Question

1. Players A and B each roll a fair 6-sided die. The player with the higher score wins £1 from the other player. If both players have equal scores, the game is a draw and no one wins anything.

i. Let X denote the winnings of player A from one round of this game. State the probability mass function of X. Calculate the expectation E(X) and variance Var(X).

ii. What is the conditional probability that player A rolls a 2 , given that player B wins.

iii. Suppose the game is repeated 10 times. Compute the probability that at least two of the rounds result in draws.

Answer 1

Solution

Back-up Theory

If a discrete random variable, X, has probability function, p(x), x = x₁, x₂, …., x_n, then

Expected value =E(X) = Σ{x.p(x)} summed over all possible values of x…..…..................................... (1)

E(X²) = Σ{x².p(x)} summed over all possible values of x………………………….......................………..(2)

Variance of X = Var(X) = E[{X – E(X)}²] = E(X²) – {E(X)}²…………………...................................……..(3)

If A and B are two events such that probability of B is influenced by occurrence

or otherwise of A, then

Conditional Probability of B given A, denoted by P(B/A) = P(B ∩ A)/P(A)…….................................….(4)

P(A/B) = P(B/A) x {P(A)/P(B)}……………………………..………..................................…………..…….(4a)

If X ~ B(n, p). i.e., X has Binomial Distribution with parameters n and p, where

n = number of trials and p = probability of one success, then,

probability mass function (pmf) of X is given by p(x) = P(X = x) = (ⁿC_x)(p^x)(1 - p)^{n – x} ..........………..(5)

This probability can also be directly obtained using Excel Function: Statistical, BINOMDIST]..........(5a)

Now to work out the solution,

Given,

X = the winnings of player A from one round of this game

Part (i)

Given,

X = the winnings of player A from one round of this game

Then,

X =   1, if A rolls a higher number than B for which the probability is 15/36

    = - 1, if B rolls a higher number than A for which the probability is 15/36

    =   1, if both roll the same number for which the probability is 6/36.

Thus, Probability mass function of X is:

X	- 1	0	1	Total
p(x)	15/36	6/36	15/36	1

Answer 1

Vide (1), E(X) = 0 Answer 2

Vide (2) and (3), Var(X) = 0.833 – 0²

= 0.8334 Answer 3

Details of calculations

x	1	0	1	Total
p(x)	15/36	6/36	15/36	1
x.p(x)	- 0.4167	0.0000	0.4167	0.000
x2.p(x)	0.4167	0.0000	0.4167	0.8334

Part (ii)

Conditional probability that player A rolls a 2 , given that player B wins

= P(player A rolls a 2/player B wins)

= P(player B wins/player A rolls a 2) x P(player A rolls a 2)/P(player B wins) [vide (4a)]

= {P(player B rolls a 3, 4, 5 or 6) x (1/6)}/(15/36) [B wins if B rolls > A’s roll]

= {(2/3) x (1/6)}/(15/36)

= 4/15 = 0.2667 Answer 4

Part (iii)

Let Y = number of rounds out of 10 rounds resulting in a draw. Then, Y ~ B(10, 1/6) ....................(6)

where 1/6 is the probability of a draw.

So, probability that at least two of the rounds result in draws

= P(Y ≥ 2)

= 1 – 0.4845 [vide (6), (5) and (5a)]

= 0.5155 Answer 5

DONE