How Does Probability Affect the Outcome of a Sudden-Death Chess Match?

[janela]

Homework Statement

Two world champion chess players play a sudden-death chess match where the first player to win a game wins the match. Each game is won by the first player with probability p and by the 2nd player with probability q and is a tie with probability (1-p-q).

a) what is the probability that the first player wins the match?
b) what is the Probability Mass Function, the mean, and the variance of the duration (number of games) of the match?

Homework Equations

The Attempt at a Solution

in progress

 

[KataKoniK]

a) The probability that the first player wins the match is equal to the probability that the first player wins the first game, plus the probability that the first player wins the second game, and so on until they win the match. This can be represented by the following equation:

P(first player wins match) = p + (1-p-q)p + (1-p-q)^2p + ...

This is an infinite geometric series with a common ratio of (1-p-q). Using the formula for the sum of an infinite geometric series, we can simplify this equation to:

P(first player wins match) = p / (1 - (1-p-q)) = p / (p+q)

b) The Probability Mass Function for the duration of the match is a discrete probability distribution that assigns a probability to each possible number of games in the match. In this case, the possible values for the duration are 1, 2, 3, ...

The PMF can be represented by the following equation:

f(x) = (1-p-q)^{x-1}p

The mean of the duration of the match can be calculated using the formula:

E[X] = \sum_{x=1}^{\infty}xf(x)

Substituting the PMF into this equation, we get:

E[X] = \sum_{x=1}^{\infty}x(1-p-q)^{x-1}p = p/(1-p-q)

The variance of the duration of the match can be calculated using the formula:

Var[X] = \sum_{x=1}^{\infty}(x-E[X])^2f(x)

Substituting the PMF and mean into this equation, we get:

Var[X] = \sum_{x=1}^{\infty}(x-p/(1-p-q))^2(1-p-q)^{x-1}p = (q+p^2)/(1-p-q)^2

 

[piercas]

...

A) The probability of the first player winning the match depends on the number of games played. If the first player wins the first game, they win the match with probability 1. If they lose the first game but win the second game, they win the match with probability p(1-p). If they lose the first two games but win the third game, they win the match with probability p^2(1-p). This pattern continues until the nth game, where the first player wins the match with probability p^(n-1)(1-p). Therefore, the overall probability of the first player winning the match is the sum of these probabilities, which can be expressed as a geometric series:

P(first player wins match) = p + p(1-p) + p^2(1-p) + ... + p^(n-1)(1-p)

Using the formula for the sum of a geometric series, this can be simplified to:

P(first player wins match) = p / (1 - p(1-p))

B) The Probability Mass Function (PMF) is a function that assigns probabilities to each possible outcome of a random variable. In this case, the random variable is the number of games played in the match. Since the match ends as soon as one player wins a game, the possible outcomes are 1, 2, 3, ..., n games played.

The PMF for this scenario is:

P(X = x) = (1-p-q)^(x-1)p for x = 1, 2, 3, ..., n

This means that the probability of x games being played is equal to the probability of the first player losing x-1 games and then winning the last game, multiplied by the probability of a tie in each game.

The mean of the duration of the match can be calculated by taking the sum of all possible outcomes multiplied by their respective probabilities:

E(X) = 1*p + 2*p(1-p-q) + 3*p(1-p-q)^2 + ... + n*p(1-p-q)^(n-1)

This can be simplified using the formula for the sum of a geometric series to:

E(X) = p / (1 - (1-p-q))

Simplifying further, we get:

E(X) = 1/(p+q)

The variance of the duration of the match can be calculated using the formula:

Var(X) = E(X