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Advanced Lottery Math

Updated on August 09, 2015
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TR Smith is a product designer and former teacher who uses math in her work every day.

Building on the previous math tutorial for calculating lotto odds, Basic Lottery Math, in this tutorial we will see how to calculate the probability of matching some of the numbers drawn and winning a lower-tier prize. These computations can be trickier than figuring the probability of winning a jackpot, since there are many ways a ticket can bear a partial match, whereas there is only one way to make a perfect match and win the jackpot. But once you know how to count the number of partial matches, you can quickly and accurately calculate the probability and odds.


General Formula for Partial Match Probability

The probability of making a partial match in a lottery is

[number of tickets with a partial match] ÷ [total number of different tickets]

The denominator in this expression, the total number of different lottery ticket combinations, was calculated in the previous tutorial. The numerator, the number of tickets bearing a partial match, depends on how many numbers match and whether the game is single-matrix or double-matrix.


Single-Matrix Lottery Partial Matches

If the game's structure is selecting N distinct numbers from 1 to P, then the number of ways to make a match only K out of N is

(N c K)*(P-N c N-K)

or in words, it's the number of ways to choose K out of N times the number of ways to choose N-K out of P-N. This is because for every way to choose K out of N correct numbers, you must also have N-K incorrect numbers drawn from a set of P-N numbers. The probability of making such a partial match is

(N c K)*(P-N c N-K)/(P c N)
= N!² * (P-N)!² / [ P! * K! * (N-K)!² * (P-2N+K)! ]

Examples of Single-Matrix Partial Matches

Consider a lottery in which you select 6 different numbers from 1 to 49. The number of ways your lottery ticket can match 4 of the 6 numbers drawn is

(6 c 4)*(43 c 2)
= 15*903
= 13545

There are 15 ways to get 4 of the 6 numbers correct, and 903 ways to choose the remaining 2 incorrect numbers. Since the total number of different lottery ticket combinations is 13983816, the probability of matching 4 out of 6 is

13545/13983816
= 0.00096861972
= 0.096861972%

To express this in the odds format "1 in X" you take the reciprocal

13545/13983816 = 1032.4

Therefore, the odds are 1 in 1,032.4. To express this in the format "Y to 1" you subtract 1 from the number, so the odds can be alternatively written as 1,031.4 to 1.


Now consider a lottery where you pick 5 distinct numbers from 1 to 72. The number of ways you can get 4 out of 5 numbers correct is

(5 c 4)*(67 c 1)
= 5*67
= 335

That's 5 ways to get a subset of 4 numbers correct, and 67 ways to pick the remaining incorrect number. This lottery has a total of 13991544 different combos. The probability is then

335/13991544
= 0.00002394303
= 0.002394303%

Or as odds, the likelihood is 1 in 41,765.8. As you can see, even though the lotteries in these two examples have roughly the same number of different ticket combinations, the probabilities of making a 4-number match are very different.

Double-Matrix Lottery Partial Matches

When a lottery has you select two groups of numbers from independent sets, you can make partial matches among both groups. Suppose the lottery is structured as choosing N distinct numbers from 1 to P and M distinct numbers from 1 to Q. The number of ways you can match K out of N and J out of M is

(N c K)*(P-N c N-K)*(M c J)*(Q-M c M-J)

And since the total number of different lottery number combinations is (P c N)*(Q c M), we get the probability expression

(N c K)*(P-N c N-K)*(M c J)*(Q-M c M-J)/[(P c N)*(Q c M)]

Oftentimes, you just pick one number out of a second set, that is, M = 1. This means you can either match 0 or 1 of the extra numbers, in other words, J = 0 or J = 1.

Examples of Double-Matrix Partial Matches

Consider a lottery where you must choose 4 distinct numbers from 1 to 36, plus 3 distinct numbers from 1 to 18, for a total of seven numbers. Since you are choosing them from independent sets, some numbers may be equal. The total number of possible lottery combinations is

(36 c 4)*(18 c 3)
= 48,066,480

To find the probability of making a partial match of 2-out-of-4 plus 2-out-of-3, we first compute the number of ways such a partial match can occur. This is

(4 c 2)*(32 c 2)*(3 c 2)(15 c 1)
= 133,920

Therefore, the probability is

133920/48066480
= 0.0027861
= 0.27861%
= odds of 1 in 359


In the Mega Millions Lottery, you select 5 distinct numbers from 1 to 75, and a megaball number from 1 to 15. This is a simpler case where you only draw one number from the second set. The total number of different lottery ticket possibilities is

(75 c 5)*15
= 258,890,850

The number of ways to match 3 out of 5 numbers and not match the megaball is

(5 c 3)(70 c 2)(1 c 0)(14 c 1)
= 338,100

The probability of making this partial match is

338100/258890850
= 0.001306
= 0.1306%
= odds of 1 in 766


What's the point of computing lottery probabilities?

The best way to not lose money in lotteries is simply to not play. However, if you can't resist spending $10 on lotto tickets every week, your best strategy is to play games that offer a higher chance of winning a prize. State and regional lotteries usually offer better odds than large national lotteries, though the jackpot prizes are also smaller.

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    • someonewhoknows profile image

      someonewhoknows 3 years ago from south and west of canada,north of ohio

      Wining the lottery! Good Luck with that!

    • calculus-geometry profile image
      Author

      TR Smith 3 years ago

      I'd get more value using a dollar bill to clean up a spill on my kitchen counter than using it to buy a lottery ticket.

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