I would like to confirm my answer for this question with stack.

The Chicago and the New York play a best-of-seven series:
These two basketball teams play against each other until 
one of them has won four games.
Assume that 1 in any game, Chicago has a probability of 
75% chance of defeating New York and the results of the 
games are independent.
Determine the probability that seven games are played in 
this series

$Pr($Chicago Win$) = 0.75$

$Pr($NY Wins$) = 0.25$

To play all seven games, we must consider two cases:

Case 1: $Pr($Chicago wins out of 7 games$)$

$Pr($Chicago wins 4) = $0.75^4$ and $Pr$(NY wins 3) = $0.25^3$

Case 2: $Pr$(NY wins out of seven games)

$Pr$(Chicago wins 3) = $0.75^3$ and $Pr$(NY wins 4) = $0.25^4$

$Pr$(plays seven games) = $Pr($Chicago wins 4) + $Pr$(Chicago wins 4)

=$\frac{81}{16384} + \frac{27}{16384}$ =0.00659

Is this correct?


1 Answer 1


The solution is not correct. It begins well, our probability is the probability Chicago wins in $7$, plus the probability New York wins in $7$.

We find the probability Chicago wins in $7$, and leave the rest to you. But in fact there is a simpler way.

Chicago wins in $7$ if (i) the teams split the first $6$ games and (ii) Chicago wins the $7$-th.

The probability of a $3$-$3$ split in the first $6$ games is $\binom{6}{3}(0.75)^3(0.25)^3$. So the probability of a Chicago win in $7$ is $$\binom{6}{3}(0.75)^4(0.25)^3.$$

Remark: We can compute less. The series lasts $7$ games precisely if the teams split the first $6$. This has probability $\binom{6}{3}(0.75)^3(0.25)^3$.

  • $\begingroup$ Oh I like this. This didn't occur to me, I was setting up a recurrence :p $\endgroup$ Mar 25, 2014 at 6:04
  • $\begingroup$ So comparing my solution, I conclude my error is that: I have to consider the order of game wins in the 3-3 game split, which rules out the possibility of Chicago or NY winning 4 out of 6 games (since games stop there). $\endgroup$
    – GivenPie
    Mar 25, 2014 at 19:32
  • $\begingroup$ A product like $(0.25)(0.75)(0.75)(0.75)(0.25)(0.25)(0.75)$ gives the probability of NCCCNNC wins in that order. The binomial coefficients in my first answer allow us to take account of all relevant orders of wins. And as you point out, we have to be careful not to count the cases where the series ends "early." $\endgroup$ Mar 25, 2014 at 20:18

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