After 9 levels of play, Day 1 of the 2014 Paddy Power Irish Open has come to a close. 411 players showed up to the Double Tree hotel in Dublin and put up their €2,250 ...
Winning Two Bracelets
by Steve Zolotow | Published: Apr 18, 2012
David Heyden recently expressed surprise at the fact that nearly every year, there is one player who wins two (or more bracelets) at the WSOP. Most of you have probably never heard of David. Although he is a quiet, unassuming player, he has been considered one of the best seven stud players for years. He has recently switched his focus to no-himit hold’em, since it is easier to find NLH games with weak players. Considering how hard it is to win one bracelet, at first glance, it seems quite remarkable, that someone can win two in the same year.
I thought it would be fun to do the math, and determine what the chances are of this happening, again. The problem is that the WSOP will have around sixty events. Some will have over four thousand entrants and others fewer than 200. Some players play only one event and others more than twenty. In general, those events with high buy-ins and small fields draw the best tournament players. These are certainly some of the most likely candidates to win multiple events. It was clearly necessary to find a way to estimate the possibility of someone winning more than one event. There will also be a super-high-roller charity tournament, with a $1 million buy-in. Along with some very wealthy players, this will attract the most successful tournament superstars. My approximation is summarized below:
To get a rough idea, let’s say that after 25 tournaments with 25 winners (some of whom are very skillful), not all of whom will play any given subsequent event. As more tournaments are played this pool of winners grows. Let’s assume that there is around a 10 percent chance that one of them will win another event. (This may be low for events with 200 entrants, but high for events with 3000.) This means the chance of no repeat in any specific event is 90 percent. Since there are more than 35 events to go, the chance that no one will repeat is .9 to the thirty-fifth power, which is less than 3 percent.
This makes there being a multiple event winner more than a 30-to-1 favorite. Note that this approximation didn’t include a factor for a repeat in the first 25 events, and uses the 90 percent figure even for the last big events which might have more than 30 previous winners playing in a field of 200. Putting all of this together 30-to-1 may be on the low side, but at least it gives some idea of why there rates to be a multiple bracelet winner.
For those who find this difficult to believe, consider the famous birthday problem: How many randomly selected people do you need to have a better than even chance of finding two with the same birthday? The amazing answer is that with twenty-four (yes, 24), a duplicated birthday becomes a favorite. By the time you reach 60 people, the chance has passed 99 percent. For those interested in more details on this problem, Wikipedia has a nice explanation (although the math does get quite complicated) at http://en.wikipedia.org/wiki/Birthday_problem. ♠
Steve “Zee” Zolotow, aka The Bald Eagle, is a successful games player. He currently devotes most of his time to poker. When escaping from poker, he hangs out in his bars on Avenue A — Nice Guy Eddie’s at Houston and Doc Holliday’s at 9th Street — in New York City.
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