In the group game \"Odd-One Out\": All players take out a fair coin, and everyon

Question

In the group game "Odd-One Out": All players take out a fair coin, and everyone flips at the same time. If one player (the "odd-one") receives heads and everyone else receives tails (or vice-versa), the game ends. Otherwise the game goes on for more rounds, until an "odd-one" is finally found.

In a group of 12 players, what is the percentage probability that the game will:

a) End in the 1st round?
b) Not end in the first 2 rounds?
c) End in the 3rd round?

[Answer is a single value for each problem part]

Explanation / Answer

Let's denote the event that the game ends in a particular round by A.

Since the coin is fair, probability of getting a head or a tail is same equal to p = 0.5

So, P(A) = 12C1p1(1-p)11-1 = 12C10.51(1-0.5)11-1 = 0.00292

So, probability that the game doesn't end on a particular round = P(A') = 1-P(A) = 0.99708

(a)

P(game ends in first round) = P(A) = 0.00292

(b)

P(game doesn't end in first 2 rounds) = 1 - P(game ends in first two rounds)

P(game ends in first 2 rounds) = P(game ends in first round) + P(game ends in second round)

P(game ends in first round) = 0.00292

P(game ends in second round) = 0.99708*0.00292 = 0.00291

So,

P(game ends in first 2 rounds) = 0.00292 + 0.00291 = 0.00583

So,

P(game doesn't end in first 2 rounds) = 1 - 0.00583 = 0.99417

(c)

P(game ends in third round) = P(game doesn't end on first round)*P(game doesn't end on second round)*P(game ends on third round) = 0.99708*0.99708*0.00292 = 0.00290

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