Two players (player A and player B) are playing a game against each other repeatedly until...

Two players (player A and player B) are playing a game against each other repeatedly until one is bankrupt. When a player wins a game they take $1 from the other player. All plays of the game are independent and identical. Suppose player A starts with $6 and player B starts with $6. If player A wins a game with probability 0.5, what is the probability the game ends (someone loses all their money) on exactly the 10th play of the game?

Solutions

Expert Solution

It's a fair and unbiased game between A and B, where -

Both A and B starts with $6 each. Each win gives $1 to the winning player and Each lose cost $1 to the losing player.

All rounds played between A and B are Independent and Identical in nature.

Probability (A wins) = Probability (B wins) = 0.50

To find out the Probability that the game ends on exactly 10th round - It means that either A or B loses exactly 8 rounds that cost them ($8) and wins 2 rounds that reward them with $2 and initial amount they start with is $6.

	A	B
Initial Amount	$ 6.00	$ 6.00
2 Win	$ 2.00	$ 2.00
8 Lose	-$ 8.00	-$ 8.00
Total after 10th round of game	$ 0.00	$ 0.00

The above table simply states that in order to end the game at 10th round, either A or B has to win exactly 2 round and lose exactly 8 round.

Therefore, Probability (Game ends at exactly 10th round) = Probability (A wins exactly 2 rounds and lose 8 rounds) +Probability (B wins exactly 2 rounds and lose 8 rounds)

Probability (Game ends at exactly 10th round) = 0.50 ^ (10) + 0.50 ^ (10) = 2 * (0.50 ^ (10)) = 2 * 0.0009765625

Probability (Game ends at exactly 10th round) = 0.001953125 = 0.20%

milcah answered 8 months ago

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