This Diwali, Google Pay came out with a scheme, collect 5 stamps and win Rs. 251. These stamps randomly occurred on payments of over Rs. 35 and on scanning Diwali items. After just 2 days, people across the country flooded social media for the exchange of Flower and Rangoli stamps. Google introduced an ad where it told that the only way of receiving the Rangoli stamp was through merchant payment.

This reminded me of the Coupon Collector Problem, the standard problem that anyone who has taken up a course in Expectations and Probability knows about. Before I explain it, let's understand the Google Pay Coupon Collector.

There were 5 stamps, namely, Diya (Earth Lamp), Jhumka (Ear-ring), Lantern, Flower, Rangoli and they could be earned through purchases over Rs 35, recharges and scanning Diwali items.





Now let us define the coupon collector problem:

A certain brand of cereal always distributes a coupon in every cereal box. The coupon chosen for each box is chosen randomly from a set of n distinct coupons. A coupon collector wishes to collect all n distinct coupons. What is the expected number of cereal boxes must the coupon collector buy so that the coupon collector collects all n distinct coupons?





Hmm... Sounds familiar right?

Now, let's see the solution:

E(x) = P(x)E(x) + P(~x)(1 + E(x+1))

Here x is the number of stamps we already have and thus E(x) is the Expected Number of moves for collecting the rest of the stamps, give we have x stamps.

What this means is that if the coupon is of type from those that we already have we still need E(x) boxes and if it is a coupon we don't have, it is basically, 1 move + Expected moves after we have x+1 type of coupons.

So, the expected number of transactions is:

E(0) + E(1) + ... + E(n)

Just that in this case these coupons aren't equally likely. Chances of getting the coupons that I calculated using the relative number of stamps with many people are as follows





P(Diya) = 0.25, P(Jhumka) = 0.25, P(Lantern) = 0.25, P(Flower) = 0.15, P(Rangoli) = 0.1

Yikes! Also, it was impossible to get Rangoli from non-payment methods so Let's see the expected number of payments for a win, assuming people got the rest of the four stamps

E(Rangoli_left) = P(Rangoli)(1) + P(Not_Rangoli)*(1 + E(Rangoli_left)

Which means, at least 1/0.1 => 10 transactions for a Rangoli, each of around 35 Rs.

So now, since we have this, we calculate, the net gain is 251 - 35*10 => -99 per win!

Damn! This means that in this lottery, the chance of losing was much more than that of a win.

The scheme hasn't ended yet and a lot of people are trying to get Rangoli stamps.

Let us summarize the benefits Google has in this:

Increased user engagement by preferring Google Pay for routine payments with the intention of getting a Rangoli Gamified the payment gateway making users expect more than a few bucks Publicity from people asking for Flower/Rangoli stamps on the social media

That is really smart and effective and it is the basic principle behind all lotteries, casinos and luck-based games.

Hats off to the Google team to pull off something really cool!

How do you think is this? Did I get the mathematics right? Do you want to know more about probability in real life? Tell me in the comments!

(PS: Also, do tag the G-Pay team)



