Question

The yearly Penn rock-paper-scissors tournament has come down to the final series between Zach and Shaya. They play a best-of-thirteen games in rock-paper-scissors to determine the final winner; in other words, they play until one player has seven wins, and that player wins the tournament. Assume that no game ends in a tie and that Zach and Shaya are equally matched, i.e. for each game, they are both equally likely to win. What is the probability that the winner of the first game between Zach and Shaya wins the tournament?

Answer 1

Answer:

p=41.79%

Step-by-step explanation:

First let us remember the calculation of a probability:

$p=\frac{number of favorable outcomes}{total number of possible outcomes}$

In this case let's calculate the number of possibilities of winning the tournament when either one of them, Zach(Z) or Shaya(S), wins the first game (number of favorable outcomes). That is:

Given that Z or S won the first game, the player has another 6 possibilities of winning among 12 remaining games, which is a combination:

number of favorable outcomes = 12C6 = 924

Therefore, the winner of the first game has 924 opportunities of winning 6 other games among 12 games and be the tournament winner.

Now, we must find the number of total possible outcomes, which we may analize as it follows:

Assuming that the first winner Z or S loses the tournament we have several outcomes:

1. After winning the first game, the player doesn't win again, then we have 0 possbilities among 7 other games, this is because the next player would win the tournament winning 7 games in a row. We then have the combination 7C0.

2. The player wins just one more game, but doesn't win again. Now we have 1 possibility among 8 games, given that the other player would win the other 7 games. We then have the combination 8C1.

3. The player wins just two more games, but doesn't win again,  now we have 2 possibilities among 9 games, given that the other player would win the other 7 games. We then have the combination 9C2.

And so on until we reach the outcome where the player wins 5 more games, but doesn't win again,  then we have 5 possibilities among 12 games, given that the other player would win the other 7 games. We then have the combination 12C5.

Now, we must also consider all the outcomes where the first winner wins the tournament, that is the first combination we calculated: 12C6

Therefore we obtain the following:

total number of possible outcomes = 7C0+8C1+9C2+10C3+11C4+12C5+12C6

total number of possible outcomes = 2,211

We have now the elements to calculate the probability:

$p=\frac{924}{2,211}=0.4179$

Hence, the probability that the winner of the first game wins the tournament is 41.79%