Bounty Hunter

Tonight I shall make another attempt at winning my first tournament at The International Club. I only had to wait one event to score a win at the old Gutshot Club, but at this venue I have final tabled seven times without taking the last, vital step.

It was (and is) one of my poker targets to score a win this year and - in reality - I have less than a handful of attempts left (not counting thumbs or little fingers, either).

I plan to play the £300 at the end of the month, and tonight’s tournament, but that’s probably about it.

We’re going away early for Christmas so I will miss GOSCARs night, and I have never dared to leave the family behind to play the boxing day event !

Sometimes, such pressure has translated into me playing my A-game. With my A-game and a few H-bombs, I always stand a chance.

On the other hand, tonight is a £25 freezeout with a fast structure so although that sort of structure suits me, it is of course a bit of a luckament.

The added sparkle tonight is the bounty element being thrown in by Gutshot.com. $100 each for knocking out any of the 4 named Gutshot pros is a nice freebie, but they’ve also decided to throw in some serious cash in case anyone manages a multiple knockout.

$1,000 for getting two of the blighters. $10k for a hattrick and $100k for the holy grail of a grand slam of knockouts !

Like finding the holy grail, it is widely assumed that this feat is impossible. To be sure, it is so unlikely that it might as well be impossible, but it makes for a great headline !

I was wondering how impossible it is. Difficult to model realistically, but here’s my first and only attempt at it:

So ....... someone has to win the tournament and I will take it as a given that this person will have knocked out four or more players on the way.

This is not certain to be the case of course; theoretically the winner may knock out only the runner-up, but I have to start somewhere. I will further simplify by assuming they knock out exactly four players.

So, assuming a nice amount of randomness and independence and all that good stuff (definitely not true; there are some clear dependencies, but bear with me) then what is the chance that those four are in fact exactly the fantastic four?

This is a bit like playing the tournament with four bounties but only revealing the names upon the opening of a sealed envelope after the event.

Seems to me that this will be of the right order of magnitude.

There will be at least 100 runners tonight, so I will take 100 as a nice round number.

How many combinations of 4 from 100 are there ?

It’s 100!/(4!x96!) I believe, which is about four million to 1 against. It’s got that “million to one shot” about it, so I think it’s a very suitable answer.

The chances of me winning the tournament are definitely better than a million-to-one, so that’s something to feel good about.


Something else that is better than a million-to-1 is the chance of me picking out the winner of the WSOP main event in a sweepstake that is running on the Gutshot forum.

I don’t know who my "pick" is yet, but I think it’s a fair bet that most people reading this would love to see James Akenhead take the big one down. Good luck James !

If best wishes were Aces, then I think he’s due a sick run of cards.

I have a clear hierarchy of favourites (in the sense of who I’d like to see win, not who I think is most likely to win). If it isn’t to be James then I’ll be supporting Steve Begleiter, as we worked together for many years at Bear, Stearns.

I don’t feel any link to any of the other players, so failing these two, I am on the Phil Ivey bandwagon. I support the view that Ivey for WSOP champion is good for poker.

However it finishes, it promises to be a fascinating final table. A whole bunch of people from the club are in Vegas to see it, I know. If James is out first hand, I'm sure they will find something to do !