A gambler has $2 and needs to increase it to $10 in a hurry. He can play a game

Question

A gambler has $2 and needs to increase it to $10 in a hurry. He can play a game with the following rules: a fair coin is tossed; if a player bets on the right side, he wins a sum equal to his stake, and his stake is returned; otherwise he loses his stake. The gambler decides to use a bold strategy in which he stakes all his money if he has $5 or less, and otherwise stakes just enough to increase his capital, if he wins, to $10. Let X0 = 2 and let Xn be his capital after n throws. Prove that the gambler will achieve his aim with probability 1/5. What is the expected number of tosses until the gambler either achieves his aim or loses his capital?

Explanation / Answer

By the optional stopping theorem, the expected value when the game is finished is equal to the expected value when the game starts. He ends with either $0 or $10, and if p is the probability that he ends in $10, then

0(1p)+10p = E(0) = 2, or 10p = 2

p=1/5.

Using the values h0 = 0, h10 = 1 the system has a unique solution, which is easily found by simple substitution. We obtain h2 = 1/5. Now denote by µi the expected time before reaching either 0 or 10 given that the process starts at state i. Then µ0 = 0, µ10 = 0, and the following system holds:
µ2 = 1+ 1/2µ4 +1/2µ0
µ4 = 1+1/2µ8 +1/2µ0
µ6 = 1+1/2µ2 +1/2µ10
µ8 = 1+1/2µ6 +1/2µ10.

This is also easily solved for µ2. The result is µ2 = 2.

The program below can be used for the simulation.

-------------------------------------------------------------------------

%I will enumerate the states in the following

%order: 0 2 6 10 4 8, and use the program

%stop_at.

rand(’seed’,324)

p=[0 1 0 0 0 0];

P=zeros(6,6);

P(2,[1,5])=1/2;

P(3,[2,4])=1/2;

P(5,[1,6])=1/2;

P(6,[3,4])=1/2;

P(1,1)=1;

P(4,4)=1;

A=[1 4];

a=0;

for j=1:1000

x=stop_at_set(p,P,A,100000);

a=a+length(x)-1;

end a=a/1000
The above gave the simulated value µ2 = 1.9760.

cheers!

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.