Concepts: The Math Behind The Lotto Max

On 22 Jun 2021, Canada’s Interprovincial Lottery Corporation awarded its largest ever Lotto Max prize pool in the amount of $140 million ($70 million for the grand prize and 70 $1 million MaxMillion prizes).

It thus seems timely to have a look at the math behind lotteries.

The website states that the odds of winning the Lotto Max are 1 in 33,294,800. This is correct to a point, but misleading.

Let’s have a look at the rules of the Lotto Max:

  1. Players choose 7 numbers out of 50
  2. Numbers cannot be repeated
  3. Numbers are automatically sorted into ascending order
  4. Each play buys 3 lines
  5. Each play costs $5

Seeing that players choose 7 out of 50 numbers that are non-repeating, the equation for the total number of possible combinations (this is different from permutations where the order in which the numbers appear is significant) when playing the Lotto Max is:

50! / (7! x 43!)

The ! sign is a factorial. For example, 3! would be 3 X 2 X 1 = 6.

We get 50! as there are 50 numbers to choose from (1 to 50). We get 7! as there are 7 numbers per play. As these 7 numbers are non-repeating, this means that should the 7 numbers be chosen at random, there are 7! ways in which they can be arranged. Hence, the total possible 7-number permutations would be divided by this.

We get 43! (which is really 50-7) as since only 7 numbers are played, the other 43 are discarded.

This gives us 99,884,400. So where does the official 33,294,800 come from?

This is a product of algebraic reduction. We see that though the total possible combinations in the Lotto Max is 99,884,400, each play gets a player 3 combinations of 7 numbers (this is assuming that none of the 3 plays ends up being identical with anyone else’s 3 plays – though that does happen, as it did for this draw, with there being 2 winners).

3 / 99,884,400 gets reduced to 1 / 33,294,800 or 1 in 33,294,800 or 33,294,799 to 1.

This looks like it makes sense till you see that by giving a reduced figure, the real numbers are being obscured from the players. One is not just up against 33 million other numbers, but 99 million. That is a big difference.

It is also interesting that the odds of about 33 million, correspond to the population of Canada at the time the draw was started in 2009. Perhaps to show that everyone in Canada had a chance to win?

So, how does one get the amount of $50 million for the Lotto Max? In order to get an idea of what the Lotto Max can pay out without resulting in a loss of revenue, we would need to look at the breakeven payout.

This is how we calculate the breakeven payout:

33,294,800 – 1 = 33,294,799

($5 x 33,294,799) / 1 = $166,473,995.

So, assuming all things being equal and random, so long as the grand prize does not reach $166,473,995, the house is guaranteed a profit.

We can prove this by calculating the expectation for these values:

3 / 99,884,400 = 0.000000030035 (this is the probability of winning the prize)

99,884,397 / 99,884,400 = 0.999999969965 (this is the probability of losing your $5 play wager)

(166,473,995 x 0.000000030035) – (5 x 0.999999969965) = 0 (this means a breakeven expectation)

One might wonder, does the number of MaxMillion prizes affect my chances of winning? The answer is no. This is because there is nothing that excludes the possibility of the same numbers being drawn for each prize. This means that in essence, all prize draws have the same probability, being independent events. Having more MaxMillion prizes just means that one would be repeating the same draw, again and again with the same probability of winning.

This is how we calculate the expectation – which means how much a player is expected to lose per play, assuming a $70 million prize. This time, we will use the reduced figures and you will see we get the same probabilities. Try it out for yourself:

($70,000,000 X (1/33,294,800)) – ($5 X (33,294,799/33,294,800))

= 2.10 – 4.99

= -2.89

This means on average, every play is expected to gain the house $2.89. This in effect means that most players would in the long run be out $2.89 for every $5 play.

And of course, this wouldn’t be complete without considering the standard deviation of expectation.

PaymentProbabilityExpectation WorkingExpectationStd. Dev. Of Expectation WorkingStandard Deviation of Expectation
-50.99999997-5  x 0.99999996996528 =-4.99999980.99999996996528 x ((-5-(-2.89)))^2 =4.42
70,000,0000.0000000370,000,000 x 0.00000003003 =2.10243040.00000003003 x ((-5-(-2.89)))^2 =147170140.90
   -4.9999998 + 2.1024304 = sqrt(4.42021412571428 + 147170140.852618)
   $                                  (2.898) $                                                                    12,131.37
      
 N =1,000,000-2.89 x 1,000,000 = 12,131.37 x sqrt(1,000,000) =
   $                    (2,897,569.44) $                                                           12,131,370.30

This means that on average, with a sample size of 1,000,000, meaning 1 million plays, the average revenue for the lottery hosting company would be $2,897,569.44. The standard deviation of expectation for 1 million plays is $12,131,370.30.

The table below shows what this translates to in practice:

ZPercentile+ve-ve
0.6775%5,230,448.66(11,025,587.54)
1.0385%9,597,741.97(15,392,880.85)
1.2890%12,630,584.54(18,425,723.42)
1.64595%17,058,534.70(22,853,673.58)
2.3299%25,247,209.65(31,042,348.53)
3.0999.9%34,588,364.78(40,383,503.66)

We get the figures above by taking the mean of $2,897,569.44 and adding or subtracting, depending on whether you are measuring the positive or negative results, by the product of the standard deviation of expectation of $12,131,370.30 multiplied by the Z-score.

For example, let us take a Z-score of 2.32 at 99%. This would mean that the sum of 99% of all winning results, all things being equal, would not exceed $34,588,364.78 while 99% of all losing results, all things being equal, would not exceed -$40,383,503.66.

I hope you enjoyed learning about this. Take care!

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