A minor debate in this forums riggie thread did get me curious on the math of this situation.

Assume a player plays in 100 MTTs with an average of 6,000 players.

Assume that the odds of finishing in every position within a tournament are the same (obviously not the case in the real world).

What are the chances that a player will have 3 or more finishes within those 100 tournaments that are within 5 places of each other (ie: 55th 57th 59th).


Thanks in advance!

This isn't particularly simple. For now I'll give an upper bound.

There are:
6000 ways to have the same place 3 times.
2*5999 ways to have a range of 1.
3*5998 ways to have a range of 2.
4*5997 ways to have a range of 3.
5*5996 ways to have a range of 4.

The sum of all that is 89960

89960 * C(100,3) * (1/6000)^3 = 6.7345 %

That over-counts the ways you can be within the range >3 times, and also the ways you can have separate clusters (e.g. 1,2,3 and 7,8,9 are two separate clusters).

I made a simulation, obtaining about ~ 22%. Here is the sim in R:

And here are the results:

My first interpretation of OP was that something like 1,5,9 counted as "all within 5 places of each other". Nick, is that how you interpreted it? (I then realized OP probably meant the places had to be pairwise-within 5 of each other.)

Under that first interpretation, my upper bound would be 24.68% and the result of nick's sim would make sense.

But if nick's sim did constrain to "pairwise-within", that means one of us made a gross error somewhere (probably me).

Also, nick, can you explain the reasoning behind: You compare finish #1 to #3, then #2 to #4, then #3 to #5 and so on? But why?

Sorry if I was not clear. The person who was going on about it finished 10th 10th and 11th, but he would have similarly complained even if it was 10th 15th and 18th, so I arbitrarily chose 5 spots, meaning that all the finishes in the 3 examples would be within 5 spaces- so an 11th and 15th would count, but 11th 14th and 18th would not (as 18 vs 11 would be too far).

I suppose I would also ask what the odds would be for 3 finishes to be within 1 spot of each other (ie: 10 10 11 , not 10 11 12) just in case he decides to be very results oriented, which no doubt he would be in this case.

Thanks for all the help!

This is what i do in my simulation:

1) Extract 100 random numbers between 1 and 6000.
2) Sort those 100 numbers.
3) Take the differences with a 2-lag. That's the command that you outlined. Since the positions are sorted, in this way I can see if there are 3 tourneys within a 5 interval. So I compare 3rd best finish with the first, the 4th best with the 2nd best and so on. If any of these differences are lower than 5, this means that the condition in the OP is met.
4) Repeat the above nsim times.

As far as I understood, the three tourneys doesn't have to be consecutive; am I wrong? Do a 13th position in the first tournament, a 16th position in the 55th tournament and a 17th position in the 96th tournament for instance qualify? However, if that's not the case, It suffices to not sort the 100 numbers and check if the absolute value of the 2-lag differences is lower than 5.

The 3 tournaments do not have to be consecutive, they can take place at any point in the 100 tournament sample.

If you do not mind can you also do a quick run with an average field size of 10,000 and three tournaments falling within 1 spot of each other (ie: 9th 10th 10th count but 9th 10th and 11th do not count).

Thanks in advance!

Sure, I can add another argument (named Within) to my function that indicates the range we are interested in. Here is the code:

And the results for your case:

You can download and install R pretty easily from here. Then you can copy/paste the function I wrote on the R console and execute the threeFinishes function varying the parameters at your wish (as I did above).

This makes sense. Seems like I'm the one who did something wrong then. I thought 6.7% was too high an answer, but your sim says it's 22%.

I still don't see where I went wrong. I thought I over-counted the cases of multiple clusters, but maybe I under-counted them?

Oh wait, I see my mistake. It's not (1/6000)^3. The order doesn't matter.

If it's 3 different places, it's 3! * (1/6000)^3
If it's 2 different places, it's 3 * (1/6000)^3
If it's all the same place, it's (1/6000)^3.

In a sec I'll recalculate it.

Edit:

C(100,3) * (1/6000)^3 *
[
6000 +
3 * 2(5999) +
5998(2*3 + 3!) +
5997(2*3 + 2*3!) +
5996(2*3 + 3*3!)
]

= 27.3857 %

Now that makes sense as an upper bound. The part in brackets equals 365820, which is a little more than 4x as large as 89960.