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

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. Compute the expected number of coins in the box after turn n Compute the probability that the game will stop eventually.

Explanation / Answer

(1.)According to the given condition in the question, we have the following probabilities

P(Drawing one coin from the box i.e -1) =2/6 =1/3

P(No coin is drawn i.e 0) = 1/6

P(Putting one coin in the box i.e 1) =1/6

P(Putting two coin in the box i.e 2) =2/6=1/3

Let X be a random variable the number added or subtracted after a particular throw of the die.So, we have the following table:

The expected value is given by the formula:

E(X) = xp(x)

Putting the values from the above table we have

E(X)= -1.(1/3)+0.(1/6)+1.(1/6)+2.(1/3) =2/3

Now, let S be a random variable which denotes the number of balls inn the box after n turns

S= X1 + X2 + ......+ Xn = ni=1 Xi

So. E(S) = E(ni=1 Xi) = ni=1 E(Xi) =n*2/3

Thus the expected number of coins in the box after n trials is 2n/3.

(2.) Logically, we can see that the game cannot stop in the first throw as we start with three coins so, the minimum number of throws needed for the game to stop is 3 (provided we get either1 or 2 in the first three throws)

Now, the only way the game will stop is when we get either 1 or 2 when the number of coins in box is 1i.e

the probability of the game stopping in the nth throw is P=2/3*(Pn-1)

where Pn-1 is the probability of 1 ball in the box in the n-1th trial.

Pn-2 is the probability of 2 ball in the box in the n-2th trial and so on.

Now for the game to end in the third trial we need to have a ball left after the second throw which means that one ball is drawn in the first throw. So the probability of the game ending in the third throw is 2/3 *2/3*2/3=8/27

So the probability of the game eventualyy ending is

P =2/3*(Pn-1)*(Pn-2).....so on.

X -1 0 1 2 P(X=x) 1/3 1/6 1/6 1/3

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