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 1 round? b) Not end in the first 2 rounds? c) End in the round? [Note: Answer is a single value for each problem part.]

Answer 1

For 12 simultaneous coin flips by 12 people, the number of possible outcomes is 2¹² = 4096. The event of 1 head and 11 tails is . Likewise, there are 12 outcomes with 1 tail and 11 heads. So the total number of possible ‘successes’ is 24.

(a) For the game to end in a single round, the probability is that of a single success in one trial:

24/4096 = 0.00585 = 0.59%.

B) For the game to not end in the first two rounds, we must have two non-successes in a row. Each non-success has a probability of 1 – 24/4096 . Two in a row have a probability of (1 – 24/4096)² = 0.9883 = 98.83%.

C) I think the question is misprinted. It should be end in the 3rd round. Solution is done according to this problem. For this we must have two non-successes, followed by one success. The probability of this is

(1 – 24/4096)² · 24/4096 = 0.00579 = 0.58%.