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Phil Hellmuth and Expected Number of WSOP Bracelets Phil Hellmuth and Expected Number of WSOP Bracelets

10-31-2015 , 05:00 PM
Phil Hellmuth and the Expected Number of WSOP Bracelets


Introduction
------------

As we are on the precipice of the 2015 WSOP Main Event Final Table, I thought it would be an opportune time to investigate a topic that has been bandied about over the years regarding how easy it is for a top player to win a WSOP bracelet.

My focus will be on the mathematical question of how many WSOP bracelets should the top pros be expected to compile over their careers. Put another way, how impressive is Phil Hellmuth's record of 14 WSOP bracelets, and how safe is Hellmuth's record?

The WSOP was a small baby born in 1970 and has grown over the years until it is now in middle-age quite immense. In fact, the WSOP has recently outgrown the US shores and has expanded to Europe and Asia-Pacific. Below I summarize the growth in the WSOP by presenting the total number of open tournaments in every five-year interval (skipping the very first year when there was only one "tournament" (an extended cash game)) along with the total number of entrants, and the average number of entrants per tournament.


Quint......Total........Total...Average
Years....Tourneys..Entrants.Entrants

1971-1975__25_____373____15
1976-1980__56___2,503____45
1981-1985__66___6,854___104
1986-1990__60__10,978___183
1991-1995__99__17,017___172
1996-2000_100__20,074___201
2001-2005_157__61,169___390
2006-2010_271_277,866_1,025
2011-2015_339_397,951_1,174

The table reflects that the WSOP has witnessed steady growth since its inception, and has experienced rapid growth in the 21st century, in the total number of tournaments, total entrants, and the average number of entrants per tournament.


Equal Skill
-----------

In thinking about the prospects of a poker player winning a WSOP bracelet, let's start simple. Suppose everyone who enters each and every WSOP tourney has an equal likelihood of winning the tournament. Of course, this assumption is unrealistic but we have to start somewhere and this is as good as place to start as any.

Since we are starting simple and unrealistic, let's also assume that a player can play in every tournament in each year. There are logistical reasons why this may not be realistic as well as poker variant reasons, but let's start here. Then, it is rather straightforward to calculate how many WSOP bracelets any one player can be expected to win in each year of the WSOP's existence.

Again it may prove interesting to present these expected WSOP bracelet totals in five-year intervals, given below.


Quint.......Total_Expected
Years.....WSOP_Bracelets*

1971-1975 3.0
1976-1980 2.7
1981-1985 1.1
1986-1990 0.5
1991-1995 0.8
1996-2000 0.8
2001-2005 0.9
2006-2010 0.7
2011-2015 1.0
* assuming each player entering a tourney has equal chance of winning, and our hero plays in each and every tournament each year.


The table reflects that WSOP bracelets, purely on the number of entrants basis, were easier to win in the very early years. Under our assumptions, a player able to play every tournament and capable in every poker variant played, could be expected to win 3 total WSOP bracelets in 1971-1975.

As the field sizes steadily increased since those early years, the number of WSOP bracelets our hero could be expected to win declined sharply. The decline eventually leveled off and has even seen a slight uptick in recent years as the total number of tourneys offered has grown significantly (including going to Europe and Asia-Pacific).

This table also affords us our first look at how many total WSOP bracelets a top pro could be expected to accumulate in his lifetime. If a player started in 1970 and played in every WSOP tourney all the way through 2015 (let's call him Robot Doyle), then he would be expected to accumulate 11.6 WSOP bracelets.

Of course, that would require super-human longevity and stamina and an impressive breadth of poker expertise. Not even the great Doyle Brunson comes close to our hypothetical Robot Doyle's attendance record (remember Doyle boycotted the WSOP for a few years in the aftermath of the Binion family squabble). In the real world, the top pros seeking bracelets construct their playing schedule to maximize their chances of winning tourneys which sometimes means skipping tourneys with large fields or in a poker variant not of the player's particular liking.

In total there have been 1,174 total WSOP open tournaments since its inception, and this "equal-skill" assumption gives us a simple way to derive our hero's likelihood of winning any tournament. Since the number of WSOP bracelets our hero is expected to win in any year (or five-year interval or over the entire 46 years of WSOP history) is simply the sum of the expectations (likelihoods) of each and every tournament, this figure is a simple sum of the 1/N's.

Similarly, since the variance of the sum of independent random variables is the sum of the underlying variances (we assume that the results of each tournament is independent of the results of all the others), the variance too is a simple tally. We are often interested in the standard deviation which, of course, is simply the square root of the variance.

We see that our Robot Doyle (under the equal-skill assumption) would be expected to win a total of 11.6 WSOP bracelets in the 1,174 WSOP tourneys in 1970-2015 with a standard deviation of 3.2 bracelets. Also, for further reference, let's note that a Robot Hellmuth (under equal-skill) would be expected to win 5.0 WSOP bracelets in the 1,039 WSOP tourneys in 1985-2015 with a standard deviation of 2.2 bracelets. I chose 1985 as the starting year for Robot Hellmuth since that is the year Hellmuth turned 21.


Adding Skill Edge
-----------------

The keen reader will be quick to point out, the best players surely have an edge on the field. Guys like Doyle, Ivey, Negreanu, and Hellmuth (among many others) have a significant edge in winning any tournament they enter. For the purpose of this exercise, we will set aside the differential edges players have across the vast array of poker variants offered at the WSOP over the years (and in the future). In this first foray, we will assume that "poker is poker" and that our hero has the same edge whether he is playing NLHE, PLO, PLO8, triple-draw, 7-card stud, etc. Of course, this is unrealistic but it is a convenient place from which to start.

How can we think about a player's edge based upon the number of players entered in a tournament? We need a simple model to start with. Let's suppose all poker players have differential abilities to win a tournament ranging from the very best player (our hero) down to the very worst player. Further assume that the differences between any two players consecutively ranked is a constant. Finally, suppose that all players who have a chance to win a tournament enter the tournament, meaning that the next person "in line" to enter the tournament (but does not enter) has zero chance to win.

Under these pristine mathematical assumptions, it is fairly straightforward to show that the top player (our hero) has a 2/(N+1) chance of winning a tournament with N entrants. For example, if 9 players enter a tourney, our hero would have a 20% chance of winning. If 99 players enter a tourney, our hero would have a 2% chance of winning. Obviously, this assumption is for mathematical convenience but let's see where it leads us.

But, wait a minute, you might be saying, the modern large fields have a significant portion of "dead money" having no realistic chance of winning the tourney. Consider the 2015 main event field size of 6,420. Surely many of those "recreational" players have no chance, right? The cutoff where the person has zero chance of winning in these large field events is very likely not the "next person not to enter" but somewhere within the field itself.

I have tried to model this as follows. I have assumed that every player in the first 100 players has a chance at winning; 10% of players between 101 and 200 have no chance; 12.5% of players between 201 and 550 have no chance; 15% of players between 501 and 1,000 have no chance; 17.5% of players between 1,001 and 2,000 have no chance; 20% of players between 2,001 and 3,000 have no chance; 25% of players between 3,001 and 4,000 have no chance; 30% of players between 4,001 and 5,000 have no chance; and 50% of players beyond 5,000 have no chance.

These figures reflect my views on the strength of today's huge field sizes, taking into account poker's global reach, the influence of online poker, etc. In all honesty, if anything I probably erred on the side of conservatism in that there is probably more dead money than reflected in these figures.

Using these conservative dead-money cutoffs, it is straightforward to derive the probability of a top player (our hero) would have in winning any WSOP tournament based upon the field size. Again, we are assuming our hero can play every event on the schedule (logistically all-but-impossible) and has the same edge no matter what the tournament's poker variant. What effect does adding an "edge" have on how many WSOP bracelets Robot Doyle or Robot Hellmuth would be expected to win?


Quint.....Total_Expected
Years.....WSOP_Bracelets**

1971-1975 4.9
1976-1980 4.7
1981-1985 2.1
1986-1990 1.1
1991-1995 1.7
1996-2000 1.7
2001-2005 1.9
2006-2010 1.5
2011-2015 2.1
** assuming our hero plays with a significant skill edge in each and every tournament each year.


As the above table shows, straightforward calculations find that Robot Doyle+ (with skill-edge) who played every WSOP tournament in 1970-2015 would be expected to win 21.8 WSOP bracelets with a standard deviation of 4.3 bracelets. Robot Hellmuth+ (with skill-edge) who played every WSOP tourney in 1985-2015 would be expected to win 10.2 WSOP bracelets with a standard deviation of 3.1 bracelets.

Since the real Phil Hellmuth has won 14 WSOP bracelets, it may be of interest to see how often Robot Hellmuth+ would have won 14 bracelets. Unfortunately, there does not appear to be any simple way to derive the probability distribution for this total since the underlying win probabilities associated with the respective tournaments are not identical as they are in the binomial case. I performed a simulation of 10,000 Robot Hellmuth+'s and found that around 15% of the time it won 14 or more bracelets.


Winner's Curse
--------------

That's all well and good but perhaps something else might be going on in the real world to cause a player named Phil Hellmuth to have won 14 WSOP bracelets (leaving aside the obvious skill edge question). What other purely "statistical" factors could be at play here? Well, what if there were more than one player who has this edge in every tournament, and they competed against each other in every tournament?

Suppose, first, that there are two of these Robots contemporaneously entering each and every WSOP tourney. Who won each tourney would be partly random luck but over the years each of these Robots (call them Robot Hellmuth+ and Robot Ivey+) would each win a significant number of bracelets. And we would fete the Robot with the most bracelets. That is, from a purely statistical perspective, the expectation of the maximum of two random variables is greater than the expectation of either one of the random variables themselves.

Readers will probably be familiar with the "winner's curse" phenomenon in auctions in which the winning bidder on an untested oil field will often be disappointed in the field's ultimate production since the winning bidder had the highest (often unrealistic) expectation of the amount of oil in the field among all the bidders in the auction.

Here, if there are several top players with a similar skill edge and enter all of the tournaments, the expected number of total bracelets won by the player with the most bracelets at the end of their careers will be greater than the number of bracelets each of the players individually would be expected to win at the beginning of their respective careers.

To make this point even clearer, consider a simple coin-flipping exercise. Suppose we flip 10 different pennies 100 times each and keep track of the number of heads for each penny. Here are the number of heads I got the first time I performed this experiment (51, 53, 60, 48, 42, 48, 53, 45, 51, 52). Penny #3 with 60 Heads would appear to be the Olivier Busquet of the penny universe! Even though all of the pennies are identical, and each is expected to win 50 of its 100 "matches", just by random luck some will likely win more than 50 and some will likely win fewer than 50.

Suppose we run the experiment 1,000 times and track the number of winning matches the penny with the most wins in each experiment (that experiment's Olivier Busquet). I find that the average number of wins for the "Uber Penny" is around 58. Of course this is significantly greater than 50, the expected number of wins each identical penny is expected to win. It is greater by construction since it is the tally of the 1,000 different maximum win totals across the 10 pennies in each experiment.

This phenomenon is fairly easy to see when it comes to oil fields or pennies, but is sometimes harder to see when it comes to poker players.


More Robots
-----------

We saw in the previous section that "biasing" statistical phenomenon can occur when randomness is spread across multiple participants. Let's see what is the impact of adding another Robot to our poker universe.

Performing a simulation of 10,000 dual-careers with Robot Ivey+ joining Robot Hellmuth+ in playing every WSOP tourney in 1985-2015 (it seems that Robot Ivey+ started when it was nine years old!), we find that each Robot would be expected to win the same 10.2 WSOP bracelets, but the number of bracelets the Robot with the most career bracelets (Uber Robot+ #2) would be expected to have won 12.0 bracelets with a standard deviation of 2.7 bracelets. Each individual Robot would be expected to win at least 14 bracelets around 15% of the time, whereas Uber Robot+ #2 would be expected to have won at least 14 bracelets around 27% of the time.

Since it is easy to add Robots (they don't seem to mind), let's see the impact of adding a third Robot called Robot Negreanu+ (who started playing when it was eleven years old). Performing another simulation of 10,000 tri-careers of our three Robots, we find that each Robot would be expected to win 10.2 WSOP bracelets, but that Uber Robot+ #3 (the Robot with the most career bracelets) would be expected to have won 12.9 bracelets with a standard deviation of 2.5 bracelets.

Uber Robot+ #3 would be expected to have won at least 14 bracelets over the course of its career 39% of the time. Clearly, the more Robots we add, the higher will we drive the number of bracelets the Uber Robot+ would be expected to have won over its career and the frequency in which it would have won at least 14 bracelets (to match the real Phil Hellmuth).


Concluding Remarks
------------------

I have attempted to come at the WSOP bracelet issue from a purely statistical perspective. We have seen that WSOP bracelets have generally been increasingly difficult to win since the inception of the WSOP in 1970 with a slight leveling off in recent years due to the increase in the total number of tourneys offered (including expanding the WSOP's imprimatur to Europe and Asia-Pacific).

Bracelet hunters will be wise to seek out the smallest fields they can find where they have a significant edge on the field. This typically can only be found in specialty events such as triple draw, razz, stud, or the like. Edges are probably slimmer in NLHE, and in any event no-limit holdem attracts such large fields that even players with significant edges cannot realistically be expected to win a single bracelet, let alone multiple bracelets.

It is conceivable that NLHE bracelet hunters can have a first-mover's advantage if they are among the first players to enter tournaments in new locales that the WSOP ventures to in the future. But this would likely be only a short-term phenomenon as other players would make the journey in subsequent years and the local poker playing population would be expected to grow in both quantity and quality.

Regarding Phil Hellmuth's current record of 14 WSOP bracelets, we have seen that this is indeed an impressive record. Even when we constructed a hypothetical Robot which had a significant skill edge and was able to play in each of the over 1,000 WSOP tourneys in 1985-2015 (the Hellmuth Era), play its best in each tourney, and have that significant edge over the field no matter what the tourney's poker variant, this robot was "only" expected to win around 10 bracelets.

It was only when we supposed that there were multiple Robots in our midst that the 14 bracelet haul became more likely. When we created three Robots (named for Hellmuth, Ivey, and Negreanu), we found that the Uber Robot+ would be expected to win around 13 bracelets and would be expected to win 14 or more bracelets around 39% of the time.

All in all, it seems it will be very difficult for any player to surpass Hellmuth's WSOP bracelet record.

(Comments welcome on any aspects of this investigation.)
Phil Hellmuth and Expected Number of WSOP Bracelets Quote
11-02-2015 , 01:18 PM
interesting stuff. i've thought about this myself so it's nice to see someone do the research and come up with the probabilities
Phil Hellmuth and Expected Number of WSOP Bracelets Quote
11-02-2015 , 05:58 PM
Quote:
Originally Posted by whosnext
Phil Hellmuth and the Expected Number of WSOP Bracelets

Since the real Phil Hellmuth has won 14 WSOP bracelets, it may be of interest to see how often Robot Hellmuth+ would have won 14 bracelets. Unfortunately, there does not appear to be any simple way to derive the probability distribution for this total since the underlying win probabilities associated with the respective tournaments are not identical as they are in the binomial case. I performed a simulation of 10,000 Robot Hellmuth+'s and found that around 15% of the time it won 14 or more bracelets.
Upon further investigation, the analog to the Binomial distribution except when the underlying success probabilities are not identical has been called the Poisson Binomial Distribution. Generally speaking, there is no straightforward way to calculate the probabilities associated with the sum (in our case the total number of Robot Hellmuth+'s WSOP bracelets).

However, the following recursive formula can be used to derive these probabilities. Let P(i) be the underlying probability our hero has in winning Tournament i (i goes from 1 to T). Let B(k) be the probability our hero has in winning k bracelets over the T tournaments (k goes from 0 to T).

B(0) = Product of [1-P(i)] where the product goes from i=1 to T. This is the standard formula for B(0) of course.

Then B(k) = (1/k)*Sum of ((-1)^(i-1))*B(k-i)*T(i) where the sum goes from i = 1 to k, and T(i) = Sum of ((P(j)/(1-(P(j)))^i where the sum goes from j = 1 to T, for k>0.

I find this recursive formula pretty interesting. Upon first glance it looks like it may have elements of Pascal's triangle and Inclusion-Exclusion in it.

Anyway, using this formula I am able to derive the probabilities that Robot Hellmuth+ (who played each and every WSOP tournament offered in 1985-2015 with a skill edge) would win any number of tournaments from 0 to 1,139.

It turns out the simulation I reported on in the original post was sufficiently accurate. The true average (mean) number of bracelets Robot Hellmuth+ would win is 10.19 bracelets, with a true standard deviation of 3.15 bracelets.

The true probability that Robot Hellmuth+ would win 14 or more bracelets (given the assumptions described in the original post) is 14.68%.

For anyone interested in the shape of the distribution (and who isn't), here are the relevant B(k) probabilities of how many WSOP bracelets our friendly robot would have been expected to win in 1985-2015.

B(0) = 0.000033
B(1) = 0.000345
B(2) = 0.001798
B(3) = 0.006236
B(4) = 0.016176

B(5) = 0.033483
B(6) = 0.057606
B(7) = 0.084728
B(8) = 0.108762
B(9) = 0.123782

B(10) = 0.126465
B(11) = 0.117161
B(12) = 0.099246
B(13) = 0.077408
B(14) = 0.055922

B(15) = 0.037613
B(16) = 0.023659
B(17) = 0.013971
B(18) = 0.007773
B(19) = 0.004087

B(20) = 0.002036
B(21) = 0.000964
B(22) = 0.000435
B(23) = 0.000187
B(24) = 0.000077

B(25) = 0.000030
B(26) = 0.000011
B(27) = 0.000004
B(28) = 0.000001
B(29) = 0.000000

and so on.

Curious readers may want to delve into the derivation of the recursive formula for the Poisson Binomial Distribution.
Phil Hellmuth and Expected Number of WSOP Bracelets Quote

      
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