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On the National Mexican Lottery, II

The jackpot this week is up at 300 million (MXP)! A few million more and this game turns into a fair or favorable game, how cool!

In my last post, I mentioned that some of my friends and students said that one could be really lucky and win the jackpot if one bought the first few tickets and won: 

"Unless he got truly lucky and won the jackpot before he spent too much, as in buying the first few tickets, and then quit, they argued!"

And there wasn't really much I could say about it.  It is true after all that one could be so lucky, with minuscule probability.  However, "what is the average number of tickets you have to buy before you win the first time, if you buy the 6-choice/7-choice/etc. repeatedly?" was a question that people kept asking me with some insistence.  Although to me it seemed somewhat evident that you needed to buy about  \binom{56}{6} tickets on average for the 6-choice, my friends and students weren't convinced until I showed them the mathematics that supported this.

It is really not difficult to calculate such if one understands what expected value is.  So let us assume the words





etc., are Bernoulli-trial strings (there really are two possibilities, the binary win or lose), and they are allowable if we stop after the first win.  For the 6-choice, each word has probability  (1 / \binom{56}{6}) \cdot (1 - (1 / \binom{56}{6}))^{n-1}  because the nth win is preceded by  n - 1 losses.

The expected value is:

 (1 / \binom{56}{6}) \sum_{n=1}^{\infty} n \cdot (1 - (1 / \binom{56}{6}))^{n-1}

One recognizes this as a convergent geometric series* (all probabilities are less than one so they lie inside the radius of convergence), and thus the above equals

 (1 / \binom{56}{6}) \cdot \frac{1}{(1 / \binom{56}{6})^2} = \binom{56}{6} ,

the sum having been substituted adequately.  Confirming my "far-out" claim (to me really unsurprisingly), you have to wait an average of  \binom{56}{6} or about thirty-two million tickets before you'll see the first win.

This idea can be extended for the 7, 8, 9, 10-choice and so on.


The series representation of the function

 \frac{1}{(1-x)^2} = \sum_{n=0}^{\infty} n \cdot x^{n-1} with radius of convergence  -1 < x < 1 .  

This is obtainable by taking the derivative of the series representation for:

 \frac{1}{1-x} = \sum_{n=0}^{\infty} x^n with radius of convergence  -1 < x < 1 .

  1. Ben
    January 3rd, 2009 at 16:52 | #1

    The formula you give for the expected value assumes the bettor isn't keeping track of what tickets s/he's buying and therefore might buy the same combination multiple times. More properly, the  n-1 losses should be expressed as  (1 - 1 / \binom{56}{6}) \cdot (1-1/[\binom{56}{6} - 1]) \cdot \ldots \cdot (1-1/[\binom{56}{6} - (n-2)]) , and the summation only needs to be from  n=1 to  \binom{56}{6} (since the total number of losses can range from 0 to  \binom{56}{6} - 1 , because if you buy  \binom{56}{6} - 1 losing tickets, the  \binom{56}{6} th is guaranteed to be a winner). Likewise, to express the win, instead of putting  1/\binom{56}{6} on the outside, we need a  1/[\binom{56}{6} - (n-1)] term inside the sum, because by the time the bettor wins on the nth ticket, s/he's already exhausted  n - 1 possible other combinations.

    I'll leave it to you to compute the actual numerical value of that sum, unless I feel like doing the math later.

  2. Ben
    January 3rd, 2009 at 20:03 | #2

    Okay. I did the math. The equation simplifies to  \sum_{n=1}^{\binom{56}{6}} n/\binom{56}{6} . The numerator sums to  \binom{56}{6}(\binom{56}{6} + 1)/2 , so the whole thing is  (\binom{56}{6} + 1)/2 , which makes sense, because if you're randomly picking combinations of 6 numbers from 56 but never repeating a combination, on expectation you have to go through half of the possible combinations before you hit the winning one.

  3. Carlos
    January 4th, 2009 at 11:27 | #3

    Good man! You are entirely right. Actually what I was thinking was choosing the same combination over and over again over a period of many same-games (or, put another way, many different people picking tickets, possibly repeated, for that one game). However, one may think one saves a lot (nearly half the money?) by keeping track of the different combinations, as you propose, but one must not forget that the expected profit is the same in either circumstance! So either way you play it you end up losing the same amount of money on average if the jackpot is not fair, keeping track on that one-game or not keeping track on many same-type games. Thanks for taking the time to do the math, my friend!

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