I was discussing optimal lottery ticket purchasing strategies with a friend, and an interesting question came up.

Suppose you doing the following:

  • You purchase multiple tickets for one draw
  • You select the option to pick the numbers at random for all tickets

It occurred to me that if the numbers are selected at random, then it would be possible - indeed quite likely if you buy several tickets - that some number(s) may appear on multiple tickets. A quick Google confirms what I expected - that the random number selection process for my local lottery is independent for each ticket even when you buy them together and for the same draw, so this would be entirely possible.

This had me wondering, does this factor decrease your odds at all, and if it does, could one improve upon the process of randomly selecting each ticket independently to improve things? Perhaps this is just a more specific version of the general question - should you avoid repeatedly selecting the same number across multiple tickets on the same draw?

The parameters of the draw are:

  • Numbers are 1-59
  • Six numbers are drawn
  • Prizes start at three numbers, increasing in size up to all six

Having not studied maths in any depth since my college days, I'm unsure how to frame the problem mathematically, so I'm interested both from a mathematical point of view and practically.

  • 5
    $\begingroup$ It's been proven mathematically that the best loterry strategy is not to buy any tickets $\endgroup$
    – Asinomás
    Dec 30 '16 at 16:48
  • $\begingroup$ Invest the money you would spend on tickets and you'll make more. $\endgroup$ Dec 30 '16 at 16:49
  • $\begingroup$ The only odds you can really affect are the odds that someone else will share the pot with you (e.g., people love to choose birthdays, etc). You need to make some behavioral guesses, though, to have a model in order to do some math. $\endgroup$
    – pjs36
    Dec 30 '16 at 16:55
  • $\begingroup$ @pjs36 That also came up in the discussion actually. It's something that makes intuitive sense to me and didn't really need any explaining $\endgroup$
    – dbr
    Dec 30 '16 at 16:56
  • $\begingroup$ @pjs36 This is a good one. It reminds me, people like to pick things like "1, 2, 3, 4, 5, 6" thinking that if they win, they'll be the only ones who picked that, but actually A LOT of people pick those numbers. $\endgroup$ Dec 30 '16 at 16:56

The first thing to do is determine the number of possible tickets, in this case ${59 \choose 6}=45057474$, and the number of ways of winning each of the prizes:

  1. Matching 3: ${6 \choose 3} \times {53 \choose 3} =20\times 23426 =468520$ (The number of ways of choosing 3 from the winning set and 3 from the loosing set.)
  2. Matching 4: ${6 \choose 4} \times {53 \choose 2} =15\times 1378=20670$
  3. Matching 5: ${6 \choose 5} \times {53 \choose 1} = 6\times 53 =318$
  4. Matching 6: ${6 \choose 6} \times {53 \choose 0} = 1$

The probability of winning in each of these ways is determined by dividing by ${59 \choose 6}$. (The probability of winning anything at all is approximately $0.01$.) The expected value of a single random ticket depends on what the actual prizes are. As a function of these prizes we get: $$\frac{234260}{22528737}E_3 + \frac{3445}{7509579}E_4 + \frac{53}{7509579}E_5 + \frac{1}{45057474}E_6$$ Let's assume that the prizes are $E_3=\$5$, $E_4=\$100$, $E_5=\$10000$, $E_6=\$1000000$. Then we get an expected value of about $\$0.19$.

If you do this twice, where the two tickets are independent of each other, you simply get twice the expected value, about $\$0.38$.

Now, if the second ticket has distinct numbers in it, the first ticket proceeds the same way but for the second, we get a minor plague of cases. First, there are ${53 \choose 6}$ ways of choosing the second ticket in each of the cases. Then, there are seven cases (each with subcases) depending on how many of the winning digits were matched by the first ticket:

  1. The first ticket matched 0 numbers. $P={53 \choose 6}/{59 \choose 6}$

Then the second ticket could match any number of winning numbers. There are 6 unpicked winning numbers and 47 unpicked losing numbers.

Number of ways of matching 3: ${6 \choose 3}\times {47 \choose 3}$

Number of ways of matching 4: ${6 \choose 4}\times {47 \choose 2}$

Number of ways of matching 5: ${6 \choose 5}\times {47 \choose 1}$

Number of ways of matching 6: ${6 \choose 6}\times {47 \choose 0}$

  1. The first ticket matched 1 number. $P={6 \choose 1}{53\choose 5}/{59 \choose 6}$

Then the second ticket could match any number of winning numbers except all of them. There are 5 unpicked winning numbers and 48 unpicked losing numbers.

Number of ways of matching 3: ${5 \choose 3}\times {48 \choose 3}$

Number of ways of matching 4: ${5 \choose 4}\times {48 \choose 2}$

Number of ways of matching 5: ${5 \choose 5}\times {48 \choose 1}$

  1. The first ticket matched 2 numbers. $P={6 \choose 2}{53\choose 4}/{59 \choose 6}$

Then the second ticket could match up to 4 winning numbers. There are 4 unpicked winning numbers and 49 unpicked losing numbers.

Number of ways of matching 3: ${4 \choose 3}\times {49 \choose 3}$

Number of ways of matching 4: ${4 \choose 4}\times {49 \choose 2}$

  1. The first ticket matched 3 numbers. $P={6 \choose 3}{53\choose 3}/{59 \choose 6}$

Then the second ticket could match up to 3 winning numbers. There are 3 unpicked winning numbers and 50 unpicked losing numbers.

Note that, in this case, we have a win even if the second ticket doesn't win so this is really two cases: One where you win twice, one where you only win once.

Number of ways of matching 3: ${3 \choose 3}\times {50 \choose 3}$

  1. The first ticket matched 4 numbers.
  2. The first ticket matched 5 numbers.
  3. The first ticket matched 6 numbers.

The second ticket cannot win in these cases.

That's 14 cases where you win something. The expected value of playing this way is then $P(T_1=0,T_2=3)(0+E_3)+P(T_1=0,T_2=4)(0+E_4)+P(T_1=0,T_2=5)(0+E_5)+P(T_1=0,T_2=6)(0+E_6) +P(T_1=1,T_2=3)(0+E_3)+P(T_1=1,T_2=4)(0+E_4)+P(T_1=1,T_2=5)(0+E_5) +P(T_1=2,T_2=3)(0+E_3)+P(T_1=2,T_2=4)(0+E_4) +P(T_1=3,T_2=3)(E_3+E_3)+P(T_1=3,T_2<3)(E_3+0) +P(T_1=4)E_4 +P(T_1=5)E_5 +P(T_1=6)E_6$

I made Sage do the algebra on this to find the expected value as a function of the prizes: $$\frac{268671400163}{25860151455138}E_3 + \frac{6890}{7509579}E_4 + \frac{106}{7509579}E_5 + \frac{1}{22528737}E_6$$ This is $\frac{1573862727}{151228955878}E_3$ less than the expected value for two independent tickets. For the prize amounts I gave above, the expected value is about $\$0.33$ or about five cents less than the expected value for independent tickets.


For the jackpot, you only care if the set of six numbers is different between your tickets. Having tickets 1,2,3,4,5,6 and 1,2,,4,5,7 gives you the jackpot on two different draws and gives you twice the chance of winning that you would have from buying only one ticket.

The downside of these two tickets is that many combinations of three numbers are repeated. You are not increasing your chances of the smaller prizes as much by buying these two tickets as you would by buying two tickets that do not share numbers. Even for this, you only care if at least three numbers match between two tickets, so having 1,2,3,4,5,6 and 1,2,7,8,9,10 is as good as having two tickets that disagree completely.

Presumably if you get a set of three on multiple tickets you get paid multiple times. That means the expected value of two tickets with overlap is the same as two without overlap. You will win less often, but some of the time you do win you win more money.

The arguments of picking unpopular numbers only matter if there is a jackpot that is divided among the winners. In that case you want unpopular numbers so you share less. If the payout even for six of six is a fixed amount, you don't care about the popularity of the numbers, but the operators do.


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