## What’s the chance that you ever take the lead?

Here’s one of an occasional series of cute problems, which I offer as a challenge for whiling away a wild and rainy Saturday afternoon..

You enter a competition which consists of a never-ending series of contests. The probability that you win any single contest is p, and the outcomes of the contests are independent of one another.

Let X be the probability that you ever take the lead in the competition. What is X in terms of p, for any value of p?

UPDATE: Since a correct answer has now been posted, here is my solution:

Consider the first contest: the probability that you win it is p and if you do you take the lead straight off.

If you lose the first one, with probability (1-p), you are down by 1. Now you must (a) make up the deficit and (b) go on to take the lead. Clearly the probability of (a) is just X (the same as getting ahead from a level start). The probability of (b) is also X.
Hence X=p+(1-p)X^2.

There are two solutions of this quadratic equation: X=1 and X=p/(1-p). But the answer must be a probability so cannot exceed unity. Hence if p>1/2 then X=1, in accord with intuition: in the long run you’d expect to lead sometime if p>1/2. If p<1/2 then the other solution is correct. The two solutions match at p=1/2.

Simples.

### 7 Responses to “What’s the chance that you ever take the lead?”

1. drewancameron Says:

Probability of absorption at a fixed barrier in a one-dimensional random walk. [I assume there’s only one other player.]

• telescoper Says:

2. drewancameron Says:

p/(1-p) for p<1/2, 1 otherwise

• telescoper Says:

That’s the right answer. Now for a full worked solution..

• drewancameron Says:

I taught the tutorial class for “Probability and Stochastic Modelling II” last year at QUT where the simple random walk and branching processes are two key pieces of the curriculum because both use the same type of solution method: set up a recurrence relation, apply induction, apply boundary conditions. In this case we start with the two-sided boundary case, find the solution, and let the lefthand boundary tend to infinity.

• telescoper Says:

There’s actually a very easy way to tackle this which just requires solving a quadratic equation..

• Anton Garrett Says:

I was going to do it by the method of induction; supposing not in the lead after the nth game, what’s the prob of still not being in the lead after the (n+1)th?