Do I Use the Binomial or the Negative Binomial?

[number0]

Homework Statement

Two teams, A and B, play a series of games. If team A has probability .4 of
winning each game, is it to its advantage to play the best three out of five games
or the best four out of seven? Assume the outcomes of successive games are
independent.

Homework Equations

http://en.wikipedia.org/wiki/Binomial_distribution

vs.

http://en.wikipedia.org/wiki/Negative_binomial_distribution

The Attempt at a Solution

This problem is basically a plug and chug problem. However, I do have some difficulties interpreting this problem.

When the problem states "the best three out of five games" and "the best four out of seven games", does it mean that when Team A wins three times or four times (given the respective parameters), does the game end?Example 1: Team A wins three times -> game is over.
Example 2: Team A wins three times -> game continues until 5 games are played.Since a person can interpret this many ways, I decided to just write out the solution for each example.Solution 1: Take the summation of the negative binomial distribution of 3 successes, p = 0.4, and the number of trials from i = 3 to i = 5.

Solution 2: Take the summation of the binomial distribution from i number of successes from i = 3 to i = 5, p = 0.4, and the number of trials is 5. Assuming that if both the cases were true, is my solution or "method" for each one correct? I know that 3 out of 5 game is the correct answer; nevertheless, I am just curious at the process to reach it because many people I met have different interpretations and methods to answer this problem.

Thanks.

 

[Ray Vickson]

Look at A winning the best 3 out of 5. In the actual tournament, play stops as soon as A wins 3. If A wins the first 3, play stops, but we could, instead, let play continue for two more games and just ignore the results. In that case, in the full 5 games, A would win >= 3 games. If A wins the tournament at game 4, we could go on to play game 5 and just ignore the result. In that case A wins >= 3 games out of 5. Finally, if A wins for the third time at game 5, no extra games are needed. Altogether, the probability that A wins the best 3 of 5 and the tournament then stops right away is the SAME as the probability that A wins at least 3 games in a full 5 games. What probability distribution would you use to compute that probability?

RGV

 

[number0]

Look at A winning the best 3 out of 5. In the actual tournament, play stops as soon as A wins 3. If A wins the first 3, play stops, but we could, instead, let play continue for two more games and just ignore the results. In that case, in the full 5 games, A would win >= 3 games. If A wins the tournament at game 4, we could go on to play game 5 and just ignore the result. In that case A wins >= 3 games out of 5. Finally, if A wins for the third time at game 5, no extra games are needed. Altogether, the probability that A wins the best 3 of 5 and the tournament then stops right away is the SAME as the probability that A wins at least 3 games in a full 5 games. What probability distribution would you use to compute that probability?

RGV

You could use the binomial. I should've done this a second time just to make sure; but I just used both the binomial and negative binomial distribution, and I reached the same answer. The first time I did it both the probabilities were way off of each other. Nevertheless, what do you mean by "we can just ignore the result"? The result is sort of built into the answer.