Interesting Gambling Game

Mar 2009
184
4
A gambler bets repeatedly on the toss of an unfair coin which lands heads with probability, p and tails with probability, q=1-p (p can not = 1/2). If the coin lands heads, the gambler wins £1, whereas if the coin lands tails, the gambler loses £1.
Starting with £k, the gambler plays the game repeatedly until he/she either goes bust or reaches £N (N>=k). For k=0,1,2,....N let Ak be the event that the gambler starts with £k and eventually goes bankrupt. Assume that the outcomes of the tosses are independent of each other.

(i) What are the values of P(A0) and P(AN)?

(ii) For k=1,2,....N-1, express P(Ak+1) in terms of P(Ak) and P(Ak-1)
 
Jun 2009
33
0
Hello.

For the first question you want the probability of going bankrupt when you start with zero pounds - but zero pounds is bankrupt already so this probability is 1.

You also want the probability of going bankrupt starting with N pounds. But when you're at N pounds you've reached your goal and don't bet any more, so you can't go bankrupt. So this probability is 0.

For the second question you have k pounds, and this is strictly between bankrupt and the goal, so you are going to bet. You want the probability of going bankrupt from k pounds. This can happen in two mutually exclusive ways. Either you win the first hand (with probability p) and you're up one pound, then go bankrupt from k+1 pounds (with probability \(\displaystyle P(A_{k+1})\)), or you lose the first hand (with probability q) and then go bankrupt from k-1 pounds (with probability \(\displaystyle P(A_{k-1})\)). So, we have:

\(\displaystyle P(A_k)=pP(A_{k+1})+qP(A_{k-1})\)

Now just solve for \(\displaystyle P(A_{k+1})\)
 
Mar 2009
184
4
Thank you, that makes sense :)

Also....

Let Mk be the number of coins tossed until the gambler is bankrupt or has £N. For k=1,2,....N-1, express E(Mk+1) in terms of E(Mk) and E(Mk-1)

Would this be similar thinking?
 
Jun 2009
33
0
Yes, quite similar. From k either you win the first hand (probability p) then ask "how long do I expect to wait til the game finishes from k+1?", or lose the first hand (probability q) and then ask "how long do I expect to wait til the game finishes from k-1?". BUT, don't forget - in this first hand you have flipped one coin already. So it's the same equation as before with expectations in place of probabilities, but plus one on the RHS.
 
Mar 2009
184
4
Thanks, i think i get that.... just checking, what do you mean by +1 on the RHS?
 
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