# Interesting Gambling Game

#### sirellwood

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)

#### Aileys.

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 $$\displaystyle P(A_{k+1})$$

#### sirellwood

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?

#### Aileys.

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.

#### sirellwood

Thanks, i think i get that.... just checking, what do you mean by +1 on the RHS?

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