H2 Math Qn

Pinkbow6826 SG

Probabilities of the objects that Alan and Ian would indicate are given below:

Alan

Scissors 0.5

Paper 0.1

Stone 0.4

Ian

Scissors 0.6

Paper 0.2

Stone 0.2

If the objects indicated are the same, the result is a draw.

In a contest consisting of at most three games, any player who attains two wins is declared the champion. It may be assumed that the object that each player indicates is independent of both his and his opponents’ previous indications.

If Ian indicates “scissors” in the first two games, what is the probability that he is declared the champion?

Pinkbow6826 SG

Probabilities of the objects that Alan and Ian would indicate are given below:

Alan

Scissors 0.5

Paper 0.1

Stone 0.4

Ian

Scissors 0.6

Paper 0.2

Stone 0.2

If the objects indicated are the same, the result is a draw.

If Ian indicates “scissors” in the first two games, what is the probability that he is declared the champion?

wee_ws SG

Hi,

This is a question on conditional probability, i.e. P(A | B).

1. Identify the events A and B.

A: Event that Ian is the winner; B: Event that Ian indicates "scissors" in first 2 games.

2. Find P(B), which is usually easy.

P(B) = 1.

3. Find P(A and B), which requires one to list cases.

Case 1: Ian wins in 2 games.

Probability = (0.1)(0.1).

Case 2: Ian wins in 3 games.

Possibilities (Ian-Alan) are:

(a) Ian lost one of the 1st 2 games and won 2 games, e.g.
Scissors-Stone, Scissors-Paper, Scissors-Paper
Scissors-Stone, Scissors-Paper, Paper-Stone
Scissors-Stone, Scissors-Paper, Stone-Scissors

Probability = 2(0.4)(0.1)[ (0.6)(0.1) + (0.2)(0.4) + (0.2)(0.5) ]

(b) Ian drew one of the 1st 2 games and won 2 games, e.g.
Scissors-Scissors, Scissors-Paper, Scissors-Paper
Scissors-Scissors, Scissors-Paper, Paper-Stone
Scissors-Scissors, Scissors-Paper, Stone-Scissors

Probability = 2(0.5)(0.1)[ (0.6)(0.1) + (0.2)(0.4) + (0.2)(0.5) ]

4. Required probability = P(A and B) / P(B) = 0.0532.

Thanks.

Cheers,
Wen Shih

eagle SG

Alternatively, we could also list out the scenarios of winning since it is "at most three".

P(Ian won 1st game) = 1 * 0.1 = 0.1
P(Ian won 2nd game) = 1 * 0.1 = 0.1
P(Ian won 3rd game) = 0.6 * 0.1 + 0.2 * 0.4 + 0.2 * 0.5 = 0.24

P(Ian Won) = P(win 1,2) + P(win 1,3) + P(win 2,3)
= 0.1*0.1 + 0.1*0.9*0.24 + 0.9*0.1*0.24 =0.0532

wee_ws SG

Hi eagle,

Correct, it is a good approach :)

Cheers,
Wen Shih

eagle SG

yup, mistake on my part. Editing now