[BruceZ]

[ 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%.