%

Switching gears, it is fairly straightforward to simply write a program to tally all the relevant board/hero/villain rank combos which yield two high card hands. It is then pretty straightforward to tally how many suit combos are possible (taking into account the number of rank overlaps between hero and villain's hole cards).

I did it two ways: (1) we need a winner; (2) chops are counted too.

(1) High-card winner (no chops):

I get 153,253,591,200 / 2,781,381,002,400 = 5.509982%

(2) Both players have high-card hands (chops are counted):

I get 164,934,908,760 / 2,781,381,002,400 = 5.929965%.


Of course, there is always the possibility that I made a mistake somewhere along the line here, but it is all pretty straightforward tallying by a computer. And these pcts seem to match the simulations presented above (within the standard error ranges).

Thanks guys. I thought high card winners would occur more frequently than 5.5% - 6% or so. I now realize that it is higher but that’s because many such winnings do not go to showdown, something I don’t see how one could realistically simulate without using historical data. So I essentially asked a question that is only partially relevant to the issue I was considering

Don't let this comment stop the simulation discussion however.

The majority of times when there is a question on a poker percentage, it is usually related to the numerator (the tally of the phenomenon in question) rather than the denominator (the entire universe of possibilities) of the occurrence fraction.

Of course, when conducting a simulation, the overall numerator/denominator issue is mostly hidden since random outcomes are drawn and you keep track of the percentage of "successes" out of the total number of samples drawn.

In the case of heads-up hold-em phenomenon, it is clear that there are a total of C(52,9)*C(9,5)*C(4,2)*C(2,2) = 2,781,381,002,400 complete deals possible where Hero is distinguished from Villain.

I hope that the total number of possible heads-up hold-em deals did not contribute to any "confusion" in this thread.

I probably should have done this the other night, but I went back and utilized my tallying program for a single player. I guess this is a check on whether my two-player results posted above are correct.

For a single player, my program finds that there are 489,183,660 deals for which a single player makes a high-card hand in hold-em.

Of course, there are a total of C(52,7)*C(7,5)*C(2,2) = 2,809,475,760 possible deals of one-player hold-em.

489,183,660 / 2,809,475,760 = 17.411920%.

This percentage agrees with OP and with Wizard of Odds.

It seems nearly impossible that we would independently get the same wrong answer.