Consider a lottery with L unique numbers. In the lottery one of the numbers is randomly drawn and grants a grand prize. n participants purchase a lottery ticket and let's say they select their numbers randomly and independently.

Say that I won the lottery. What is the probability I would have to share the grand prize? If there was just a single other participant in the lottery, the probability of them guessing the same number I did is $\frac{1}{L}$, and hence the chance I'm a sole winner is $1-\frac{1}{L}$. If there were two other participants, the probability the first of them didn't guess as I did, and that neither did the second participant is $(1-\frac{1}{L})^2$. To be a sole winner then, it would seem that the probability is $(1-\frac{1}{L})^n$.

Just for sports, the probability of winning one of the big American lotteries is ~1:300M. If there are a million tickets bought, then the probability I would have to share the winning prize (conditioned on me winning) is ~1:1000. A small number, but still several orders of magnitude greater than the chance I was taking with purchasing the original ticket.

The greater the number of other participants in the lottery (n) the smaller my chance of being a sole winner. Say I bought a ticket and I'm determined to purchase a second. If I select a different number on my second ticket, my chance of winning the grand prize doubles. If I select the same number again, if I end up sharing the grand prize, I would increase my share in the winning (say it's proportional to the number of tickets I have with the winning number).

Under what conditions should a second lottery ticket be selected with the same number as the first?


2 Answers 2


On the assumption that you are going to buy two tickets and the question is whether they should match, I can't see a case for matching them. From an expected value point of view, each ticket that doesn't match has an expected value of $\frac 1{(k+1)L}$ where there are $k$ other tickets that match those numbers. If you don't match your tickets you get two of these. If you match them you get less than twice the value of one ticket. The only way it makes sense is if all sets are already picked but there is one particularly unpopular set of numbers and you buy it twice. The disaster is if you buy a set that nobody else has bought twice. The second ticket has not increased your winnings at all.

If you use a more realistic model that says winning twice the money is not twice as nice as winning the basic value it is even more so. If the amount is large you probably don't care about doubling it. If the amount is small you don't care about the money, you are just happy if you win, so win twice as often.


Consider the possibility of just a single additional winner. Notation:

  • $v :=$ the value of the grand prize.
  • $p_{single} :=$ the probability of being a single winner.
  • $p_{win} :=$ the probability of any ticket to win the lottery.

The expected value of winning in the case of two different lotteries is the weighted average of the total value in the case of a single winner and the value shared with an additional winner, twice (since there are two tickets):

$L_{different} = 2p_{win}\left[p_{single} \times v + (1-p_{single})\dfrac{v}{2}\right]$

The expected value of winning in the case of two identical lotteries is the weighted average of the full value in the case of a single winner and a greater share of a the grand prize (now two-thirds rather than a half):

$L_{same} = p_{win}\left[p_{single} \times v + (1-p_{single})\dfrac{2v}{3}\right]$

The OP's question is under which conditions $L_{different} < L_{same}$.

$\iff 2p_{win}\left[p_{single}v + (1-p_{single}) \dfrac{v}{2}\right]< p_{win} \left[p_{single}v + (1-p_{single})\dfrac{2v}{3}\right]$

$\iff 2\left[p_{single}v + (1-p_{single})\dfrac{v}{2}\right]< p_{single}v + (1-p_{single})\dfrac{2v}{3}$

$\iff 2p_{single} + (1-p_{single})< p_{single} + (1-p_{single})\dfrac{2}{3}$

$\iff p_{single} + (1-p_{single})< (1-p_{single})\dfrac{2}{3}$

$\iff p_{single} + \dfrac{1}{3}(1-p_{single}) < 0$

$\iff 2p_{single} + 1 < 0.$

Since $p_{single}$ is positive, as Ross Millikan answered, this last inequality can never be true. In other words, buying the same ticket twice is never advised.


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