Best of N Heads-up series. How many to "prove" a winner?
In the starting embers of a flamewar over in the home games section, Haddon made an interesting comment. (Please note he was referring to a Head-up match
So I thought... should the best player win really? If so, how often. Obviously, if they only play one match, the 60% favorite will win 60% of the time. What if it's 2 of 3? 3 of 5? etc.?
My math skills are a little slower than they used to be so, while I work on generalizing the problem for any % or best of series, I thought I'd throw that out there, and let the younger mathematicians get a crack at it.
Haddon wrote:It isn't one game where someone can get lucky, it is 3/5, the best heads up player SHOULD win.
So I thought... should the best player win really? If so, how often. Obviously, if they only play one match, the 60% favorite will win 60% of the time. What if it's 2 of 3? 3 of 5? etc.?
My math skills are a little slower than they used to be so, while I work on generalizing the problem for any % or best of series, I thought I'd throw that out there, and let the younger mathematicians get a crack at it.
Comments
Because it's more profitable than 3 out of 5. MLB used to do a 3/5 in the first round of the play offs.
I don't actually believe you can "prove" any thing regardless of the size of the best of series. (My belief is that as the number of games increase, the winning percentage approaches a limit.)
What I'm trying to answer is "Given Player A as a x% chance to win a Heads-up match against Player B, what is the chance that Player A will win a Best of n series"?
The easy case is a Best of 1 series. Player A will win 60% of the time.
If we look at a Best of 3 series, then Player wins if he loses 0 or 1 match at the most. I've come up with a forumla, but I don't think it's at all correct. I know there's a flaw here somewhere...
E is the event given above.
P(E) = P(Player A wins exactley two matchs)
= P(A loses at most 1 match)
= P(A loses exactly 0 matches) + P(A loses exactly 1 match)
= (.6x.6) + P(lose one of the first two matches and wins the last match)
= 0.36 + 2x(.6x.4)x.6
= .36 + .288
= 64.8?
(This number feels right, but when I extend my logic to a best of 5 series, it all falls apart.)
There's a big flaw in the way I'm computing this, but I can't figure out where I'm going wrong. I think there must a better way to state the event to make the math clearer. I don't I'm determining "A loses exactley 1 match" correctly because I don't think that WLW and LWW are mutuall exclusive events.
To provide some anecdotal evidence, Jenn is pretty good HU, and she has had double-digit winning AND losing streaks during a 50-game run (approximate/from memory). Considering Jenn is better than her average opponent ($5 online), she should win a good portion, so the huge losing streak is strange (and shows the huge variance HU).Â
My gut guess would be a lot of games (hundreds), so nothing realistic if you're trying to have a tournament.
You would need to define what it means to be a "winner" against the relevant opponent. For the sake of using numbers instead of x, y, and z's, let's say that you would call Player A a winner against Player Z if A would beat Z in the long run 60% of the time.1
Suppose that you observe a series of 10 matches and Player A (as expected) wins 6. The binomial distribution governs these kinds of repeated independent events. Using a binomial calculator,2 the probability of observing at least 6 wins for Player A is about 63%. On the other hand, under the contrary hypothesis that Player A is a (40%-60%) losing player, observing 6 (or more) match wins has probability about 17%.
Most statisticians would happily accept a 95% confidence in one hypothesis over an alternative as "proof". So, a 10 match series is nowhere even close to this.
We are looking for the value of N (number of matches) that sets the probability of Player A being a losing player equal to 5%. This roughly occurs when Player A wins 12 matches out of 20. If Player A was a losing player, this would occur only 5.65% of the time. Still possible, but generally good enough to be accepted as statistical "proof".
The key logical idea here is that it is easier to disprove (statistically) that Player A is a losing player. This is a very common approach in statistics when testing hypothesis. You often suppose the opposite hypothesis to the one you want, and then show that it is highly improbable.
For someone who demands 99% confidence, it turns out that3 (at least) 24 wins in 40 matches would be enough to "disprove" that Player A is a losing player. (Alternatively, 8 wins in 10 matches would show that a player is probably not a 40%-60% losing player.)
The real difficulty in the analysis I've done is the pegging down of winning-losing player as exactly 60 : 40. Someone who is more of an expert at Bayesian stats might be able to consider the whole range of winning-losing scenarios as p : (1-p), but that is beyond the scope of this forum post.
Finally, there is typically a great deal of observer bias in experiments like this one. For example, if two players happen to sit down and play 10 heads-up matches and one player won all 10 matches, then it would be natural to wake up and try to compute probabilites and play with binomial calculators.4 However, to be a fair statistical test, besides being repeated several times if possible, the statistical experiment should not be triggered by the events that you are experimenting on. (A classic example of this is somone observing pocket Aces dealt twice during the same hand, and then wondering what the probability of this occurring is.)
ScottyZ
1In other words, Player A would have a 60% chance of winning each individual match.
2For example, http://www.stat.tamu.edu/~west/applets/binomialdemo.html
3There is no fancy mathematiccs going on here. I'm just going all trial-and-error on the binomial calculator's ass.
4Well, natural for statisticians maybe.
With no offense intended to either group, I'm technically 66% mathematician and 33% statistician.
And 1% poker player.
ScottyZ
For example, not necessarily against one another, but if you were to have both players play an identical AI program, where hole cards and board cards are predetermined to be identical, both enduring the same bad beats and successes, you could therefore imply which of the two is "better"
However, given the nature of poker, you could not then simply say "Player A defeated the AI in 132 hands, while Player B took only 99, therefore B is the superior player", or even the win/loss records against the computer. If you could then somehow ensure that both players receive identical hands while heads up (at different times) and see who best exploits their opponent, that could be interesting, however, the problem there of course is that they will get suspicious of the board / hole cards being identical - even if say Player A experience Player B's hands in a completely differnt game....
Hmmm... wish I was still in university research methods course... this could be interesting.
Mark
Zithal, your formula is correct. Â Given your simplifying assumptions, the better Player A has a 64.8% chance of winning a best-of-three heads-up series, while Player B has a 35.2% chance.
For a best-of-five series, here is one formula.
P (Player A wins best of 5 series) = P (A wins in exactly 3 games) + P (A wins in exactly 4 games) + P (A wins in exactly 5 games)
= (.6 * .6 * .6) + [(.4 * .6 * .6 * .6) * (LWWW + WLWW + WWLW)] + [(.4 * .4 * .6 * .6 * .6) * (WW LLW + LL WWW + WLLWW Â + WLWLW + LWWLW + LWLWW)
= .216 + (.864 * 3) + (.03456 * 6)
= 68.256%
In other words, the probablility of the slightly better player winning does not increase that much with a longer series. Â NBC Heads Up Poker Championship and Poker Superstars Invitational have the correct format of just having a best-of-3 finale.
You aren't suggesting that best of 3 is "sufficient" to determine the best player or that NBC's motivation for a best of 3 format is "determining the best player" I hope?
Incidently, I don't really think a 60-40 advantage describes a "slightly better player" either. I think this is a fairly large edge. I'd guess the edges you probably are looking at in the HU Championship are more along the lines of 52-48. I think a 55-45 edge would be fairly large amongst pros.
And as Beanie said wrt variance, tilt etc. this model is incredibly simplistic (for calculations sake of course). Beanie's suggestion of a balance between luck and cards is probably valid on some level, but quantifying it is hard. And I would guess a deep stack, slow progression tournament structure is probably less dependant on variance than the turbo-like structure of Poker Superstars.
/g2
Like most things about poker, I believe the answer is counter-intuitive. One cannot play against one opponent to determine who is best. For a simple example you may have a great pre-flop game and you're playing against an opponent who has a horrible pre-flop game but does have a superior post-flop game. With not so deep stacks and escallating blinds -- it will favour pre-flop guy and he will prevail no matter how many matches.
Poker is about having many, many different skills. To get to the real answer you will need to play somewhere near 1,000 matches against different opponents and exact same structure. Then you can only say you're king of the $500 sitngo with the pokerstars structure.
Even then the conditions won't be the same. One player may be playing for leisure and the other for a living. Or one opponent may be a millionaire and the other a regular working dude.
So, can you ever say one opponent is better than the other? I really don't think so. Now, if you could introduce something like Ozzie-rules-heads-up-matches, then I think you could get to the heart of the matter.
Cheers
Magi
I don't think that's what Daniel N had in mind when he called his site Full Contact Poker.
Welcome back Scotty and a few other posters who had been hiding. It's great to try and digest your greater mathematical skills than mine.
for a 3 games series, either player can win by either (win, win), (win, loss, win), (loss, win, win). If you add the probabilities of each of these three events happening you get a 64.8% chance that a player thats a 60% favorite to win an individual game will win the series. Without actually having a proof, I'm pretty sure that in a 5 game series, the same pattern would continue and the winning chance of the better player would keep increasing, and that the probablility of the better player winning will approach 100% as the number of games approaches infiniti.
A way to answer your original question would be to take an example game, say our 60:40 matchup, and decide how much of the time you would like the favorite to win, say 80%, and then solve for the number of games needed to be played to achieve this. Then you could at least have a better idea who the better player truely is.
(Note: This has nothing to do with Brent and Shannon, just wanted to join the discussion.)
Luck is a factor, I know. But it is cannot be nearly as high as 50%, how could it be if there are consistant winning players?
In heads up with lots of play, as the NBC structure offers, the better player should win a best 3/5 or 4/7 series. I am confident that if I played an inferior player they could not/I would allow them to suck out on me with luck 3/5 or 4/7 times when I had all or most of my chips in the middle. A good heads up player does not overuse the 'all in' so the above is a given. The odds of a Set over set, flush over flush, straight over bigger straight coming on the flop in heads up poker is not large enough to worry about an inferior player beating you 3 or 4 times in these situations. With good reads, controlled play and timely bluff I believe I could beat a player I thought was inferior pretty much every time in a Heads Up best of 5 or 7 match.
When people run bad they blame cards, but knowing that most of us on this forum are students of the game, think back to Doyle's book where he talked about running bad. He said his play was great but he was getting drawn out on, not hitting flops and getting cold cards. Then he had someone (I forget who) critique his play and they pointed out holes in his game that had developed. I believe in luck, I believe in hot streaks, but I do not believe that people lose for weeks/months at a time because of the deck. A week maybe I can understand, maybe two. But I have had weeks with no cards, and when you have no cards you need to up your attentiveness and strike when you know the other guy/girl is weak. I can see running bad online hurting a lot more than live. But in a live game, the cards don't matter as much as the people.
In heads up if you really don't respect someone's game and you know you are better than them, how can you not beat them in a series? You only have them to worry about, every hand, every word, every bet...you have to track only ONE player's patterns and tendancies. If you can't figure them out after 100 or so hands, they a) aren't nearly as bad as you thought they were or b) are just a better heads up player than you. I'd play anyone heads up in a best of 5 or 7 series and gladly tell them they are better than me at heads up if they one. Hell, Trevor schooled me at KWGM2 and on any day I would be quick to say I one hundred percent got outplayed.
Most of you won't respect what I've said for the most part because you think I'm a luck box. But I have a great gut (no joke intended) and trust my reads, which are usually correct. I know I've strayed from the topic at hand...and I know there isn't any math in this post, but it is 4AM almost and thought I'd throw in my take on heads up/poker in general. I am going on to conclude with: Luck is not that big a part of poker if your game is sharp.
Me being flamed for my 'inexperienced' beliefs about poker ---->
Thanks, BlondeFish! I know exactly what I did wrong in my best of 5 calculation.
I really think a 60% edge is a HUGE edge when it comes to the short term luck factor of poker., especially when it comes to two player who understand the game and get their fair share of wins.
Just for fun, I created a quick and dirty program to simulate playing 1,000,000 million different Best Of matches to see how close we got to our calculated numbers. Results below...
Best Of
3
5
7
9
11
21
51
101
1001
%
64.2
67.6
70.2
72.5
74.4
81.5
91.8
97.4
99.99999
So we can see as the number of games approaches infinity, the probability of the player with the greater edge winning the match approaches 1.
If we drop the Edge down to a more realistic 55% edge, playing a best of 5 series will only result in a 58.5% win rate for the player with the edge. Even playing a Best of seven series only brings this number of to almost 60%.
As everyone's been saying, the big problem with the calculation is the % edge that one player has over another. It's impossible to determine as there hundreds of different factors which will contribute to this number.
But I do think they we've shown that even if someone has a large edge (ie. 60% over an opponent), playing a Best of 5 series, will have no relation on who the better player is.