Question

The Cardinals lead the division by three games over the Brewers and by four games over the Cubs. The Cubs and Cardinals play each other six more times and each plays other teams three times, while the Brewers play nine remaining games against other teams. Assuming each game's result is a 50/50 coin flip, what is the probability of a three-way tie? Please round your answer to the nearest tenth of a percent.

Answer 1

Current situation -

Cardinals are 4 games ahead of Cubs

Cardinals are 3 games ahead of Brewers

Lets first consider Cardinals and Brewers.

For brewers to win 3 games more than Cardinals, following winning possibilities can occur -

(C0,B3) , (C1,B4) , (C2,B5) , (C3,B6) , (C4,B7) , (C5,B8) , (C6,B9)

Now the least three and first three possibilities can be eliminated because Out of 9 matches, 6 are between them and will results in 3 wins. the rest three matches can result in 6 wins (3 for both), so we take possibilities where we have total of 9 wins -

(C3,B6)

For this case, we will have a case for Cubs as well. Since, cubs play all games with different team other than cardinals and brewers, the possibility for cubs become -

(7W,2L)

This result in a probability of -

Cardinals winning 3 game and Brewers winning 6 games and Cubs winning 7 games.

For cardinals to win 3 games, it must be from the games with Brewers. This will also include Brewers' probability of winning three matches, and then Cardinals must lose other 3 games, so, probability is-

(6C3 * 0.5 ^ 3 * 0.5 ^ 3) * (3C3 * 0.5 ^ 3)

And, the probability of brewers winning other three matches is -

(3C3 * 0.5 ^ 3)

the probability of Cubs winning seven matches is -

9C7 * 0.5 ^ 7 * 0.5 ^ 2

So, the probability becomes -

6C3 * 9C7 * 0.5 ^ 21

= 0.00034