Each morning Joe leaves his house and goes for a run Hes equ

Each morning, Joe leaves his house and goes for a run. He’s equally likely to leave either from the front or back door. Upon leaving the house, he chooses a pair of running shoes (or goes barefoot if there are no shoes at the door from which he departed). On his return, is equally likely to enter, and leave his running shoes, either by the front or back door. If he owns a total of k pairs of shoes, what’s the long-run proportion of time that he runs barefooted?

Solution

The long run proportion of time that he runs barefooted= 1 / (1+2^k)

Method:Let A_n be the event that in the nth step,he was barefoot and let P_n be the corresponding probability.

P(A_n)=P(A_n|A_n-1)P(A_n-1)+P(A_n|A_n-1^c)P(A_n-1^c)

P_n=(1/2)*P_n-1+1/2 *1/2^k *(1-P_n-1) =a+b P_n-1 (say)

=a+b(a+bP_n-2)=....= a(1+b+...+b^b-2)+b^n-1 P₁ =a(1-b^n-1)/(1-b )+b^n-1 * P₁ =a/1-b (as n tends to infinity)

=1/(2^k+1)