3. In the game of Patience, you roll a fair, 6-sided die repeatedly. If the die

Question

3. In the game of Patience, you roll a fair, 6-sided die repeatedly. If the die ever shows 1, you lose the game and get S0. You may also choose to stop at any time before rolling a 1, in which case you win the amount shown on the die after the last roll (so if you choose to stop after rolling a 3, then you get $3). One strategy is to wait until a roll of k or greater before cashing out (with k being 2, 3, 4, 5, or 6). Find the expected value of the game when you follow this strategy with k = 2, k 3, k 4, k-5, and k = 6 (that's a total of 5 expected values to calculate). Which choice of k is best?

Explanation / Answer

For a fair die, probability of each number from 1 to 6 = 1/6

Let;s start with k=6. The player quits only if 6 shows up.Otherwise, he waits till a 1 shows.

Probability of win = 1/6, probability of loss = 5/6For K = 5

Expected value of the game = 6 x (1/6) + 0 x (5/6) = $1

For k = 5, Probability of win = Probability of seeing a 5 or a 6 = (1/6)+ (1/6) = 2/3

Expected value from the game = 5 x 1/6 + 6 x 1/6 + 0 x 4/6 = 11/6

For k = 4, Probability of win = Probability of seeing a 4 or a 5 or a 6 = 1/6 + 1/6 + 1/6

Expected value from the game = 4 x 1/6 + 5 x 1/6 + 6 x 1/6 + 0 x 3/6 = 15/6

On similar lines, for k= 3, Expected value from the game = 3 x 1/6 + 4 x 1/6 + 5 x 1/6 + 6 x 1/6 + 0 x 2/6 = 18/6 = $3

and for k= 2, Expected value from the game = 2 x 1/6 + 3 x 1/6 + 4 x 1/6 + 5 x 1/6 + 6 x 1/6 + 0 x 2/6 = 20/6

So the choice of k=2 provides the highest expected value from the game.

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