### Mathematics of Housie/Tambola

My first original question :P

Source of inspiration: Discussion with one of the seniors (CSE, IITB Alumnus 05-09 & Quant Analyst) during Pre-Placement Talk of a firm during Campus Placements - "I have seen your blog. Its very good. But its not original. Try and make your own questions."

Problem:
Ever played housie/tambola? I played housie once every month for 6 good years of my life. Won some prizes. Each coupon (a ticket with 15 numbers) cost Rs. 10 then. Full Housie was nearly 150 Rs, exact number depending upon the number of tickets sold.

I played one such game once again after a lapse of 6 years I think. At the end of the game I saw that 76 numbers out of 90 had been cut on the board. I couldn't help but to think what is the expected number of numbers I expect to be crossed if N (say 100) tickets have been sold. Also, I wanted to find out what is a good number of people who should be there to play housie. Of course if there are zillions of people, the game would be over in approximately 15 calls. If there is one person, we expect the game to go on very close to 90 calls. What is the number of people playing the game so that the expected number of calls would be say 70.

Of course, I played the game in a homely environment where the "dealer" did not keep any money. So, all the money collected was given back as prizes. Since every person has equal expectation to win, I should expect to get back my investment.

The questions posed are:
1) What is the expected number of numbers I expect to be crossed if 100 tickets have been sold?
2)  What is the number of people playing the game, i.e. the number of tickets sold so that the expected number of calls would be 70?

Cheers to my first original question :P

Solution: Analysis done by me and posted in comments!

1. Not many people are interested in doing the calculations.. :P
I hope my analysis is correct :)

Problem 1:
Probability that all the numbers on a given ticket are crossed in <= k calls (k>=15) is
binomial(90-15, k-15)/binomial(90,k) = {75!k!(90-k)!}/{(k-15)!(90-k)!90!} = {k.(k-1).(k-2) .. (k-14)}/{90.89.88. .. .76}
(say p(k))

Probability that at least one number is left on a given ticket in k calls is 1 - p(k)

Probability that at least one number is left in all 100 tickets after k calls = (1-p(k))^100

Probability that the game finishes in greater than k calls = (1-p(k))^100

Probability that the game finishes in greater than equal to k calls = (1-p(k-1))^100

So, expected number of calls = sigma over k from 15 to 90 (1-p(k-1))^100
= sigma over k from 15 to 90 (1-[(k-1)(k-2)..(k-15)/(90.89..76)])^100

After doing calculations for this expression in MAPLE, I will add the answer in the next comment.

Observations: Clearly, the expected value is less than 76. When the number of tickets sold is exactly one, then also, we expect the number of numbers crossed to be < 76. So, the game we played was indeed "unexpectedly" long :)

Problem 2:
The expression formed using logic as in the previous problem is as follows:
70 = sigma over k from 15 to 90 (1-[(k-1)(k-2)..(k-15)/(90.89..76)])^N
Solve for N.

2. Doing calculations using MAPLE

So, If 100 people are playing or 100 tickets are sold, we expect 67.52 numbers to be crossed.

If 53 people would have played, we expect 70 numbers to be crossed.

3. @ Pratik: I like this problem. I have a question regarding your solution. I think in your solution you have implicitly assumed that K calls for each player are independent, only then you can write Probability that at least one number is left in all 100 tickets after k calls = (1-p(k))^100. I think all we can assume is that N tickets are independent. For example consider following 2 events:

E1 = At least one number is left in ticket 1 after 15 calls
E2 = At least one number is left in ticket 2 after 15 calls
Since these 15 calls are 'same' for player 1 and player 2, I think E1 and E2 are not independent. Please correct me if I am wrong.