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marbles
10-31-2002, 03:44 PM
Scenario: One table tournaments, buy-in is \$11 (\$1 for the house). Payouts are \$50 for 1st, \$30 for 2nd, \$20 for 3rd.

Assumption: Hero will finish in 3rd twice as often as 2nd, and finish in 2nd twice as often as 1st. This is based on a strategy that rarely puts the hero in strong chip position once all remaining players are in the money.

Question: Bankroll issues aside, how often would our hero have to finish in the money in order to break even LR? I've been getting 40.5%, which seems awfully low to me. Here is my calc:
p=probability of 3rd place.
E.V.=0=-11+20(p)+30(.5)(p)+50(.25)(p)
p=.232
prob (in the money)=p+.5p+.25p=.232+.116+.058=.405 (rounded)

If this is right, wouldn't a player finishing ITM 50% of the time have a huge +EV?

lorinda
10-31-2002, 06:35 PM
When our hero finishes in the money he recieves:
4*20 + 2*30 + 1*50 / 7 =
80+60+50 / 7 =
\$27.143

his outlay is \$11, so to be break even, he needs odds of 27.143:11 or 2.468-1
or 40.53%

If this is right, wouldn't a player finishing ITM 50% of the time have a huge +EV?

This is correct. The tournament forum often quotes (and my figures fall into the right ball park) 50% ROI (Return on Investment) for a strong player in these tournaments, which is unsurprising considering you are only being raked \$1 for the play of 80 hands or so when you win.

In the \$11 tourneys a strike rate of 50% is not uncommon amongst pro players, which is why tourney players can afford to play at lower limits than ring players.

Good Luck.

Lori

marbles
10-31-2002, 07:07 PM
Thanks for the confirmation. So far, I've played in 11 pay tourneys, with one win, one place, and two shows (ITM 36.3%). Now, of course those numbers aren't statistically significant, but if 40.5% is the break-even bar, I'll take my chances!

Bozeman
11-01-2002, 10:58 PM
BTW, Marbles, I think that a strategy that places 1st 1/4 of the time that it places 3rd is profoundly suboptimal. You may reduce your fluctuations, but you are missing out on a lot of payoffs. Experimentally, my finish probability is a monotonically decreasing function of finish position.

Good Luck,
Craig