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KHALI
04-20-2005, 10:24 AM
Need a little help here from you guys with a hypothetical question.

Say 5 friends all play in a 10 tournament series. Each tournament has 200 players. All 200 players have equal ability. What is probability of any 2 of the 5 making a final table(top ten)? What is probability of any combination of 2 making 2 or more final tables?

How would you start to calculate this probablility?

BruceZ
04-20-2005, 01:20 PM
[ QUOTE ]
Need a little help here from you guys with a hypothetical question.

Say 5 friends all play in a 10 tournament series. Each tournament has 200 players. All 200 players have equal ability. What is probability of any 2 of the 5 making a final table(top ten)? What is probability of any combination of 2 making 2 or more final tables?

How would you start to calculate this probablility?

[/ QUOTE ]

There are 2 ways to interpret this question. Do you mean that 2 players make the same final table, or can they make 2 different final tables?

Greeksquared
04-20-2005, 01:30 PM
Intuitively each player has a 1/20 chance at getting to the final table. But since, they are all playing the same tournament the chances of a second player getting to a final table is less because one spot is already taken.

The chances of 0 getting to the final table is the chance that they all don't make the final table.
(19/20)(189/199)(188/198)(187/197)(186/196)= .7717

The chance of the team getting one person to the final table of the first tourney is just

(1/20)(190/199)(189/198)(188/197)(187/196) but there are 5 combinations of this so we multiply this by 5 and get .2031

Similarly

for 2 (1/20)(9/199)(190/198)(189/197)(188/196)*10= .0193
for 3 (1/20)(9/199)(8/198)(190/197)(189/196)*10= .00085
for 4 (1/20)(9/199)(8/198)(7/197)(190/196)*5= .0000157
for 5 (1/20)(9/199)(8/198)(7/197)(6/196)= .0000000993

To find the chance that there will be 2 of the friends at atleast 2 or more of the final tables we see that there is .0208 chance at 2 or more members getting to the final table.

So the chance that this happens 0 or 1 times out of ten is just a binomial distribution. P(0)=.8104, P(1)=.1721.

So the probability of this happening 2 or more times is 1-(P(0)+P(1))=.0175

Greeksquared
04-20-2005, 01:38 PM
You can also use the hypergeometric distribution to count the ways to reach a final table. Looks nicer than my original way.

For example getting three to the final table would be equal to

(10c3)*(190c2)/(200c5)

BruceZ
04-20-2005, 02:07 PM
[ QUOTE ]
Intuitively each player has a 1/20 chance at getting to the final table. But since, they are all playing the same tournament the chances of a second player getting to a final table is less because one spot is already taken.

The chances of 0 getting to the final table is the chance that they all don't make the final table.
(19/20)(189/199)(188/198)(187/197)(186/196)= .7717

The chance of the team getting one person to the final table of the first tourney is just

(1/20)(190/199)(189/198)(188/197)(187/196) but there are 5 combinations of this so we multiply this by 5 and get .2031

Similarly

for 2 (1/20)(9/199)(190/198)(189/197)(188/196)*10= .0193
for 3 (1/20)(9/199)(8/198)(190/197)(189/196)*10= .00085
for 4 (1/20)(9/199)(8/198)(7/197)(190/196)*5= .0000157
for 5 (1/20)(9/199)(8/198)(7/197)(6/196)= .0000000993

To find the chance that there will be 2 of the friends at atleast 2 or more of the final tables we see that there is .0208 chance at 2 or more members getting to the final table.

So the chance that this happens 0 or 1 times out of ten is just a binomial distribution. P(0)=.8104, P(1)=.1721.

So the probability of this happening 2 or more times is 1-(P(0)+P(1))=.0175

[/ QUOTE ]

You are interpreting the problem as asking for the probability of 2 or more friends facing each other at the same final table. I agree with your numbers. Here is my solution for that interpretation:

Probability of 2 or more friends making the final table at a particular tournament:

P(2) = 1 - [C(195,10) + C(5,1)*C(195,9)] / C(200,10) = 2.08%

The value in brackets is the probability of 0 or 1 of the friends making the final table, so 1 minus this is the probability of 2 or more making it. Now the probability that this happens at least once in 10 tournaments is 1 minus the probabilty that it does not occur for 10 tournaments, which is:

1 - [1 - P(2)]^10 = 19.0%

The probability that this happens in 2 or more tournaments is 1 minus the probabilty that it happens 0 or 1 time. This is:

1 - [ (1-P(2))^10 + 10*P(2)*(1-P(2))^9 ] = 1.75%.

KHALI
04-20-2005, 02:53 PM
I was interested in the probability of any two of the players making any final table out of the ten opportunities. It could be player A in week 2 and player C in week 7.

BruceZ
04-25-2005, 09:09 AM
[ QUOTE ]
I was interested in the probability of any two of the players making any final table out of the ten opportunities. It could be player A in week 2 and player C in week 7.

[/ QUOTE ]

OK, this is a different question from the one we solved previously. It is extremely complicated to compute the exact answer because the result for each player slightly changes the probabilities for the other players, i.e., the friends are not independent. However, if we make the simplifying assumption that the friends are independent, the calculation becomes quite simple, and it turns out that this gives a result within 0.2% of the exact answer. I know this because I performed the exact calculation first, before doing the approximate solution as a check. I will show both solutions here.

First we’ll do the exact solution. We can take 1 minus the probability of no player placing, minus the probability of exactly 1 player placing. Since 1 player can place in anywhere from 1 to 10 tournaments, we must sum the 10 probabilities corresponding to each of these cases. These will be for one specific player placing, and the other four losing all 10 tournaments, so we will multiply these 10 probabilities by 5 since any of the friends can be the winner. Note that the tournaments are independent, but as we consider each friend in turn, the probability for each friend is dependent on the other friends, and this dependence greatly complicates the terms.

1 -

(190/200)^10*(189/199)^10*(188/198)^10*(187/197)^10*(186/196)^10 -

5*[
10*(10/200)*(190/200)^9*(190/199)*(189/199)^9*(189/198)*(188/198)^9*(188/197)*(187/197)^9*(187/196)*(186/196)^9 +

C(10,2)*(10/200)^2*(190/200)^8*(190/199)^2*(189/199)^8*(189/198)^2*(188/198)^8*(188/197)^2*(187/197)^8*(187/196)^2*(186/196)^8 +

C(10,3)*(10/200)^3*(190/200)^7*(190/199)^3*(189/199)^7*(189/198)^3*(188/198)^7*(188/197)^3*(187/197)^7*(187/196)^3*(186/196)^7 +

C(10,4)*(10/200)^4*(190/200)^6*(190/199)^4*(189/199)^6*(189/198)^4*(188/198)^6*(188/197)^4*(187/197)^6*(187/196)^4*(186/196)^6 +

C(10,5)*(10/200)^5*(190/200)^5*(190/199)^5*(189/199)^5*(189/198)^5*(188/198)^5*(188/197)^5*(187/197)^5*(187/196)^5*(186/196)^5 +

C(10,6)*(10/200)^6*(190/200)^4*(190/199)^6*(189/199)^4*(189/198)^6*(188/198)^4*(188/197)^6*(187/197)^4*(187/196)^6*(186/196)^4 ] +

C(10,7)*(10/200)^7*(190/200)^3*(190/199)^7*(189/199)^3*(189/198)^7*(188/198)^3*(188/197)^7*(187/197)^3*(187/196)^7*(186/196)^3 +

C(10,8)*(10/200)^8*(190/200)^2*(190/199)^8*(189/199)^2*(189/198)^8*(188/198)^2*(188/197)^8*(187/197)^2*(187/196)^8*(186/196)^2 +

C(10,9)*(10/200)^9*(190/200)*(190/199)^9*(189/199)*(189/198)^9*(188/198)*(188/197)^9*(187/197)*(187/196)^9*(186/196) +

C(10,10)*(10/200)^10*(190/199)^10*(189/198)^10*(188/197)^10*(187/196)^10
]

=~ 66.7%

Now let’s make the approximation that the friends are independent. This allows us to replace all of the above mess with just the following which gives almost the same result:

1 – [(19/20)^10]^5 - 5*[1 - (19/20)^10]*[(19/20)^10]^4

= ~ 66.5%.

[ QUOTE ]
What is probability of any combination of 2 making 2 or more final tables?

[/ QUOTE ]

For this case, we must subtract from the previous result the probability of 1 player placing in 1 or more tournaments, while another player places in exactly 1 tournament. The first player can place in anywhere from 1 to 10 tournaments. Also, we must distinguish between the cases where the second player places in one of the same tournaments in which the first player places from the case where he places in a different tournament from one that the first player placed. This will require two separate sets of probabilities

This calculation is twice as complicated as the previous one. If you are interested in seeing these equations, PM me, and I can email you the Excel spreadsheet. The result came to 33.4%.

Again, we can check this by assuming that the friends are independent, and again this gives almost the same result:

66.5% (from part 1) - C(5,2)*[10*(1/20)*(19/20)^9]^2*[(19/20)^10]^3 –

5*4*[1 - (19/20)^10 - 10*(1/20)*(19/20)^9]*[10*(1/20)*(19/20)^9]*[(19/20)^10]^3

=~ 33.6%.