If you play around with a commonly-referenced MTT simulator (this one: https://www.primedope.com/tournament...ce-calculator/) by simming just one event a bunch of times (say 1,000,000 times), you'll generally see the simulator producing the following results:

Let N=# of entrants

C=# of players cashing

1/N=Random-chance win probability (prob of winning if all players have 0% ROI)

C/N=Random-chance cashing probability (prob of cashing if all players have 0% ROI)

R=ROI


The simulation results consistently imply that for a given R, that player will cash at a rate of (1+R)*C/N and will win at a rate of (1+R)/N

I'm trying to explicitly do the EV calculation to prove those results to myself but I haven't been able to match those results.

Is the following an accurate statement? If so, can anyone offer a proof?

For a given ROI R and a given field size of N, and a given number of places paid C, a good estimate of that player's winning percentage is (1+R)/N a good estimate of that player's cashing percentage is (1+R)*C/N

The calculations you set out will only be accurate if the player has an equiprobable distribution of cashed places. Say, if there are 10 places in the money, the player would have to on average finish as frequently in first place as in the following 9 places. I'm also assuming that an assumption is that the ROI doesn't include fees paid to site/casino, as otherwise, 1/N and C/N would not be an accurate measure of the probability of winning if all players have 0 ROI.

In effect, what the calculation is doing is assuming that the ROI is the probability of the player finishing in the cashing places (again, in an equiprobable distribution over all cashing places) compared to the non-cashing places. So, for a 10% ROI in a 100 person freezeout with 10 places paid, it means that instead of having an equal probability of finishing in each possible finishing place, you have a 10% higher probability of finishing in each individual place 1-10 compared to finishing in each individual place 11-100.

I haven't really thought this through, so there may be aspects which aren't completely accurate, but hopefully that should be the jist of it.

I understand the equiprobable part

in practice, are marginal gains/losses over the random-chance in probability of finishing in i-th place unevely distributed across different values of i?

Compared to that sim, in practice for a given ROI, players will cash less often but finish in the higher paying positions more often. Basically less cashes, more first place finishes. The end result is the same ROI. The difference will be that the swings in reality will be bigger than what that sim suggests.

Why would the aggregate results distribution have heavier tails than a normal distribution? Some sort of compounding effect whereby skill differentials become more pronounced as the event approaches the endgame?

Shouldn't the central limit theorem come into play once you get a sufficient sample?

Isn't it reasonable to assume that over a really long sample the results will be normally distributed like the simulator is doing?

Never gave it much thought tbh. It probably has to do with the psychology of weaker players around money jumps and better players generally playing more so for the top spots rather than just cashing. Could also be just a sampling bias whereby the biggest winners will usually also be the luckiest. But generally if you look at a solid winning regs they will have a skewed finish distribution.

I fooled around a bit with this and can suggest a math “proof” for the cashing percentage.

ROI is defined as Profit/Investment. Consider a tourney with a total buy-in of B consisting of the amount for the prize pool, P, plus an entry fee. N players enter and C cash.

Then the ROI is as follows:

ROI =(Pc*(NP/C)-B)/B,

where NP/C is the average return of a cashing player given no other data.

After a little algebra ,

Pc= (1+ROI)*CB/NP [ ~ (1+ROI)*C/N (fee close to 0)]

This reduces to the stated approximation if the entry fee is close to 0. Why the N and C factors are used but not B and P is something I can’t explain other than for simplification.

For a randomly sampled player, it should, but with different skill levels and different levels of risk aversion, you would expect the distribution of paid places to be skewed towards 1st places for some players, and towards Min cashes for others.

Is there any empirical research about just how much a really good, risk-seeking, winning player is going to improve their chances of winning a given event beyond the random-chance probability? Like can we say that a 30-40% ROI crusher is generally giving themselves a 2/N or 3/N or even better likelihood of winning an event with N entries (where N is large)?

Would rebuys magnify this effect (because then the strategy of infinite rebuys allows a very risk-seeking player to flex that behavior to the max)? How would rebuys affect the skewness and kurtosis of the results distribution?


Is it possible to be a really good yet relatively risk-averse player who also achieves a 30-40% ROI by trying to maximize the probability of merely cashing? Or should we expect only those players who are really good and risk-seeking of being able to achieve such high ROIs?

I don't know if there has been any empirical research on this. It may have been feasible back in the days when you couldn't opt out of sharkscope, but nowadays, any such research would be a little biased, at least when using that data.

I'm not too sure what impact having rebuys would have. In terms of the risk aversion, an optimally good player will be risk neutral, so the level of risk involved in the decision should be irrelevant, only the EV should be a factor. Trying to maximise the probability of cashing is by definition a risk averse strategy, which will not maximise EV. Of course, it will be possible for such a strategy to have a high ROI, but it will have a lower expected ROI than a risk neutral strategy. However, it will have a lower variance/standard deviation.