Chapter 3, Problem 1

[dstefan]

Chapter 3, Problem 1:

Let k be a positive integer with 2 \leq k \leq 12. You throw two fair dice. If the sum of the dice is k, you win $w(k), or lose $1 otherwise.

Find the smallest value of w(k) which makes the game worth playing.

Solution:

If 2 \le k \le 7, then P(x+y=k) = \frac{k-1}{36} and

E[X] = w(k)\frac{k-1}{36} - \frac{36 - (k-1)}{36} = \frac{w(k)(k-1) - (37-k)}{36}

The game is worth playing if E[X] \ge 0, i.e., if w(k) \ge \frac{37-k}{k-1}, for 2 \le k \le 7.

If 8 \le k \le 12, then P(x+y=k) = \frac{13-k}{36} and

E[X] = w(k)\frac{13-k}{36} - \frac{36 - (13-k)}{36} = \frac{w(k)(13-k)-(23+k)}{36}

The game is worth playing if w(k) \ge \frac{23+k}{13-k}, for 8 \le k \le 12


The corresponding values of w(k) for 2 \leq k \leq 12 are

w(2) = w(12) = 35

w(3) = w(11) = 17

w(4) = w(10) = 11

w(5) = w(9) = 8

w(6) = w(8) = 6.2

w(7) = 5

[quantyst]

To incorporate two or more cases into a single formula, I'd like to suggest the following:

P{X+Y=k} = (6 - abs(k-7)) / 36.

So, w(k) = (30 + abs(k-7)) / (6 - abs(k-7)).