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 k = 2, we stop as soon as we roll the dice once, no matter what the output is, Therefore now as all faces are equally likely, we get the expected value here as:

E(X) = 1/6*(0 + 2 + 3 + 4 + 5 + 6) = 20/6 = 3.3333

For k = 3, we stop when we get anything except 2. Therefore if the expected value here is E(X), then, we get here:

E(X) = (1/6)*0 + (1/6)*E + (1/6)*(3 + 4 + 5 + 6)

beucase in case we get a 2 we rool it again, therefore in that case again the expected value is E(X)

(5/6)E = 3

E = 18/5 = 3.6

For k = 4, we stop as soon as we roll the dice once, no matter what the output is, Therefore now as all faces are equally likely, we get the expected value here as:

E(X) = (1/6)*0 + (2/6)*E(X) + (1/6)*( 4 + 5 + 6) because here the dice is tossed again in case we get a 2 or a 3

(2/3)E(X) = 2.5

E(X) = 3.75

For k = 5, we stop as soon as we roll the dice once, no matter what the output is, Therefore now as all faces are equally likely, we get the expected value here as:

E(X) = (1/6)*0 + (3/6)*E(X) + (1/6)*( 5 + 6) because here the dice is tossed again in case we get a 2 or a 3 or a 4

0.5E(X) = 11/6

E(X) = 3.6667

For k = 6, we stop as soon as we roll the dice once, no matter what the output is, Therefore now as all faces are equally likely, we get the expected value here as:

E(X) = (1/6)*0 + (4/6)*E(X) + (1/6)*( 6) because here the dice is tossed again in case we get a 2 or a 3 or a 4 or a 5

(1/3)E(X) = 1

E(X) = 3

Therefore the expected values for different K here are computed to be:

Therefore K = 3.75 should be used here.

K E(X) 2 3.3333 3 3.6 4 3.75 5 3.6667 6 3

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