Theoretical lottery question

[SGspecial]

Imagine a lottery. Six numbers, 00-99. To win you must match all six numbers in the right order. There is no limit to the number of tickets that can be sold. You can choose your own numbers or generate them randomly and duplicate tickets are permitted. If a number is drawn for which no tickets have been sold, that number is discarded and another number is drawn so there is always at least one winner and no rollover.

Assume that each ticket costs $1.00 and the lottery pays back 100% of the ticket sales, with the winnings split equally among winning tickets. Also assume that there is no tax liability.

Assuming that you have an unlimited bankroll and nothing better to do with your money, how do you maximize your EV?

[WhiteWolf]

My thoughts:

Never buy numbers that duplicate your own holdings. That only costs you double without increasing your chances of winning. This alone won't be enough to give you an edge, though.

I think your only edge will come if you somehow could ensure that you never buy a ticket that duplicates someone else's numbers. If you could do this, your best strategy would be to buy up every number combination that has not been picked by someone else.

Your only edge here is picking up the overlay from other players who have duplicate picks. When someone picks a holding that has already been choosen, they just increase the pool for all other players without that pick, while not decreasing the chance for the other players to hit.

[WhiteWolf]

Upon further reflection, there may be a better strategy than my original one - simply buying one of every possible combination. Once you've done this, you've made it a -EV proposition for anyone else who also wants to buy a ticket - they'll be laying X-to-1 odds for a payout slightly higher than .5X-to-1. Since they are making a -EV bet, the gain accrues to everyone else already in the lottery, and since you hold the great majority of the tickets, you are the one who benefits the most.

[Siegmund]

Since every combination is equally likely to win, your only concern is with finding numbers which are relatively unlikely to be bought by anyone else. (In the real world, people like to play birthdays and other dates so much that I would bet that a random selection of 6 numbers 32 to 45 would be among the most profitable to play.) Should be easy enough to use a combination of published studies of past lottery ticket sales, and the records of how many ways the prizes of this game are split, to find the best numbers.

If I have perfect information about people's purchasing habits, I buy at least one of every ticket combination that the general public buys less often than it should (each such ticket is +EV).

No, buying one of every ticket isn't a better strategy, and no, it doesn't make it a -EV for anyone else who wants to buy a ticket: he too can buy one of every number and achieve the same EV0 position as you hold.

And no, buying multiple tickets of the same number is not, automatically, a bad strategy. (Usually, yes, but not always.) Here's a simple contrived example: suppose that 100C6*10-8 tickets have already been bought by the other players: ten of every possible combination except 1-2-3-4-5-6, for which only two tickets have been sold. If you buy one 1-2-3-4-5-6 ticket, you are paying $1 for a 1/100C6 chance of winning 1/3 of the prize pool, for an return of $3.33. Buying two tickets means paying $2 for a chance of winning 1/2 the prize pool, for a return of $5.00. Buying three tickets means paying $3 for a chance of winning 3/5 of the prize pool for a return of $6.
In this example, your first ticket has an EV of $2.33, your second ticket with the same number adds another $0.67, and your third ticket is EV-neutral. (Fourth and subsequent tickets would be -EV.)

[SuperPokerJedi]

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Assuming that you have an unlimited bankroll and nothing better to do with your money, how do you maximize your EV?

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This makes the question almost impossible and quite unecessary to look to invest in anything at all!

FYP

[AaronBrown]

A couple of quick comments on Siegmund's excellent post. OP specified you have to match the numbers in order, so there are 858,277,728,000 (100 x 99 x 98 x 97 x 96 x 95) tickets, not C(100,6) = 1,192,052,400.

Let P be the prize pool and N be the number of distinct combinations held by other people (N <= P). Your best investments are to buy all the combinations held by no one else, there are (858,277,728,000 - N) of them. Your expectation on each one is (P - N) / 858,277,728,000. You only have positive expectation to the extent the public duplicates tickets.

In order to buy a ticket for a number that someone already holds, you need the prize pool to be larger than 2 x 858,277,728,000, or $1.7 trillion. That seems unlikely.

[WhiteWolf]

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In order to buy a ticket for a number that someone already holds, you need the prize pool to be larger than 2 x 858,277,728,000, or $1.7 trillion. That seems unlikely.

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I originally thought along the same lines. However, I did some calculations, and if you are able to buy up a large number of tickets, you do have a positive expectation by buying up a number already held by someone else. The pool doesn't even have to be particularly large.

To illustrate, let's run a simplified lottery where you only have to pick a single number between 1-100. You know the numbers 1-10 have been picked by 2 people each, and the numbers 11-40 have been picked by 1 person each. The initial prize pool, before you invest, is $50.

If you just take the 60 unused numbers, you automatically show a profit because of the duplicated numbers. You will have a 60% shot of winning a $110 pool, giving you an EV of $6 (.6*110 - 60)

However, if you spend $100 to cover every number, you have a 60% chance of winning a $150 prize, a 30% chance of winning a $75 prize, and a 10% chance of winning a $50 prize. Your EV is now 17.5 (.6*150 + .3*75 + .1*50 - 100), a big improvement over the original strategy. Obviously a surprising result, but the math doesn't lie.

I think the interesting insight is that, while buying a previously picked number is generally -EV when looked at in and of itself, this is a zero sum game and that -EV must flow somewhere. If you already hold a large majority of the numbers, most of that -EV is returned to you. Combine that with the fact that your buying of the duplicated number hurts the EV of the player who already holds that number (where, once again, you will be in the position of picking up most of that lost EV), and you have a surprisingly better strategy than just looking for unbought numbers.

Another interesting result is that this strategy works even if there are no already-duplicated numbers, so long as there are unchosen numbers left for you to pick. I'll leave it as an exercise for the reader to show this.

However, I do suspect that there will be situations where particular numbers are so overbought that you would want to avoid them while still buying other duplicated numbers. I'm also not sure how this strategy would work if you're not able to buy up a large number of unduplicated numbers, so I wouldn't make any general claims that this strategy is always the "best."

[AaronBrown]

I stand corrected. You can capture a maximum of $30 of expected value in your example by buying one of all the untaken numbers and very large numbers of tickets for all the numbers that have been taken.

For example, if you buy 1,000 tickets for numbers 1 to 40, and one ticket each for 41 to 100, you spend $40,060. 10% of the time you win $40,030, 30% you win $40,070 and 60% you win $40,110. That means 10% of the time you lose $30, 30% you win $10 and 60% you win $50 for an expected value of $30 ($29.97, but if you buy enough tickets you can get as close as you want to $30).

[WhiteWolf]

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I stand corrected. You can capture a maximum of $30 of expected value in your example by buying one of all the untaken numbers and very large numbers of tickets for all the numbers that have been taken.

For example, if you buy 1,000 tickets for numbers 1 to 40, and one ticket each for 41 to 100, you spend $40,060. 10% of the time you win $40,030, 30% you win $40,070 and 60% you win $40,110. That means 10% of the time you lose $30, 30% you win $10 and 60% you win $50 for an expected value of $30 ($29.97, but if you buy enough tickets you can get as close as you want to $30).

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Another counter-intuitive result. This problem gets more interesting by the minute....

[SGspecial]

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Another counter-intuitive result. This problem gets more interesting by the minute....

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And the answers do too! The point of making a Trillion possible combinations tho (and it IS a trillion since I never said the numbers would be excluded from the pool once they were selected in a previous position) was to make the percentage of combinations with a logical pattern VERY small. I like the answers you have so far, but what if you have no reliable information about which combinations may have been picked already?