Problem
Consider an experiment of tossing a coin repeatedly until the outcomes of two consecutive tosses are same. If the probability of a random toss resulting in head is 1/3, then find the probability that the experiment stops with head.
Solution
To find the probability, let’s list down a few sequences of tosses:
- HH
- THH
- HTHH
- THTHH
- HTHTHH
- THTHTHH
- … and so on
Now, we simply need to add the individual probabilities of each case, and we’re done.
Let p the probability of a head and q be the probability of a tail. Let’s list down the probability of each sequence above.
| Sequence | Probability |
|---|---|
| HH | p² |
| THH | p²q |
| HTHH | p³q |
| THTHH | p³q² |
| HTHTHH | p⁴q² |
| THTHTHH | p⁴q³ |
| … | … |
So, the required probability is p² + p²q + p³q + p³q² + p⁴q² + p⁴q³ + …, i.e. we get an infinite sum.
Now, this can be expressed as a sum of two infinite geometric progressions (GP).
Here’s how: (p² + p³q + p⁴q² + …) + (p²q + p³q² + p⁴q³ + …)
Each GP has the common ratio pq. Using the standard formula for the sum of an infinite GP, the above expression becomes p²/(1 – pq) + p²q/(1 – pq).
This can be simplified further as p²(1 + q)/(1 – pq). Now, we’ll substitute the values: p = 1/3 and q = 2/3.
We’ll get the required probability as (1/3)²(1 + 2/3)/(1 – (1/3)(2/3)), which simplifies to 5/21.
Comments
Not much meat in the problem. Such probability problems based on infinite sums are quite common in various textbooks.