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Problem

In the World Series of baseball, two teams (call them A and B) play a sequence of games against each other, and the first team to win four games wins the series. Let p be the probability that A wins an individual game, and assume that the games are independent. What is the probability that team A wins the series?

There will be at maximum seven games played to decide a clear winner.

This is a practice problem from a course on Probability. Its solution provided in the course gives two approaches.

First approach:

The Game will stop as any one of team wins.

$$ P(A) = P(\text{A winning in 4 games}) + P(\text{A winning in 5 games}) + P(\text{A winning in 6 games}) + P(\text{A winning in 7 games}) $$

For A to win, the last game must be won by team A.

$$ \Rightarrow P(A) = p^4 + { 4 \choose 3}p^4q + { 5 \choose 3}p^4q^2 + { 6 \choose 3}p^4q^3 $$

Second approach

Imagine telling the players to continue playing the games even after the match has been decided, then the outcome of the match won’t be affected by this, and this also means that the probability that A wins the match won’t be affected by assuming that the teams always play 7 games.

$$ P(A) = P(\text{A winning 4 times in 7 games}) + P(\text{A winning 5 times in 7 games}) + P(\text{A winning 6 times in 7 games}) + P(\text{A winning 7 times in 7 games}) $$

I am not able to follow the second approach. Why is second approach correct?

Update:

I am looking for an intuitive explanation for the second approach because in the second approach it appears that probabilities for winning 5, 6 and 7 matches are used which were not required for team A to win.

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  • $\begingroup$ Why is the second approach correct? In the same way that if I flip a coin one time the probability it is heads is the same as the probability the first coin flipped is heads when flipping a coin seven times in succession. If I ask what is the probability I win the very next game against my friend where I play only one game, the probability is the same as if i ask what is the probability I win the very next game against my friend when I play fifty games after the first only important game. $\endgroup$ – JMoravitz Mar 15 '17 at 22:40
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    $\begingroup$ Because a team winning in the second version will also win in the first, and vice versa $\endgroup$ – Henry Mar 15 '17 at 22:40
  • $\begingroup$ @JMoravitz Thanks. The second approach uses winning 5, 6, as well as 7 games. If I am correct then in the coin example, only the probability of getting heads the first time is considered even if we flipped it 7 times. $\endgroup$ – ovais Mar 15 '17 at 22:50
  • $\begingroup$ Both ways to compute $P(A)$ produce a polynomial in $p$. Since they are both correct, the the two methods should produce the same polynomial; and they do: $-20p^7 + 70p^6 -84p^5 + 35p^4$. $\endgroup$ – Fabio Somenzi Mar 16 '17 at 0:51
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Team $A$ wins the game if they win four matches times before $B$ wins four matches.   Thus the first approach measures the probability of team A doing so when team $B$ wins zero, one, two, or three matches before $A$'s fourth victory.

$$P(A) = \left(\binom 30 p^3q^0+\binom 41 p^3q^1+\binom 52 p^3q^2+\binom 63 p^3q^3\right)p$$

Now imagine the teams kept playing for a full seven matches even after one of them wins the game.   Team $A$ wins the game if they win at least four matches among those seven, since if they do so then team $B$ can win at most three matches before $A$ wins their fourth.

$$\begin{align}P(A) & = \binom 77 p^7q^0+\binom 76 p^6q^1+\binom 75 p^5q^2+\binom 74p^4q^3\end{align}$$

We can see these are equal by taking the terms of the first equation, and including the imagined games after victory.   Then if you use the binomial theorem to expand...

$$\begin{align}P(A) &= \binom 30 p^4q^0(p+q)^3+\binom 41 p^4q^1(p+q)^2+\binom 52 p^4q^2(p+q)+\binom 63 p^4q^3 \\[1ex] &= \binom 30 p^4q^0(p+q)^3+\binom 41 p^4q^1(p+q)^2+\binom 52 p^5q^2+\left(\binom 52+\binom 63\right) p^4q^3\\[1ex] &\vdots \\[1ex] &= \binom 77 p^7q^0+\binom 76 p^6q^1+\binom 75 p^5q^2+\left(\binom 30+\binom 41+\binom 52+\binom 63\right) p^4q^3 \\[1ex]& = \binom 77 p^7q^0+\binom 76 p^6q^1+\binom 75 p^5q^2+\binom 74p^4q^3\end{align}$$

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  • $\begingroup$ Thanks, Graham, I am looking for an intuition that why the second approach is using the probability of winning 5,6 or 7 matches in computing the result. It appears that it may increase $ P(A) $ but it does not as both answers are same. $\endgroup$ – ovais Mar 16 '17 at 10:59
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As an addendum to Graham's answer, after suitable massaging of the formulae, we get, for a best-of-$n$ tournament with odd $n$:

$$ \sum_{0 \leq k \leq \frac{n-1}{2}} \binom{n}{\frac{n+1}{2}+k} \sum_{0 \leq j \leq k} (-1)^j \binom{\frac{n+1}{2}+k}{j} p^{\frac{n+1}{2}+k} \enspace. $$

We can then generate plots like this for $P(A)$:

enter image description here

Our intuition that a larger $n$ favors the stronger team is confirmed.

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My question was to understand the intuition behind why the second approach is correct, especially why the second approach is apparently using probability for winning in 5, 6 or 7 matches while computing $ P(A) $.

To figure out I tried to understand the cases which first event - P(A wins 4 times in 7 games) - of the second approach missed because that can give some clue over why we were adding the probabilities for winning 5,6 or 7 matches.

In this approach, we imagined the game is continued for 7 matches. For simplicity let's assume that $ p = q = \frac 1 2 $.

In the first approach, we added probabilities of four disjoint events - P(A winning in 4 games), P(A winning in 5), .. , P(A winning in 7 games). The denominators of each of these cases - representing the total number of possible cases - are different - $ 2^4, 2^5, 2^6, 2^7 $.

In the second approach, denominators in the four disjoint events - P(A winning 4 times in 7 games), P(A winning 5 times in 7 games), ... P(A winning 7 times in 7 games) - are same and equal to $ 2^7 $ representing the total number of possible outcomes.

Since the winner is decided as soon as one of the team wins 4 games, in the second approach, we have over counted the number of possible outcomes.

To reach the correct answer, either we decrease or adjust the number of possible outcomes from over counted outcomes or adjust it by increasing the number of favourable outcomes.

Adjusting the denominator will transform the second approach into the first approach. Thus let's try adjusting the favourable outcomes.

To put it accurately, we are not adjusting the favourable outcomes but after adding total number of ways A can win 4 matches, all the other outcomes of the games irrespective of success or failure of A will be considered as favourable.

This is equivalent to: P(A wins 4 matches and loses 3) + P(A wins 5 matches and loses 2) + P(A wins 6 matches and loses 1) + P(A wins 7 matches and loses 0).

Thus we are adding probabilities of winning 5, 6, and 7 matches because now the numerator contains all possible outcomes for losing 0, 1,2 or 3 matches once team A has won and the denominator already counts all the possible outcomes in the 7 games.

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