# Expected Number of Trials until Success

Consider the following famous puzzle.

*In a country, all families want a boy. They keep having babies till a boy is born. What is the expected ratio of boys and girls in the country?*

This puzzle can be easily solved if we know following interesting result in probability and expectation.

**If probability of success is p in every trial, then expected number of trials until success is 1/p**

**Proof:** Let R be a random variable that indicates number of trials until success.

The expected value of R is sum of following infinite series E[R] = 1*p + 2*(1-p)*p + 3*(1-p)^{2}*p + 4*(1-p)^{3}*p + ........ Taking 'p' out E[R] = p[1 + 2*(1-p) + 3*(1-p)^{2}+ 4*(1-p)^{3}+ .......] ---->(1) Multiplying both sides with '(1-p)' and subtracting (1-p)*E[R] = p[1*(1-p) + 2*(1-p)^{2}+ 3*(1-p)^{3}+ .......] --->(2) Subtracting (2) from (1), we get p*E[R] = p[1 + (1-p) + (1-p)^{2}+ (1-p)^{3}+ ........] Cancelling p from both sides E[R] = [1 + (1-p) + (1-p)^{2}+ (1-p)^{3}+ ........] Above is an infinite geometric progression with ratio (1-p). Since (1-p) is less than, we can apply sum formula. E[R] = 1/[1 - (1-p)] = 1/p

**Solution of Boys/Girls ratio puzzle:**

Let us use the above result to solve the puzzle. In the given puzzle, probability of success in every trial is 1/2 (assuming that girls and boys are equally likely).

Let p be probability of having a baby boy. Number of kids until a baby boy is born = 1/p = 1/(1/2) = 2 Since expected number of kids in a family is 2, ratio of boys and girls is 50:50.

Let us discuss another problem that uses above result.

**Coupon Collector Problem:** *Suppose there are n types of coupons in a lottery and each lot contains one coupon (with probability 1 = n each). How many lots have to be bought (in expectation) until we have at least one coupon of each type.*

The solution of this problem is also based on above result.

Let X_{i} be the number of lots bought before i’th new coupon is collected.

Note that X_{1} is 1 as the first coupon is always a new coupon (not collected before).

Let ‘p’ be probability that 2nd coupon is collected in next buy. The value of p is (n-1)/n. So the number of trials needed before 2nd new coupon is picked is 1/p which means n/(n-1). [This is where we use above result]

Similarly, the number of trials needed before 3rd new coupon is collected is n/(n-2)

Using Linearity of expectation, we can say that the total number of expected trials = 1 + n/(n-1) + n/(n-2) + n/(n-3) + .... + n/2 + n/1 = n[1/n + 1/(n-1) + 1/(n-2) + 1/(n-3) + ....+ 1/2 + 1/1] = n * H_{n}Here H_{n}is n-th Harmonic number Since Logn <= H_{n}<= Logn + 1, we need to buy around nLogn lots to collect all n coupons.

**Exercise:**

1) A 6 faced fair dice is thrown until a ‘5’ is seen as result of dice throw. What is the expected number of throws?

2) What is the ratio of boys and girls in above puzzle if probability of a baby boy is 1/3?

**Reference:**

http://www.cse.iitd.ac.in/~mohanty/col106/Resources/linearity_expectation.pdf

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.