The Dice game of “Pig” can be played with the following rules. Roll two six-side

Question

The Dice game of “Pig” can be played with the following rules. Roll two six-sided dice. Add the face values together. Choose whether to roll the dice again or pass the dice to your opponent. If you pass, then you get to bank any points earned on your turn. Those points becom permanent. If you roll again, then add your result to your previous score, but you run the risk of losing all points earned since your opponent had rolled. Continue to roll as much as you want. However, once a “1” comes up on either die, your score is reduced to 0, leaving you only with point that you have previously "banked." Furthermore, you must pass the dice to your opponent.

What is the mean value of a SUCCESSFUL roll? What is the expected value if you were to only roll once?

Considering the multiplication rule, what would be the probability of rolling successfully two times in a row?

If your strategy was to roll two times, what would be your probabilty of failure? Find the expected value if you decided to roll twice.

Using the logic discussed above, find the expected value for a strategy in which you roll three times before passing. Do the same for four and five times.

What is best stategy?

Explanation / Answer

Suppose X1 = number thrown by first die and X2 = number from 2nd die

Let Y = X1 + X2

Then, to be a successfull role, neither X1 nor X2 can be 1.

So, Y must have values from the set {4, 5 , ..., 12}

Now, number of ways 4 can come up as total(Y) is: 1 since (X1,X2) cane be only (2,2)

Similarly, 5 can come up as Y is: 2 since (X1,X2) = (2,3) or (3,2) i.e. 2 ways

SImilarly if Y = 6, then (X,X2) = (2,4) or (4,2) or (3,3) i.e. 3 ways

if Y = 7, then the pair will be (2,5),(5,2),(3,4) or (4,3) i.e. 4 ways

if Y = 8, then pair will be (2,6),(6,2),(3,5),(5,3) or (4,4) i.e. 5 ways

if Y = 9, then par will be (3,6) , (6,3), (4,5) or (5,4) i.e. 4 ways

if Y = 10, then pair will be (4,6), (6,4) or (5,5) i.e. 3 ways

if Y = 11, then we get (5,6) or (6,5) i.e. 2 ways

if Y = 12, then we get (6,6) i.e. 1 way

Hence, Mean Value of a Successful Roll is average of Y from the set {4,5,...12}, which is:

= (4+5+...+12)/9 = 8

So, Mean Value = 8

For 1 successfull roll, expected value will be:

= EY

= 4*P(Y=4) + 5*P(Y=5)+....+12*P(Y=12)

=4*(number of ways Y can come up as 4 with nither of X1 X2 as 1) *P(X1=a and X2=b with a+b = 4 and neither of a and b is 1)

+....

+12*(number of ways Y can come up as 4 with nither of X1 X2 as 1) *P(X1=a and X2=b with a+b = 12 and neither of a and b is 1))

Using simple excel or simple logic of maths, this comes

= (1/36)*[ 4*1 + 5*2 + 6*3 + 7*4 + 8*5 + 9*4 + 10*3 + 11*2 +12*1)

=(1/36)*200

= 5.56

So, expected value of successfull roll= 5.56

P(Y is success in 1st roll & Y is success in 2nd roll)

= P(Y is success is 1st roll) * P(Y is success in 2nd roll)

= square(P(Y is success is 1st roll)) , using inpendence and multiplication rule

= square (P(X1 is not 1 neither X2 = 1))

= square(1- P (X1 = 1 or X2 =1))

= (1 - 11/36)^2

= (25/36)^2

= 0.48

So, probability of rolling successfully two times in a row = 0.48

Now, P(failure if 2 rolls are thrown)

=P(1st roll is failure) + P(1st roll is success and 2nd roll is failure)

=11/36 + 25/36 *11/36

= 11/36*(61/36)

= 0.52

so, P(failure if 2 rolls are thrown) = 0.52

First 4 sub-parts I have answered. Not answering more due to lack of time. Thanks.

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