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# Binomial Love

3/12/2008

Warning: This article contains some mathematics that might not be suitable for small children or those with a weak heart. If you experience symptoms of nausea, light-headedness or feel the urge to scream loudly, please discontinue reading this article and immediately seek medical attention.

Jim McManus, author of one of poker's finest books, "Positively Fifth Street," recently sent me an e-mail:

"Ladies and Gentlemen: I'm trying to put into historical context Stuey Ungar winning 10 of the 30 major [no-limit hold 'em] events that he entered. Can anyone say with authority at what rate the most successful [no-limit hold 'em] pros actually finish in first place? (I will certainly note that the fields Stuey faced were usually between 75 and 300.) Thanks for your help. Jim"

Nearly everyone who had the opportunity to play with Stu came away awestruck and poorer. Doyle Brunson stated that Stu was easily one of the best-winning players he ever played with. Howard Lederer responded that he believed that if Stu were alive today, he'd be by far the best in the world.

I never had an opportunity to test myself against him. But, I still wanted to help Jim put this kind of feat into some context. The answer: mathematics. Specifically, the very helpful Binomial formula:

P(k out of n) =- n! / k!(n-k)! * (p^k)(q^(n-k))

Here, "P" represents the probability of something happening "k" out of "n" times, given that the event has a probability of happening of "p", with the probability of it not happening equaling "q".

Here's a quick example:

Suppose you have a perfect coin, and you want to figure out how often that coin will flip exactly seven heads out of 10 tosses. You use the binomial function above:

n=10
k=7
p=0.5 (50 percent)
q=0.5 (50 percent)

After toiling with the math, you'll figure out that in this case, P is approximately 11.718 percent. In other words, if you flip a coin 10 times, it will come up heads exactly seven times 11.718 percent of the time.

If you want to answer a slightly more complicated question like, "What is the probability of flipping heads seven or more times?" you can do that with repeated applications of the binomial and then adding up the results.

So, enough tutorial. Let's use this equation to put Stu's accomplishment in some kind of mathematical, historical context.

Let's say that back in the "old days," Stu played in an average field size of 200 players or so. From everything I've been told, that is somewhat reflective of reality. Let us then try to figure out how likely it would have been for him to win 10 tournaments out of 30 given certain skill factors relative to his opponents. An average player in the field will have a 0.5 percent (or 0.005) chance to win the tournament, or 1/200.

Our first binomial: Stu was an average player.

n=30, k=10, p=1/200, q=199/200

Let me save you some work. After repeated application of the binomial, you'll find that the probability of an "average" player in a 200-player field winning 10 out of 30 tournaments is 2.6x10-16. That is 0.0000000000000026 percent. I think we're going to have to conclude that Stu was a little more skillful than the average player.

Option 2: Stu was 10 times as likely as the average player to win.

n=30, k=10, p=10/200, q=190/200
P(>=10) = 0.000116 percent

Interpretation: A player 10 times as skilled as an average player is still about 9000 to 1 against to win 10 out of 30 tournaments with 200 players in the field.

Option 3: Stu was 30 times as likely as the average player to win.
n=30, k=10, p=30/200, q=170/200
P(>=10) = 0.965 percent

Interpretation: A player 30 times as likely to win as an average player in a 200-player field will win 10 out of 30 tournaments about 1 out of 100 times.

Can you imagine walking into a tough field of 200 battle-tested players like those Stu played against and thinking, "Wow, I'm about 15 percent to win this tournament?"

So, how would Stu do today? Once again, we can turn to our friend the binomial. Let us assume that Stu played a "full schedule" of \$10,000 events over the last seven years. By my estimates, that's about 100 events total. Let's also assume that he'd be facing about 500 players on average in each tournament. How many of these events would he be expected to win given that we're assuming that he was about 30 times more skillful than the average player?

n=100, k=0, p=30/500, q=470/500
P(>0) = 99.79 percent

Interpretation: Stu is about 99.79 percent to have won one or more events.

n=100, k=5, p=30/500, q=470/500
P(>=5) = 72.3 percent

Interpretation: Stu is about 72.3 percent to have won five or more events.

n=100, k=10, p=30/500, q=470/500
P(>=10) = 7.75 percent

Interpretation: Stu is about 7.75 percent to have won 10 or more events.

How about the World Series of Poker Championship event the last three years? The event has had an average of about 7,000 players over the last three years.

n=3, k=1, p=30/7000 q=6970/7000
P=1.28 percent

Interpretation: Stu would have had a 1.28 percent chance to win the WSOP sometime in the last 3 years.

More Practical Applications

OK, enough about a guy I never met or played against. How about using the binomial for something a little more realistic.

Let's say I'm trying to figure out my chances of being a loser after playing 10 \$100 heads up sit-and-go tournaments in row. Let's say I have a slight edge on my opponent and I'm 55 percent to win each individual sit and go.

n=10, k=4, p=0.55, q=0.45
P(<=4) = 26.15 percent

Interpretation: I'll end up a loser about 26 percent of the time in this scenario.

Now I've really got the guy on the line and he agrees to play me 100 times. What are my chances of coming out a loser now?

n=100, k=49, p=0.55, q=0.45
P(<=49) = 13.45 percent

Interpretation: Even after 100 heads up games, I'll still end up a loser 13.45 percent of time even given a massive edge of 10 percent against my opponent.

You just lost with pocket aces vs. an under pair for the fourth consecutive time. Internet poker is rigged! What are the chances of that?

n=4, k=4, p=0.20, q=0.80
P=0.16 percent

Interpretation: You're going to lose with those aces vs. an under pair four times in a row about 1/625 times.

You're going to play 50 WSOP tournaments this year and you are an unbelievably excellent player with a 20 percent chance of cashing in each event. What are the chances you have a miserable year and cash only four or fewer times?

n=50, k=(04), p=0.20, q=0.80
P=1.84 percent

Interpretation: You've got about a 1/56 chance of having a really bad year.

You're an average player with a 10 percent chance of cashing in an event, and plan on playing 10 WSOP events this year. What are your chances of getting shut out?

n=10, k=0, p=0.10, q=0.90
P=34.86 percent

Interpretation: Don't be so hard on yourself if you go through a "bad streak" in tournament poker. Remember, Chris Ferguson played in about 40 WPT events before he finally cashed -- by my calculations, the chances of that were about 1/500.

Conclusion

Get comfortable with your inner binomial. You'll find that it is an invaluable tool in your poker arsenal. Anytime you are running really bad or really good, use this mathematical tool to "double-check" your reality and keep yourself grounded. And remember, this formula shows you that you need a massive edge applied over thousands of trials to "ensure" success.

For those that are interested in doing the math without doing the math, try this Web site:

http://faculty.vassar.edu/lowry/binomialX.html