# chances of making it into top 10%

• Mar 11th 2010, 01:18 PM
doolindalton
chances of making it into top 10%
There is a poker tournament. There are 100 entrants and if you finish anywhere in top 10, you win something. Let's say all players who enter have an equal chance of winning something, which would be 10% in this example.

Now, let's say 4 friends (A,B,C,D) decide to enter the same tournament. What is the probability that at least one of the four friends will win something.

My answer was 32.77% and wanted to check with the forum. This probably involves knowing some kind of permutation or combination formula, which I can't recall after 20 yrs of being away from this stuff, so I did it by drawing out a matrix of all the possible combination of events that would qualify as "at least one of the four making it into top 10".

Something like,

[P(A) * !P(B) * !P(C) * !P(D)] +
[P(A) * P(B) * !P(C) * !P(D)] +
[P(A) * P(B) * P(C) * !P(D)] +
[P(A) * P(B) * P(C) * P(D)] +
... etc until I've covered all possibilities.

Where P(X) means person X will win something, or 10% and !P(X) means the opposite, or 90%.

Am I at least correct in the approach, if not the final answer of 32.77%
• Mar 11th 2010, 01:50 PM
mr fantastic
Quote:

Originally Posted by doolindalton
There is a poker tournament. There are 100 entrants and if you finish anywhere in top 10, you win something. Let's say all players who enter have an equal chance of winning something, which would be 10% in this example.

Now, let's say 4 friends (A,B,C,D) decide to enter the same tournament. What is the probability that at least one of the four friends will win something.

My answer was 32.77% and wanted to check with the forum. This probably involves knowing some kind of permutation or combination formula, which I can't recall after 20 yrs of being away from this stuff, so I did it by drawing out a matrix of all the possible combination of events that would qualify as "at least one of the four making it into top 10".

Something like,

[P(A) * !P(B) * !P(C) * !P(D)] +
[P(A) * P(B) * !P(C) * !P(D)] +
[P(A) * P(B) * P(C) * !P(D)] +
[P(A) * P(B) * P(C) * P(D)] +
... etc until I've covered all possibilities.

Where P(X) means person X will win something, or 10% and !P(X) means the opposite, or 90%.

Am I at least correct in the approach, if not the final answer of 32.77%

Let X be the random variable 'number of friends that win something'.
X ~ Binomial(n = 4, p = 0.1)
Calculate Pr(X > 0) = 1 - Pr(X = 0).
• Mar 11th 2010, 02:13 PM
doolindalton
I don't know what Binomial is, but you proposed a simpler way to calculate this, which is

1 - P(no one gets into top 10)

or

1 - (90% * 90% * 90% * 90%) = 34.39%

Is this correct?
• Mar 11th 2010, 02:21 PM
mr fantastic
Quote:

Originally Posted by doolindalton
I don't know what Binomial is, but you proposed a simpler way to calculate this, which is

1 - P(no one gets into top 10)

or

1 - (90% * 90% * 90% * 90%) = 34.39%

Is this correct?

Yes.