This is obvious since if we reroll then we have a 1/6 chance of getting a 1,2,3,

Question

This is obvious since if we reroll then we have a 1/6 chance of getting a 1,2,3,4,5, or 6 consecutively so we'd have another reroll. Then for Y_2: (I'll make it explicit to see the idea, it's like having a nest pdf, rather, it's like flipping the coins)

2

This is obvious since we now reroll for 1 or 2.
Assuming you get the pattern

Hence:

E(x) = (1+2+...+6)/6=21/6 = 3.5...

Now, intuitively we can say thif if we roll anything below a 3 we are better rolling again because the expected of the higher greater numbers is higher than the lower numbers (1,2,3). It's obvious once we see:

E(Y_1)=1/36(1)+7/36(2+3+...+6)=141/36 3.91666666

E(Y_2)=2/36(21+2)+8/36(3+4+...+6)= 4.16666666

if you see the pattern then

E(Y_3)=17/4 = 4.25

And to ensure that we are in fact getting the best number by rerolling if we get anything less than a 3:

allows for E(Y_4)=4/36*(1+2+3+4)+10/36*(11)=4.1666666666

Y_1 P(Y_1) 1 1/36 2 7/36 3 7/36 4 7/36 5 7/36 6 7/36 Problem 24 You are offered to play the following game. You roll a fair die once and observe the result which is shown by the random variable X. At this point, you can stop the game and win X dollars. You can also choose to roll the die for the second time to observe the value Y. In this case, you will win Y dollars. Let W be the value that you win in this game. What strategy do you use to maximize EW? What is the maximum EW you can achieve using your strategy?

Explanation / Answer

P(1) = P(2) = P(3) = P(4) =P(5) = P(6) = 1/6

Expected value in first roll = (1/6)*1+(1/6)*2+(1/6)*3+(1/6)*4+(1/6)*5+(1/6)*6 = (1/6)[1+2+3+4+5+6]

= (1/6)[6(7)/2]

= 7/2 = 3.5

our statergy is simple if we get value less than 3.5 in first roll we will roll the die again

maximum expected value of W using this statergy

E(W) = (1/6)[3.5]+(1/6)[3.5]+(1/6)[3.5]+(1/6)*4+(1/6)*5+(1/6)*6 = 4.25

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