# Weird money game...You roll a fair die either 1; 2 or 3 times. When you stop, you get the latest number of dollars on the die (\$1-\$6). For example, if you have 3 in the rst try and stop, then you...

Weird money game...

You roll a fair die either 1; 2 or 3 times. When you stop, you get the latest number of dollars on the die (\$1-\$6). For example, if you have 3 in the rst try and stop, then you get \$3. If you have 3 in the rst try, 6 in the second try and (of course) stop, then you get \$6. If you get 3 in the rst try, 4 in the second, and 1 in the third, then you get only \$1 (cannot go back to 4). What is the best strategy to get the maximum amount of dollars as an average? Explain the mathematical reasoning of your strategy. (Hint: Compute the expected value of rolling once to gure out the strategy when you have another chance. Then compute the expected value of rolling twice.)

beckden | Certified Educator

E[X] = 1/6*1 + 1/6*2 + 1/6*3 + 1/6*4 + 1/6*5 + 1/6 *6 = 21/6 = \$3.5

The expected value after 1 throw is \$3.50.

Our strategy would be to throw again if we get below a certain value on the first throw.

If we throw a 1 on the first throw and throw again what is our expected value.

E[X|Y<2] = 1/6*2+1/6*3+1/6*4+1/6*5+1/6*6 + 1/36*1 + 1/36*2 + 1/36*3 + 1/36*4 + 1/36*5 + 1/36*6 = 20/6 + 1/6(7/2) = 47/12 = 3 11/12

E[X|Y<3] = 1/6*3 + 1/6*4 + 1/6*5 + 1/6*6 + 1/3(7/2) = 18/6+ 7/6 = 25/6 = 4 1/6

E[X|Y<4] = 1/6*4 + 1/6*5 + 1/6*6 + 1/2(7/2) = 15/6 + 7/4 = 17/4 = 4 1/4

E[X|Y<5] = 1/6*5 + 1/6*6 + 2/3(7/2) = 11/6 + 14/6 = 25/6 = 4 1/6

E[X|Y<6] = 1/6 * 6 + 5/6(7/2) = 6/6 + 35/12 = 47/12 = 3 11/12

So our optimal strategy would be to throw again if our first throw is a three or less and we get an expected value of \$4.25.

Now we need to find out what our expected value is if we throw again.

for E[X|Y<A and Z<B] where Y is the first throw and Z is the second throw would take 36 Computations we could assume it is around 3 so check 2,3,4 for first throw and 2,3,4 for second throw.

If you do these expectations you will find that you should again if you get a 4 or less on the first throw and a 3 on the second throw, and your expected value is \$4.666666.

Hope this helps,

Dennis Beck