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I've been doing some C & P problems and I got stuck on this one:

The Cubs are playing the Red Sox in the World Series. To win the world series, a team must win 4 games before the other team does. If the Cubs win each game with probability $\dfrac{3}{5}$ and there are no ties, what is the probability that the Cubs will win the World Series?

I came up with separate probabilities for each case (cubs win after 4 games, cubs win after 5 games...) and added them up: $$\binom30\left(\frac25\right)^0\left(\frac35\right)^4+\binom41\left(\frac25\right)^1\left(\frac35\right)^4+\binom52\left(\frac25\right)^2\left(\frac35\right)^4+\binom63\left(\frac25\right)^3\left(\frac35\right)^4$$ but it was wrong. What did I do wrong and how should this problem be done? Thanks.

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    $\begingroup$ Your formulas and final answer look right to me. $\endgroup$
    – Empy2
    Oct 5, 2015 at 0:07
  • $\begingroup$ Sorry, I made a mistake - Thanks. $\endgroup$ Oct 5, 2015 at 17:00

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Your formula and answer are right, but there is a much simpler way

Just compute bin$(7,3/5), P(4\le x\le 7)$

The logic is that games can at most extend to $7$, and once the Cubs have won $4$ games, it does not matter whether they win or lose the rest.

The answer I get is $\dfrac{11097}{15625}$, the same as yours.

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  • $\begingroup$ Wow that's much easier! Thanks a lot! So the answer is just C(7,3)*(2/5)^3*(3/5)^4 ? $\endgroup$ Oct 5, 2015 at 16:59
  • $\begingroup$ Not quite. It is $$\sum_{k=4}^7\dbinom{7}{k}0.6^k\cdot0.4^{7-k}$$ $\endgroup$ Oct 5, 2015 at 17:29
  • $\begingroup$ The gain is that it becomes a routine binomial distribution with fixed n. $\endgroup$ Oct 5, 2015 at 17:52
  • $\begingroup$ Oh okay. Thanks! $\endgroup$ Oct 5, 2015 at 18:53
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In the first term, the combinatorial coefficient should be $4\choose 0$ instead of $3\choose 0$, corresponding to the fact that we are computing the ways the Red Sox can win 0 out of the first 4 games. This is not important for the first term (both $4\choose 0$ and $3\choose 0$ equal 1) but it becomes crucial in subsequent terms: In the second term the $4\choose 1$ should be $5\choose 1$--we are now counting the ways in which the Red Sox can win 1 out of the first 5 games, etc.

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  • $\begingroup$ I was going to do that, but Red Sox can't win the last game, or else the series would have ended before that game, for example C(5,1) for the second term would imply that the cubs can win the first game and the rex sox can win the last game, which is false because then the series would have been only 4 games and the red sox couldn't have won the last game. Is there anything wrong with my logic? -thanks $\endgroup$ Oct 5, 2015 at 0:58
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    $\begingroup$ My bad...you are correct. Maybe your arithmetic is off? $\endgroup$ Oct 5, 2015 at 1:05

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