Tom goes to the casino with $5. The game consists of Tom tossing a fair coin. If

Question

Tom goes to the casino with $5. The game consists of Tom tossing a fair coin. If the result of a toss is Heads, the casino pays Tom $1 while if the result is Tails, Tom pays the casino $1. Thus, after each coin toss, the amount of money that Tom has hereinafter referred to as his wealth - either increases by $1 or decreases by $1. The (independent) coin tosses continue until one of the following two events occurs: Tom's wealth increases to $8 or Tom's wealth decreases to $0. At this point, Tom goes home (with wealth $8 or $0 as the case may be). What is the probability that Tom goes home after three coin tosses? What is the probability that Tom goes home after five coin tosses? Conditioned on the event A = {Tom tosses the coin at least 5 times}, what is the conditional pmf of X, Tom's wealth after 5 tosses have occurred? Note that the increase or decrease in Toni's wealth due to the result of the fifth toss is included in X.

Explanation / Answer

1) If he goes home after 3 tosses then all must be heads , so the prob is 1/8 2) 5 tosses means he may win or lose before going home, loses : all tosses are tails : 1/32 wins: this must be 4 heads + 1 tails , tail is not the last(or second last) since then he will get 8 before itself, so tails is one of 1,2,3 so 3 * 1/32 = 3/32 Total : 4/32 = 1/8 3) Since tom tosses the coin atleast 5 times it cannot be HHH.. all others are valid., So P(X=0 and A) = 1/32 (all tails) P(X=1 and A) = 0 (since we will require tails - heads = 4 but heads + tails = 5) P(X=2 and A) = 5/32 (now here tails-heads=3 , heads +tails = 5 -> heads = 1 tails = 4) and any order is fine P(X=3 and A) = 0 (again by the argument of x=1) P(X=4 and A) = 5C2 / 32 = 10/32 (tails-heads=1 , head+tail=5 so head =2 tail =3 again any order is fine) P(X=5 and A) = 0 again by argument of x=1 P(x=6 and A) = 9/32 (now head=3 tail=2 but first 3 cant be heads, ways = 5C2-1 (only one is possible HHHTT)) P(X=7 and A) = 0 P(X=8 and A) = 3/32 (4 heads 1 tails but since he does play for 5 turns the tail cant be 4th or 5th so 3 ways) What is P(A) = sum of the above probs = 28/32 So P(X=0|A) = 1/28 So P(X=1|A) = P(X=3|A) = P(X=5|A) = P(X=7|A) = 0 P(X=2|A)=5/28 P(X=4|A) = 10/28 P(X=6|A)=9/28 P(X=8|A)=3/28 comment if you have any doubts

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