Question

A miner is mining in his mine. Everyday he tosses a fair coin to decide whether to use a pickaxe or dynamite to mine. If he uses a pickaxe, he can get gold worth 100 dollars for the day. If he decides to use dynamite, however, he will first need to buy dynamite for 1, 000 dollars. Then, with a 1/3 chance he will get gold worth 12, 000 for that day and with a 2/3 chance he will get nothing and end up blowing up his mine which makes him unable to mine in the future. Assume the miner always has enough money to buy dynamite.

(a) Let T denote the number of days he mines before the mine collapses. What is E[T]?

(b) Let R denote the profit he makes from the mine until it collapses, i.e. the total worth of gold yield from the mine less the money he spent on dynamite. What is E[R]?

Answer 1

(a)

Everyday he tosses a fair coin to decide whether to use a pickaxe or dynamite to mine. Thus, there is 0.5 probability to use a dynamite and collapses the mine. Thus, T ~ Geometric(p = 0.5)

From Geometric distribution, E[T] = 1/p = 1/0.5 = 2

(b)

Let T denote the number of days he mines before the mine collapses. Thus there were T-1 days when the he uses a pickaxe, he can get gold worth 100 dollars for the day.

R when miner used uses a pickaxe = $100 => E[R | pickaxe] = $100

R when miner used uses a dynamite = $12, 000 - $1,000 = $11,000 for p = 1/3

= -$1,000 for p =2/3

E[R | dynamite] = (1/3) * 11,000 - (2/3) * 1,000 = $3,000

E[R] for a day = P(pickaxe) * E[R | pickaxe] + P(dynamite) * E[R | pickaxe]

= 0.5 * 100 + 0.5 * 3000

= 1550

Since the expected number of days for mining to collapse is 2.

Total worth of gold yield from the mine less the money he spent on dynamite = 2 * E[R] for a day

= 2 * 1550

= $3,100