Hard Probability Problem

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.

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?

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).

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

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

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

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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.

GA

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

Thanks guys!