Question

In a casino game, gamblers are allowed to roll n fair, 6-sided dice. If a 6 shows up on any of them, the gambler gets nothing. If no 6’s appear, the gambler is paid the sum of the values on the dice in dollars. The gambler is free to choose n, the number of dice rolls.

a) Derive a formula for the gambler’s expected payoff (the total dollars won). Plot this payoff for values of n from 1 to 20. What is the smallest n that maximizes the expected payoff ?

b) Suppose the gambler chooses to roll 10 dice (this is not necessarily the answer to part (a)). What is the expected number of distinct dice values that show up? In other words, what is the expected number of faces that are rolled at least once?

Answer 1

a)
`\bf \boxed{W(n) = 3*n*\left( (5)/(6) \right)^n\;dollars}`

the smallest n that maximizes the expected payoff is n=5.

b) 4

Explanation:

a)

The expected amount of $ won for each die would be the average of 1, 2, 3, 4 and 5 which is $3.

Let W(n) the expected money won when rolling n dice.

n =1

If the gambler rolls only one die, the expected money won would be $3 times the probability of not getting a 6, which is 5/6.

So

`\bf W(1) = 3*1*(5)/(6)`

n=2

If the gambler rolls 2 dice, the expected money won would be $3 times the probability of not getting a 6 in any of the dice. Since the outcome of the rolling does not depend on the previous rollings, the probability is

`\bf (5)/(6)*(5)/(6)=\left( (5)/(6) \right)^2`

and

`\bf W(2) = 3*2*\left( (5)/(6) \right)^2`

n=3

Similarly, since the probability of not getting a 6 in 3 dice equals

`\bf \left( (5)/(6) \right)^3`

`\bf W(3) = 3*3*\left( (5)/(6) \right)^3`

and the formula for the expected money won with n dice would be

`\bf \boxed{W(n) = 3*n*\left( (5)/(6) \right)^n\;dollars}`

In the picture attached there is a plot of the values of the expected money won for n=1 to 20 (See picture)

For n=5 and n=6 we get the maximum profit expected of $6.02816=$6 rounded to the nearest integer.

Hence, the smallest n that maximizes the expected payoff is n=5.

b)

The probability that face k (k=1,2,...or 6) shows up is 1/6,

as this face can be in any of the 10 positions of the arrangement, there are 10 ways that face k can show up.

The probability that face k (k=1,2,...or 6) shows up twice is
`\bf \left( (1)/(6) \right)^2`

as this face can be in any of the
`\bf C(10;2)=\binom{10}{2}` (combinations of 10 taken 2 at a time) positions of the arrangement, there are C(10;2) ways that face k can show up twice.

The probability that face k (k=1,2,...or 6) shows up three times is
`\bf \left( (1)/(6) \right)^3`

as this face can be in any of the
`\bf C(10;3)=\binom{10}{2}` (combinations of 10 taken 2 at a time) positions of the arrangement, there are C(10;3) ways that face k can show up twice.

So, we infer that the expectation is

`\bf \sum_(k=1)^(10)\binom{10}{k}(1/6)^k=3.6716\approx 4`

and the expected number of distinct dice values that show up is 4.