Math q.

[PartyGirlUK]

2 people are online and playing $200 SnGs at the same time. During this time 100 $200 SnGs run. Each has exactly 10 players. Player 1 plays 20 tournaments, player 2 plays 15 tournaments. I'm trying to find out the likelihood of them playing in a tournament together so I can do a binomial test.

For simplicity I've been working out the probability of them playing in a tournament together at (20/100) * (15/100), or .03

So say they actually played 6 tournaments together, the bitest would be

observations - 100
p - .03
outcomes - 6

Now this is flawed, for reasons anyone smart enough to give me the correct answer will see immediately. What is the best way to fix it so that I get a more accurate value of p?

Thanks,

Dean

[evank15]

Title should be "stats q", not "math q".

[almostbusto]

as you probably already realize the trials aren't IID so to get the actually answer would be pretty tricky (i think)

personally i can't think of a better way to go about it than what the OP suggests.

[Siegmund]

The value of p is correct; the problem, as already noted, is that the 100 SnGs aren't independent. You can set it up as you have proposed and see how things look; in the real world, the standard deviation will be a bit higher than the independent model suggests.

[thylacine]

[ QUOTE ]
2 people are online and playing $200 SnGs at the same time. During this time 100 $200 SnGs run. Each has exactly 10 players. Player 1 plays 20 tournaments, player 2 plays 15 tournaments. I'm trying to find out the likelihood of them playing in a tournament together so I can do a binomial test.

For simplicity I've been working out the probability of them playing in a tournament together at (20/100) * (15/100), or .03

So say they actually played 6 tournaments together, the bitest would be

observations - 100
p - .03
outcomes - 6

Now this is flawed, for reasons anyone smart enough to give me the correct answer will see immediately. What is the best way to fix it so that I get a more accurate value of p?

Thanks,

Dean

[/ QUOTE ]

Probability of them playing in exactly k tournaments together is

C(k,20)C(15-k,80)/C(15,100)

or eqivalently

C(k,15)C(20-k,85)/C(20,100)

or something like that.

[PartyGirlUK]

[ QUOTE ]
The value of p is correct; the problem, as already noted, is that the 100 SnGs aren't independent. You can set it up as you have proposed and see how things look; in the real world, the standard deviation will be a bit higher than the independent model suggests.

[/ QUOTE ]

Is the value of p correct? To give a simple example:

assume there are 2 tournaments going and we look at 2 players, A and B. How likely is it that they play together? Is it .5? or is it 9/19?

[51cards]

It strikes me as odd, but def 9/19.