Poker Coverage: Poker Legislation Poker Tournaments U.S. Poker Markets Sports Betting

CPPT VII - DeepStack Championship Poker Series

$5,000 CPPT Main Event No-Limit Hold'em $2 Million GTD


Stephen Chidwick Headlines Day 4 Of Card Player Poker Tour Venetian Main Event

From a field of 564 entries, the Card Player Poker Tour Venetian main event is down to its final 12 players with one of the best players on the planet near the top of the ...

A Look At The Probability Of What Poker Pro Mark Newhouse Just Did At The WSOP

Odds Were About 1 In 5.3 Million, According To MIT Ph.D Student Will Ma


Even though he’s unfathomably disappointed, poker pro Mark Newhouse made poker history—and it’s because of the math.

Early this week, the 29-year-old finished ninth in the 2014 World Series of Poker main event just a year after finishing in the same spot at the 2013 November Nine final table. It was a feat that required incredible play, but what were the odds of back-to-back ninths?

If you assume everyone in the main event field is symmetric, then the odds are 1 divided by the field size in 2013 multiplied by 1 divided by the field size in 2014 (1/6352 * 1/6683).

According to that, you could expect to finish ninth in consecutive years once in about every 42.5 million times you decide to put up $10,000 to play the main event. That’s a lot of buy-ins.

However, the odds are clearly not that low considering Newhouse as a player, according to Will Ma, CardRunners Pro and Ph.D student in mathematics at MIT.

Here’s what Ma said:

If you assume Newhouse has a 200-percent ROI (return on investment) playing the WSOP main event, which is actually a reasonable ROI since the field is loaded with amateurs and Newhouse is a seasoned professional, the odds were about 1 in 3,100 to finish ninth once and around 1 in 9.6 million to come in ninth in back-to-back Novembers in Las Vegas.

Say you are Phil Ivey in the main event, and it’s not unreasonable to think you have an ROI as high as 400 percent. Ivey would have odds of roughly 1 in 1,550 to finish ninth once and odds of about 1 in 2.4 million to finish ninth in consecutive years.

It’s also worth considering the player’s approach to the final table. If you are playing fast and loose like Newhouse did, it’s more likely you would finish in ninth or first, rather than second or third. Ma’s “crude estimate” for Newhouse (someone with a 200-percent ROI) would be about 1 in 2,300 to finish ninth once and thus roughly 1 in 5.3 million to finish ninth back-to-back.

So, there you have it. Before the 2013 main event, Newhouse would finish in ninth that year and also in 2014 about every 5.3 million times he played the main event—assuming the event remained about the same in terms of field size and strength over all the eons! Let’s hope the main event can at least return to 2006’s level within the next million years.

“As you can see, boosting someone’s ROI increases their chances of getting ninth way more than just pegging them an ‘aggressive player at the final table,’” Ma remarked.

Newhouse obviously felt safe to Tweet the following back in July. He just got really, really unlucky (about as unlucky as one can get after winning more than $1.4 million).

Image via WSOP.

Tags: Mark Newhouse,   WSOP,   2014 WSOP


over 4 years ago

You could pick any random spot to finish in and the odds would be the same. But to just make the final table two years in a row is not nearly as improbable as some people think. Definitely not as unlikely as winning the lottery .


over 4 years ago

Not really true, depending on how you want to interpret or analyze the question. Let's say we're comparing the 2 scenarios over a 1 year period of time . Let's say a good player (who is +ev) plays back to back in the WSOP main event and each time there is 5,000 players. If everyone is the same, it would be 544-1 to make a final table, but let's say with his skill he's 250-1 to make a final table. Back to back final tables at those odds has a probability of aprox. 63,000-1. Now, there are so many different types of lotteries, but let's say in that same 1 yr. time span, a player plays a lottery with odds of 5 million to 1 , and plays 20 tickets twice a week for 1 year. His odds of hitting the lottery in that year is about 2,400-1. One can take any angle they want and make the numbers come out in infinite ways.


over 4 years ago

Congrats Mark on making the final table "AGAIN" this year job well done , So what are the odds that he will make it a third time. 100+million to 1 ? No, Its Mark against the number of players who enter the field in 2015, and the lottery, you have a 50/50 chance, you either win the main prize or you don't, simple.


over 4 years ago

If you are going to compare the probability of two events then it makes sense to compare them as isolated events. If someone says what is the probability of Mark Newhouse making back to back final tables vs. winning the lottery then that seems to me to imply buying only one ticket. The odds of hitting Powerball are over 175 million to one and Mega Millions over 250 million to one. Obviously you could have Newhouse buy 10,000 tickets or any number and that would change the answer.


over 4 years ago

stupid responses....