Just as there are no "formulas" for calculating poker equities in the general case (there is no shortcut for tallying outcomes over all possibilities), run-it-twice or run-it-thrice probabilities do not automatically fall out of the run-it-once probabilities in the general case.
So let's say what we can about AKo vs QQ. For concreteness throughout this post I will assume that there is one suit overlap between the two hands. Call it AsKh vs. QhQd.
Poker equity calculators tell us the following outcome tallies:
| Hand | Tally | Percent |
|---|
| AsKh | 736,843 | 43.0323% |
| QhQd | 968,234 | 56.5457% |
| Ties | 7,227 | 0.4220% |
| . | | |
| TOTAL | 1,712,304 | 100.0000% |
The probability of most interest to us in trying to answer OP's question is the probability that AsKh will
lose to QhQd over all possible boards, which above we see is 56.5457%. That is, we "remove" the tie probability from the hand's equity since we are interested in AsKh losing all three runouts when we run-it-thrice.
As most people know, there is a rather simple formula for calculating the probability of something occurring 3 of 3 times if the one-time probability is known assuming the three trials are
independent. Like the probability of flipping 3 heads in a row with a fair coin is 50%^3 = 12.5%. Using this admittedly incorrect method would yield a prob of 56.5457%^3 = 18.0800%.
However, in a run-it-thrice situation the three boards are most definitely not independent. For example, if AsKh loses on first two boards, we would expect the prob of AsKh winning the third board to be elevated since the deck stub should now be "rich" with Aces and Kings, say. Thus, we should expect the "true" probability of AsKh losing all three runouts to be less than the 18.0800% derived above assuming independent runouts.
I will present two approaches. First, I will convert the single-win probability to "equivalent outs" and then use standard outs formulas to approximate the three-loss probability.
Method 1: The single-win probability of AsKh vs QhQd was seen above to be 43.0323%. Assuming a simple situation in which Hero will win if any of his "outs" appear, we would have the following:
EWIN = Prob of hitting any out in 5 board cards out of 48 possibilities
= 1 - Not hitting any out in 5 cards out of 48 possibilities
= 1 - C(48-EOUTS,5)/C(48,5)
where EOUTS is the number of Hero's "equivalent outs" to win the hand.
Setting EWIN = 43.0323% and using linear interpolation on the Combinatoric ratio listed above yields EOUTS = 4.90.
Now we can use this value of EOUTS to approximate the probability of losing 3-of-3 runouts:
= C(48-EOUTS,15)/C(48,15)
= 14.61%
where we used EOUTS = 4.90 and again used linear interpolation of the combinatoric ratio.
Method 2: This approach performs a simulation over many deals and tallies how often AsKh loses 3-of-3 runouts to QhQd.
The table below lists the results of a simulation over 1,000,000 run-it-thrice deals. The figures in the table are from Hero's perspective (Hero has AsKh):
| Run-it-Thrice Outcome for AsKh vs QhQd | Percent |
|---|
| 3W 0L 0T | 5.9184% |
| 2W 1L 0T | 32.6149% |
| 1W 2L 0T | 44.8261% |
| 0W 3L 0T | 15.3756% |
| 2W 0L 1T | 0.2305% |
| 1W 1L 1T | 0.6540% |
| 0W 2L 1T | 0.3762% |
| 1W 0L 2T | 0.0019% |
| 0W 1L 2T | 0.0024% |
| 0W 0L 3T | 0.0000% |
I hope the entries in this table are self-explanatory. The first entry shows that AsKh won all 3 runouts (3 wins, 0 losses, 0 ties) 5.9184% of the time. Etc.
The entry that we are most interested in is the "0W 3L 0T" entry which shows that in this simulation AsKh lost all 3 runouts 15.3756% of the time.
So it looks like the "equivalent outs" approach is decent but not great. Recall that Method 1 approximated the prob of losing all three runouts was 14.6% whereas Method 2, a simulation over 1,000,000 run-it-thrice deals, gave a more accurate approximation of 15.4%. Standard sampling theory suggests that this 15.4% estimate should be within 0.1% of the true percentage.
Last edited by whosnext; 05-19-2020 at 03:52 AM.
