Can someone please check my work?
 

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?

Re: Can someone please check my work?
 

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

Re: Can someone please check my work?
 

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!

Re: Can someone please check my work?
 

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