The Giants and Royals are playing in the World Series. In the World Series, the two teams play repeatedly until one team has won 4 games. Suppose each team has a 1/2 probability of winning each game. If the World Series lasts exactly 6 games, what is the probability the Giants win?

supermanaccz  Oct 17, 2017

The Giants and the Royals each have a 50% or 1/2 chance of winning each game. To win the series you must win 4 games out of 7. There are 2 possible outcomes for each of the individual games. 2*2*2*2*2*2*2= 128 possible outcomes.


     This may be wrong because you said 4 out of 6 games, but then it is possible for a 3-3 tie to appear, so I solved the problem with the possible games played being 7.

Mr.Owl  Oct 17, 2017

I don't understand the question.

the series last 6 games is that whether or not anyone has won yet?

So maybe a team won or maybe no-one won?

If a team did win then the chance of it being either team is 50%. I mean 50% each.........


I guess you must mean that maybe no team won?

Melody  Oct 18, 2017

My understanding of the problem is, even though the series could last 7 games, in this case it lasted only 6 because one of the teams achieved its fourth win in the sixth game.

Since each team has a 1/2 probability of winning each game, each team will have a 1/2 probability of winning the series (this does not depend upon how many games the series actually lasts).

So, the probability that the Giants win is 0.50  (one-half).

geno3141  Oct 18, 2017

Hi Geno

It is nice to see you.

That is what I thought but it could be that the 6 was meant to be a 7 in which case it would still be half.

or that we were suppose to include the possibility of a tie, whch would make the problem a tad harder.



Say we were meant to include the possibility of a tie.

Then the outcomes culd be 2:4,   4:2,   or  3:3

I do not think it could be 1:5 or 0:6 because then the series would have finished earlier.


The number of ways a team can win 2 games in 6 is 6C2  =  15

The number of ways a team can win 3 games in 6 is 6C3 =  20

The number of ways a team can win 4 games in 6 is 6C4 = 15


So the probability that they score 3 games each is      20/50 = 4/10

The remaining possibilities are split equally between A and B wining    0.5(1-0.4) = 0.3


So if they play 6 games the probabilities are 

P(tie) = 0.4

P(A wins) = P(Bwins) = 0.3

Melody  Oct 18, 2017

Here's my take on this one :


[ There aren't any tie games possible in the World Series  ]


We play 6 games  and we want the Giants  to win any 4 of them


So we have


C(6,4) (1/2)^6  =  15 / 64


Of course.......this is the same probability that the Royals win in 6 games, as well



cool cool cool

CPhill  Oct 18, 2017

This is an AOPS question. This presentation is an adaptation of Lancelot Link’s solution for a closely related question.


This is question has a subtle contingency and requires two probability equations to solve.


For the Giants to win the series in game six, they need to win three games in five trials. This is a binomial distribution.


\(P(G\text{ win series in 6 games}) = \\ P(G \text{ wins 3 in 5})*P(G\text{ wins 6th game}) =\\ \binom{5}{3}(0.5)^3(0.5)^2 *(0 .5) = 0.15625\)


Further ... For the series to end in six games, either the Giants or the Royals must win the series in six games.


\(P(\text{Series ends on game 6})=P(G\text{ wins series on game 6})+P(R\text{ wins series on game 6}) \\ = 2P(G\text{ wins series on game 6}) \\ = (2)\binom{5}{3}(0.5)^3(0.5)^2 * (0.5) \\= \frac{5}{16}\)


The probability the Giants win the series in game 6  is 5/16.





Edit: changed ASOP (the fabulist) to AOPS, (the fabled academy of higher mathematics).

GingerAle  Oct 18, 2017
edited by GingerAle  Oct 18, 2017

Thanks guys!

supermanaccz  Oct 21, 2017

17 Online Users

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.