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?
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?
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
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
Hi eagle,
Correct, it is a good approach :)
Cheers,
Wen Shih
yup, mistake on my part. Editing now