A game starts with 3 coins in a box. At each turn the number of coins in the box

ID: 3363475 • Letter: A

Question

A game starts with 3 coins in a box. At each turn the number of coins in the box is counted and the following procedure is repeated k times:

A fair die is thrown, and depending to the outcome the following four things can happen:

If the outcome is 1 or 2 the player takes 1 coin from the box.

If the outcome is 3 no action is taken.

If the outcome is 4 the player puts 1 coin in the box (assume that the player has an unlimited supply of coins).

If the outcome is 5 or 6, the player puts 2 coins in the box.

Whenever the box is empty, the game stops.

1. Compute the expected number of coins in the box after turn n

2. Compute the probability that the game will stop eventually.

1) We need to calculate the average gain or loss per toss of the die in this case

Now the individual probabilities look like this:

a) Probability of Loosing 1 coin = 2/6 ( getting 1 or 2)

b) Probability of Adding 0 coin = 1/6 ( getting a 3)

c) Probability of adding 1 Coin = 1/6 (( Getting a 4)

d) Probability of adding 2 coins = 2/6 ( Getting 5 or 6)

the average gain or loss is obtained by taking the possible outcomes, multiplying by the probability, and adding the results

The average gain or loss per toss= -1* 2/6+ 0*1/6+ 1*1/6+ 2*2/6 = 1/2 ( It is a gain)

So expected number of coins after n tosses = 3+ n* (Average gain or Loss) = 3+ n/2

2) This part of the problem can be solver with Gambler's Ruin problem

p= probability of adding 0 or more coins = 1/6+ 2/6+1/6 = 4/6= 2/3

q= probability of loosing coin= 1/3

since p>0.5 then q/p<1 (q/p in this case will be 1/2)

The probability that the infinite number of coins will be added = 1-(q/p)^i where i is the initial starting amount

So in this case P that infinite number of coins will be added = 1- (1/2)^3 = 1-1/8 = 7/8

So probability that the game will eventually stop = 1- Probability of infinite conts getting added = 1-7/8 = 1/8