Tournament Position

[Cobra]

I am not sure how to figure this out. Any help would be appreciated.

Lets say you play in a monthly tournament of NL hold-em with 26 players. You finish in the following postition on the first eight tournaments.

2,2,9,26,9,2,16,4th.

What is the probability the player would come in 9th or better in at least one of the next two tournaments.

I would like to see the work as well so I can figure other examples.

Thank you,

Cobra

[SumZero]

[ QUOTE ]

Lets say you play in a monthly tournament of NL hold-em with 26 players. You finish in the following postition on the first eight tournaments.

2,2,9,26,9,2,16,4th.

What is the probability the player would come in 9th or better in at least one of the next two tournaments.

[/ QUOTE ]

What are your assumptions? If you assume that all the players are equally good then there is a ~57.2% chance that he finishes in the top 9 in at least one of the next two tournaments. If you take his previous results and instead say he has 3/4 chance on each tournament of coming in the top 9 then the chance that he does goes up to 93.75%. He has a 56.25% chance of coming in the top 9 in both the next tournaments, roughly the same as the chance the "average" player has of coming in the top 9 in at least one of the next two tournaments.

More generally if you want to know the chances of an event with probability p occuring r times in n chances then the formula is:

Combin(n,r)*p^r*(1-p)^(n-r)

where Combin(n,r) = n!/((n-r)!*r!)

When the question is at least k times then you either count all the possibilities or count the compliment and subtract from 1. For instance you can do:

Sum(from i=k to n) {Combin(n,i)*p^i*(1-p)^(n-i)}

or

1 - Sum(from i=0 to k-1) {Combin(n,i)*p^i*(1-p)^(n-i)}

In your example n=2, k=1, and p is the only unknown that varries based on what your assumptions are.

The at least once problems are pretty much always easiest to solve as:

1 - (1-p)^n (the other factors are 1 so drop out) so here 1 - (1-p)^2 is your answer.

[Cobra]

Thank you for the response.

Cobra

[AaronBrown]

[ QUOTE ]
If you assume that all the players are equally good then there is a ~57.2% chance that he finishes in the top 9 in at least one of the next two tournaments. If you take his previous results and instead say he has 3/4 chance on each tournament of coming in the top 9 then the chance that he does goes up to 93.75%.

[/ QUOTE ]
There is another natural assumption that gives an intermediate answer. We know that the probabilities of finishing in variouis positions must add up to 1. It seems reasonable that there is some most likely position, and that the probability falls away smoothly as you go to higher or lower positions. One way to model that is to pretend you are flipping biased coins, and your position is equal to 1 plus the number of heads you get out of 25 flips. A very good player would have a low probability of getting heads, a very bad player would have a high probability.

I want to emphasize that there's no reason this model would be true, it just gives a reasonable way to estimate the chance of finishing in the top 9, using all the information from the past results. Just counting how many are in the top 9 is a more reliable method, but it throws away a lot of information (finishing 1st versus 9th makes no difference to this estimate, nor does finishing 10th versus 26th).

This model suggests the player has a 0.6361 chance of finishing in the top 9 in one tournament, so a 0.8676 chance fo finishing in the top 9 in at least one of the next two tournaments.

[Cobra]

AaronBrown,

I don't know if your calculations are complicated but could you posibbly show how you got your answers. I do not understand it from your description of your technique.

Thanks again,

Cobra

[AaronBrown]

It's not complicated, but let me emphasize again that it's just an ad hoc way to estimate the answer, not a mathematically rigorous approach.

We start by guessing the shape of the distribution of probabilities for different position finishes. It woudn't make much sense, for instance, if someone had a 10% chance of being first, 2% chance of being second and 8% chance of being third. We would expect the probabilities to change smoothly from position to position, with a single peak at the most likely outcome. For example, a person who was likely to finish 5th might have a 10% chance of being 5th, 8% of 4th or 6th, 6% of 3rd or 7th, with decreasing probabilities as you got further away from fifth.

There are lots of distributions with this general shape, but a simple one is to say the probability of finishing n-th is the probability of getting n-1 heads when you flip a biased coin 25 times. If the probability of getting heads is p, this will peak for position 25p + 1, will fall off to either side, and the total probabilities add to 1. A person with p = 0 will always finish 1st. A person with p = 1 will always finish 26th. A person with p = 0.4 will finish 11th on average, and will get that position about 1 time in 6.

The advantage to making this assumption is we can estimate the most likely p given all the tournament finishes, then we can use this p to compute the probability of finishing in the top 9. It's not guaranteed to work, but it gives pretty reasonable values in this case.

The average of the 8 finishes was 8.75, implying p = 7.75/25 = 0.31. If you flip a coin 25 times with probability 0.31 of getting heads, you'll get 8 or fewer heads with probability 0.6361.

[Cobra]

Very Cool. Thanks for the information.

Cobra