Expert: Paul Klarreich - 5/16/2008

Question
When Susan and Jessica play a card game, Susan wins 60% of the time.  If they play 9 games, what is the probability that Jessica will have won more games than Susan?

Answer
Questioner:   Wendy
Category:  Advanced Math
Private:  No
 
Subject:  Probability
Question:  When Susan and Jessica play a card game, Susan wins 60% of the time.  If they play 9 games, what is the probability that Jessica will have won more games than Susan?
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Hi, Wendy,

In general, if  w = p(Wendy wins a game), and
j = (1 - w) = p(Jessica), and  n  games are played, we have:

p(Wendy wins exactly k games) =  C(n,k) w^k (1- w)^(n-k)

Naturally,

SUM[k=0 to n] C(n,k) w^k (1- w)^(n-k) = (w + (1-w))^n = 1

In this case, n = 9, and you want Wendy to win 5 to 9 games:

SUM[k=5 to 9] C(9,k) w^k (1- w)^(9-k)


C(9,5) w^5 (1- w)^(4)
C(9,6) w^6 (1- w)^(3)
C(9,7) w^7 (1- w)^(2)
C(9,8) w^8 (1- w)^(1)
C(9,9) w^9 (1- w)^(0)

And, since  w = 3/5, 1-w = 2/5, that comes to

C(9,5)(3/5)^5 (2/5)^4 + C(9,6)(3/5)^6 (2/5)^3 + C(9,7)(3/5)^7 (2/5)^2 + C(9,8)(3/5)^8 (2/5)^1 + C(9,9)(3/5)^9 (2/5)^0 =

[ C(9,5)(3)^5 (2)^4 + C(9,6)(3)^6 (2)^3 + C(9,7)(3)^7 (2)^2 + C(9,8)(3)^8 (2)^1 + C(9,9)(3)^9 (2)^0] over 5^9

which you can work out now.  
Excel is a good tool for the calculation.  Mine looks like this:


k   C(9,k)   3^k*2^(9-k)   p[k]
5   126   3888   0.250822656
6   84   5832   0.250822656
7   36   8748   0.161243136
8   9   13122   0.060466176
9   1   19683   0.010077696

Total =  0.73343232, which is your answer.  Interesting that the first two entries in the p[k] column are identical.  See if you can prove that.

BTW, look for a recent answer in this category.