Hi all,

I’m a probability n00b so I wanted to check if I approached this calculation correctly.

I’m trying to estimate the odds of direct bubbling 3 tournaments in a row, with 54, 18, and 78 players respectively.

My lifetime ITM% for live tournaments is 25%. Let’s assume this to be accurate.

Now let’s assume that my chances of bubbling are the same as my chances of cashing (of course my chances of winning and min-cashing and bubbling are not the same, but for simplicity’s sake let’s assume they are, especially since my chances of bubbling are actually LESS than my chances of cashing, so the final result errs on the side of higher probability.)

Well, if each tournament were completely random, the odds would be 1/54 * 1/18 * 1/78 = 0.0000132. But my chances to finish in that direct spot are higher since my ITM% is higher than random. If we assume that each tournament pays out 10%, and my ITM% is 25% (including bubble), then my chance to place in exactly one of the ITM seats is .25/.1 = 2.5 times higher than random.

(2.5/54) * (2.5/18) * (2.5/78) = .00021, or roughly 1 out of 5,000.

Correct?

I'm going to say that 6 people make the money in the 54 person tournament, 2 people in the 18 person tournament, and 8 people in the 78 person tournament. If these are not correct, this analysis won't work.

That leave us with 48, 16, and 70 OOTM positions respectively. Let's assume that if you finish OOTM there is an equal chance of landing in each of those positions. So that is 1/48, 1/16, and 1/70 for each of the tournaments. If we weight those each by a 75% chance that your finish OOTM we get:

1/48*.75, 1/16*.75, and 1/70*.75

If we multiply all three of those by each other to get the combined probability we get 0.00000784738. That is the probability of finishing in the exact bubble seat of three tournaments in a row (if announced in advance and this is the only three tournaments you play and all of our assumptions are met).

However, the question you asked in the post does not seem to be the question you asked in the subject line...so I'm not really sure if that is what you wanted to know or not.

edit: This isn't right. The total probabilities do not equal 1.0 with this method. I'm doing something wrong.


Nevermind. I think I do have it right.

You have a 75% chance of not making the money and there are 48, 16, and 70 OOTM positions respectively. If they are all equally likely it is simply .75 / 48, .75 / 16, and .75 / 70. Then we multiply those three numbers to get the chances of exact bubbling all three which is the same as above.

i think it's better to assume that your chance of bubbling is closer to the average chance of finishing in a pay spot, rather than in a non-pay spot, since the bubble position is closer to the average pay spot than the average non-pay spot. so i would tend to prefer OP's calculation.

on OPR you can get your total finishes broken down into different percentile groups, so that might help you get a more accurate estimate of your bubble %.

I think the probability of bubbling has far too much correlation with specific play style to conform with an assumption.

One could argue:
> Poor players bubble infrequently by not getting there.
> Medium players bubble very infrequently, by folding to the money once they get there
> Good players bubble frequently (i.e. more often than they make last money spot) because good play is often to be aggro on the bubble
> etc

If you are a 25% ITM player, then I think you fall into Medium / Good category, but the best way to validate the assumption is this

"OP - how do you play on the bubble? Do you fold to get ITM?"

If the answer to this is yes, then I bet that drops the total independant probabilty by at least 2 OOMs.

I completely agree. The assumptions used to do the analysis above are certainly incorrect. And their degree of incorrectness is related to the degree to which such numbers can be interpreted.

For what it is worth, I also agree with DarkMagus argument. Getting ITM 25% of the time probably means there is not an equal probability of falling in each of the OOTM positions as my analysis assumed.