The Story of Golden Tickets
Probability stories — 01
Back when we were kids, I remember a certain chocolate company in Sri Lanka had a contest ( or a prize draw competition). Each chocolate wrapper had a generic figure of a family member (father, mother, uncle etc.) printed on the hidden side of it. The goal was to collect all the family member figures to become eligible for a prize. Although I do not remember many details of the contest (I never won a prize too), this is a very common type of contest all over the world, where people collect things in the hunt for an ultimate prize. But how the heart of such kind of a well-planned arrangement (made by the company) can be seen with probability theory? Here, I try to discuss that!
Mr. Wonka (yes, exactly him!!!) decided that to regain sales and popularity of his milk chocolate bar, he needs to have a contest. (As suggested by his dearest friend, Mr. Bucket). So, he devised a plan to get some golden tickets inside the chocolate bars. “Well…” he thought, “how should I design this contest…so that more people would buy my chocolate…”. As a start, Mr. Wonka decided that he would have a ticket numbered either 1, 2 or 3 inside a chocolate bar. One would have to collect a full set of tickets (one from each number) in order to become eligible for a prize. But how much of a fortune would this bring about? That would simply depend on the amount that people buy.
Let us say people would have to buy N chocolate bars to find all the 3 tickets that make the set. We may take N as the average amount of bars they have to buy for the calculation. Amidst all the rush to find the tickets, a little kid from Sri Lanka tries to make the set.
01. He starts by buying his first chocolate bar. There he finds a ticket numbered 1, 2, or 3 (with a probability of 1). Now he needs to find the rest. If he is lucky, he will find another different number on his next buy. What is the probability that he would find a new number in the next chocolate bar? We can say that the probability of getting a 1 (P(1)) = P(2) = P(3) = 1/3 = p, as we only have those 3 numbers. Then, for any number (1, 2, or 3) (where q = 1-p),
Finding the desired number on the 1st try = pFinding the desired number on the 2nd try = p*q = pq (; we fail once and succeed on the 2nd try. Failing on the 1st try does not affect the second one)" " " " on the 3rd " = p*q*q = pq^2 (;likewise…)" " " " on the 4th " = p*q^3..." " " on the nth " = p*q^(n-1)" " " on the (n+1)th " = p*q^(n)...
The average successful number of tries (m) (or expectation ) of finding the desired number would be, (recall that, E(X) = X_1 * P(X_1) + X_2 * P(X_2) + … + X_n * P(X_n) )
m = p + 2pq + 3pq² + 4pq³ + …We can find an answer by following a small trick,
mq = pq + 2pq² + 3pq³+ 4pq⁴ + …Then by subtracting the two equations,
m-mq = p + pq + pq² + pq³ + pq⁴ + …which turns out to include sum of a geometric series!
m-mq = p (1 + q + q² + q³ + q⁴ + …) = p * (1/1-q) = 1 Hence, m = 1/(1-q) = 1/p
So, we expect to find the desired number at an average of 3 tries (when p = 1/3 ). (Which equals buying 3 chocolate bars). We now go back to our little kid in Sri Lanka.
02. Since he already has a number, the probability of finding a new number (any from the remaining two) on his second buy is 2/3. That would happen on an average number of 3/2 tries (recall we got m = 1/p). But the search does not stop there. He needs the final missing number as well. That would take an average of 3/1 tries, with a probability of 1/3 (the last remaining number from the three).Now Mr. Wonka can reach the final answer. The average number (expected amount) of chocolate bars (N) a person would have to buy is,
N = 1 + 3/2 + 3 = 5.5The little kid would have to buy 5.5 chocolate bars, (which is actually 6, unless the shopkeeper sells him halves.) to make the complete set. But, remember this is an average. If he is lucky he would find the set in a minimum of 3 bars. Or else, he may go up to 7 bars, 8 bars, 9, 10, or even higher.
One may think that this is not a big number. Yet, Mr. Wonka has managed people to buy more already (or at least motivated to buy more). If Mr. Wonka had 1000 customers who used to buy barely 2000 chocolate bars monthly, now he may have an additional 1000 sales to the least (assuming everyone would be interested in collecting the tickets) and an average increase of about 4000 chocolate bars. After all, Mr. Wonka could surely distribute only a handful of complete sets which would make it even harder to find a complete set. You can imagine now where this would be heading…
But that is not the most interesting part. Suppose Mr. Wonka decides to include 4 numbers (A complete set would now be tickets numbered 1, 2, 3 and 4). Then N would be 8.33. For 5 numbers it would be ~11.4. If you could work it out in a similar fashion as above, we can see that it follows a pattern; the Harmonic Series. It is considered to have a sum that does not have a finite limit. The first n terms of the series have an approximation of nln(n) + nγ + 0.5 (where γ ~ 0.577). If we visualize how the results would turn out as the ticket numbers go higher as below, we can see that at the number of 20, the average buys would be about 70 which definitely would create good amounts of income. I personally think if the number is too high, then the motivation to buy more would drop a bit. But, if the company is creative they could come up with a good maneuver. (I faintly recall that in the contest we had in Sri Lanka that I mentioned in the beginning, there were at least 10 figures to make a full set).
What if Mr. Wonka comes up with the idea of including unnumbered tickets (like, “Try Again!” ones) as well. There could be other such interesting variables to this problem. What do you think? Leave a comment about what you think too! Thank you for reading.
Reference: 50 Challenging Problems in Probability — Frederick Mosteller
