- #1

- 261

- 26

<Moderator's note: Moved from a technical forum and thus no template.>

Hi all,

I'm looking at an exercise in probability and I have a little doubt. So the exercise goes like this:

"Two players are playing against each other. They have the same probability of winning a single game. In order to win the match a player needs to win ten games. What is the probability that one of the players wins in 12 or less games?"

The probability that one of the player wins 10 out of 12 matches is pretty easy, but I think that this problem is more complex than that, although the final answer is hardly different.

The number of outcomes in 12 matches is [itex]2^{12}[/itex]. However, I think that some of these outcomes should be discarded, on account that there is no need of further games since one of the players already won.

Specifically the one instance where one of the players wins all 12 games, and the 12 instances where one of the players wins 11 out of 12 matches.

Thus, one should take into account only [itex]2^{12}-13[/itex] possible outcomes.

In

$$\binom{12}{10}= 66$$

of these outcomes one player wins 10 games and loses two.

So, I'd say that the probability the problem is asking for is

$$\frac{1}{2^{12}-13}\binom{12}{10}$$

which is almost the same as the fraction of 10 victories in 12 games, i.e.

$$\frac{1}{2^{12}}\binom{12}{10}$$

Does this make sense?

I also obtained the number of successful games as the number of 10 victories in 10 games (1) plus the number of 9 victories in 10 games, assuming that the 11th is a victory (10) plus the number of 9 victories in 11 games (assuming the 12th is a victory): [itex]1+10+ 11\cdot 10/2=66[/itex].

I'm not sure on how counting the total number of games in a similar fashion.

Any insight is appreciated.

Thanks a lot!

Hi all,

I'm looking at an exercise in probability and I have a little doubt. So the exercise goes like this:

"Two players are playing against each other. They have the same probability of winning a single game. In order to win the match a player needs to win ten games. What is the probability that one of the players wins in 12 or less games?"

The probability that one of the player wins 10 out of 12 matches is pretty easy, but I think that this problem is more complex than that, although the final answer is hardly different.

The number of outcomes in 12 matches is [itex]2^{12}[/itex]. However, I think that some of these outcomes should be discarded, on account that there is no need of further games since one of the players already won.

Specifically the one instance where one of the players wins all 12 games, and the 12 instances where one of the players wins 11 out of 12 matches.

Thus, one should take into account only [itex]2^{12}-13[/itex] possible outcomes.

In

$$\binom{12}{10}= 66$$

of these outcomes one player wins 10 games and loses two.

So, I'd say that the probability the problem is asking for is

$$\frac{1}{2^{12}-13}\binom{12}{10}$$

which is almost the same as the fraction of 10 victories in 12 games, i.e.

$$\frac{1}{2^{12}}\binom{12}{10}$$

Does this make sense?

I also obtained the number of successful games as the number of 10 victories in 10 games (1) plus the number of 9 victories in 10 games, assuming that the 11th is a victory (10) plus the number of 9 victories in 11 games (assuming the 12th is a victory): [itex]1+10+ 11\cdot 10/2=66[/itex].

I'm not sure on how counting the total number of games in a similar fashion.

Any insight is appreciated.

Thanks a lot!

Last edited by a moderator: