Gambler’s Ruin: A gambler starts with k units of money. After each play of the g

Question

Gambler’s Ruin: A gambler starts with k units of money. After each play of the game, he either wins 1 unit with probability p, or loses 1 unit with probability 1 p. Each play is independent of any other play. The gambler continues until he has N units of money (N > k), or he goes broke. 1. Suppose k = 4 and N = 7. What is the probability that he has 5 units of money after 5 plays? 2. Let pk denote the probability of going broke after starting with k units. Derive an equation expressing pk in terms of pk 1, pk+1, and p. What are the values of p0 and pN? 3. (Optional) Solve the above equation for pk in terms of N, k and p.

Explanation / Answer

Solution

Part 1

Given k = 4 and N = 7, P(he has 5 units of money after 5 plays)

= P(he gains in 3 plays and loses in 2 plays) [Note that this is only possibility]

= 5!p3(1 - p)2 ANSWER [Note: Factor 5! is necessary because he can win and lose in any order. Also note that there is no possibility of his going broke before the 5th play since this can happen only if he loses all the first 4 plays, but he can lose only 2 plays. ]

Part 2

Given pk = P(of going broke after starting with k units).

Going broke after starting with k units is possible as follows:

(Win 0 and lose k) or (Win 1 and lose k + 1) or (Win 2 and lose k + 2) or …… this will go on indefinitely.

So, pk = p0(1 - p)k + p1(1 - p)k+1 + p2(1 - p)k+2 + …….. to infinity

= (1 - p)k/{1 – p(1 - p)} [the above series is a Infinite Geometric series with

a = (1 - p)k and r = p(1 - p)]

So, pk = (1 - p)k/{1 – p(1 - p)}

Similarly, pk+1 = (1 - p)k+1/{1 – p(1 - p)} and pk-1 = (1 - p)k-1/{1 – p(1 - p)}.

Connecting the above three and simplifying,

pk = (pk+1 + pk-1)/(2 + p2 – 2p) ANSWER

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