Can you buy ever lottery ticket combination?

[just_some_guy5]

@brickbacon

Also, for an investor, you would need to include that you would have several other winning tickets with 4 and 5 correct numbers.

Someone posted these amounts earlier but I did the math and don’t believe their estimate of $66mm for non-jackpot prizes was accurate.

Looking at the frequency of wins and payouts (which aren’t split) for non-jackpot prizes I get a total return of $32mm (pre-tax). This would obviously be in addition to the jackpot prize, which if taken as a lump-sum would be $359mm (also pre-tax). Clearly the tax rates impact the economics of purchasing all possible tickets, but a pre-tax win of $391mm for a cost of $176mm seems pretty enticing.

Of course, as others have mentioned the key question is whether you’ll have to split the jackpot, as your total winnings with a jackpot split 2 or 3 ways reduces your total payout to $211mm and $152mm, respectively. I suspect there’s actually a reasonable way to approximate the odds of having to split your jackpot (based on estimates of ticket purchases, for example), which could probably yield a more definitive answer.

[just_some_guy5]

Saw a statistic that said 137mm tickets were purchased before the last mega millions drawing. If we assume that at least 150mm tickets will be purchased this time around (could certainly be a higher number given all the publicity), seems like there would be an 85% chance someone should win (of course, since lots of people play set number combos like 1-2-3-4-5 the actual probability may be lower).

Another issue with this strategy would be signalling. If 5 individuals / firms independently decide to pursue a strategy of buying up all ticket combinations, they will most certainly lose money.

[brickbacon]

@hajario

Of course not. Each draw is a totally independent event.

What you might be able to do is to pick numbers that are less popular with the masses (7, 21…) so that if you do win, you are less likely to have a split pot.

Incorrect. You are right that the odds a number comes up are independent, but the chances you split are not. The odds of splitting change because people don’t make independent choices. For example, the chances that someone picks 1-2-3-4-5 and a megaball of 6 are very low. Not because that couldn’t happen, or because it’s less likely to arise, but because people are irrational. Therefore, a winning ticket with those numbers is less likely to split than one with a few common numbers that people play. This is also now magicians guess a number or card you are thinking of.

By looking at trends of numbers people play, you can guard against irrationality. Let’s say for example that people often play numbers 1-12 and 1-31 more often because they are using family birthdays. You can buy multipliers for the most common numbers so that the split is more advantageous to you. That calculus is obviously very complex, which is why it would be a nice thing to investigate.

[ultrafilter]

@lazybratsche

How do you get that number? I think you’re overcounting somewhere… (44 choose 6) is 7 million, which is what Wallenstein’s quote says. And 7 million seconds is only a few months…

ETA: I believe lotteries use combinations, not permutations.

That’ll teach me to do combinatorics early in the morning.

[hajario]

@brickbacon

Incorrect. You are right that the odds a number comes up are independent, but the chances you split are not. The odds of splitting change because people don’t make independent choices. For example, the chances that someone picks 1-2-3-4-5 and a megaball of 6 are very low. Not because that couldn’t happen, or because it’s less likely to arise, but because people are irrational. Therefore, a winning ticket with those numbers is less likely to split than one with a few common numbers that people play. This is also now magicians guess a number or card you are thinking of.

By looking at trends of numbers people play, you can guard against irrationality. Let’s say for example that people often play numbers 1-12 and 1-31 more often because they are using family birthdays. You can buy multipliers for the most common numbers so that the split is more advantageous to you. That calculus is obviously very complex, which is why it would be a nice thing to investigate.

Actually, we are saying the same exact thing but you explained it better.

[Asympotically_fat]

I believe it happened in Ireland once. Not every combination was bought, but enough were to make it a good investment for the syndicate who did indeed win the lottery. Sorry no cite, but I saw a TV documentary on it about 20 years ago.

[tim-n-va]

@just_some_guy5

Someone posted these amounts earlier but I did the math and don’t believe their estimate of $66mm for non-jackpot prizes was accurate.

Looking at the frequency of wins and payouts (which aren’t split) for non-jackpot prizes I get a total return of $32mm (pre-tax). This would obviously be in addition to the jackpot prize, which if taken as a lump-sum would be $359mm (also pre-tax). Clearly the tax rates impact the economics of purchasing all possible tickets, but a pre-tax win of $391mm for a cost of $176mm seems pretty enticing.

Of course, as others have mentioned the key question is whether you’ll have to split the jackpot, as your total winnings with a jackpot split 2 or 3 ways reduces your total payout to $211mm and $152mm, respectively. I suspect there’s actually a reasonable way to approximate the odds of having to split your jackpot (based on estimates of ticket purchases, for example), which could probably yield a more definitive answer.

You are correct. I dropped a digit copying from wikipedia and vastly over-counted $3 winners. (1 in 14 versus 1 in 141)

[brickbacon]

@hajario

Actually, we are saying the same exact thing but you explained it better.

Fair enough. I didn’t mean to come across as snarky.

I think it would be a fun exercise to talk through the issue to try to model the data better. For example, I would bet numbers 1-12 are far more common than most other numbers on people’s tickets.

[Meatros]

@VinylTurnip

Wasn’t there a documentary that followed a group that did this? I can’t seem to find it now—as I recall they were not able to get all the tickets bought and printed in time, so it was very tense at the end, but the winning number turned out to be one of the ones they had purchased. Anyone else remember seeing it?

I think we saw the same thing, since I specifically remember that they didn’t get to buy all of the tickets for some reason and still won.

[Wallenstein]

@VinylTurnip

Actually, reading Wallenstein’s link, I assume it’s talking about the same group— but I could swear I remember seeing something that actually followed and filmed them as they planned the thing.

I was thinking of exactly the same one… I can’t remember if it was the one I linked to, I don’t recall them being Australian, but it seems pretty close.

[just_some_guy5]

@brickbacon

By looking at trends of numbers people play, you can guard against irrationality. Let’s say for example that people often play numbers 1-12 and 1-31 more often because they are using family birthdays. You can buy multipliers for the most common numbers so that the split is more advantageous to you. That calculus is obviously very complex, which is why it would be a nice thing to investigate.

I don’t think this is an optimal strategy. While buying multipliers for more common numbers would boost returns in the scenario where 1-2-3-4-5 is a winner, it seems like you’d be better off simply not buying common number combos in the first place. For example, if 1-2-3-4-5 is played 3x as often as other combinations, buying this combo with a 3x multiplier would cost $3. At the end of the day, you’re paying 3 to collect the same winnings you’d receive for an equally probable win on a non-popular combination without a multiplier (which would only cost $1).

However, I wonder how much of an issue common number combinations would really be if you were planning to buy the vast majority of all lotto ticket combinations. Seems like there are very few common number sets relative to the universe of possible combinations.

[tim-n-va]

Not vouching for the credibility but I’ve seen offers to sell the data on the numbers played in a lottery. The idea was that by using a “contrarian” approach you could lower the odds of sharing. No effect on the odds of actually winning.

[si_blakely]

@brickbacon

As an aside, what would be the sweet spot in terms of betting? Obviously having all the tickets would give you 100% chance of winning, but it also gives you the most maximum outlay of money. I wonder if, by looking at historic trends, you could boost the odds somewhat.

I once worked out a strategy based on covering every 3 and 4 ball combination, giving an income marginally greater than investment, plus a possible bonus on any higher level prizes. The ticket purchase number was pretty reasonable, certainly not as bad as buying all tickets, but you would still need lots of people. Unfortunately, I’ve forgotten the math I used at the time. And the high level of investment skews the output, so you do need to be pretty careful.

Si

[brickbacon]

@just_some_guy5

I don’t think this is an optimal strategy. While buying multipliers for more common numbers would boost returns in the scenario where 1-2-3-4-5 is a winner, it seems like you’d be better off simply not buying common number combos in the first place. For example, if 1-2-3-4-5 is played 3x as often as other combinations, buying this combo with a 3x multiplier would cost $3. At the end of the day, you’re paying 3 to collect the same winnings you’d receive for an equally probable win on a non-popular combination without a multiplier (which would only cost $1).

However, I wonder how much of an issue common number combinations would really be if you were planning to buy the vast majority of all lotto ticket combinations. Seems like there are very few common number sets relative to the universe of possible combinations.

I misunderstood, the multiplier is only for non-jackpot prizes, and only requires an extra dollar commitment (eg. $2 ticket). The multiplier number is randomly selected as either 2,3, or 4. So any doubling up would just be the result of buying more tickets.

That said, I am not sure your strategy would be any better. Obviously, avoiding all common numbers would leave you too exposed to losing. Splitting a jackpot would suck, but it sucks much less than losing outright.

What if we model this using a simpler system where 100 people guess a number between 1 and 100. I think this is fairly appropriate since they will likely sell close to 175 million tickets. Lets say the winner gets $150, and each ticket costs $1. What would be the best way to allocate up to $150 to maximize your chances of winning money?

Feel free to add other assumptions, or correct any I have made.

[just_some_guy5]

@brickbacon

What if we model this using a simpler system where 100 people guess a number between 1 and 100. I think this is fairly appropriate since they will likely sell close to 175 million tickets. Lets say the winner gets $150, and each ticket costs $1. What would be the best way to allocate up to $150 to maximize your chances of winning money?

If it’s allright with you, I’ll slightly modify the parameters and try to explain my reasoning for avoiding popular numbers. As you said, let’s assume a $150 jackpot awarded for a random number drawn between 1 and 100. Let’s further assume that the “popular numbers” are 1 through 10, and they happen to be 3x as popular as numbers 11 through 100. To keep it as simple as possible, let’s assume that if we select a number between 1 and 10 we’ll split it with 2 other people, while a win with numbers 11-100 will not be split (this occurs b/c the numbers 1-10 are 3x as popular).

Given the above parameters, one option would be to simply buy all 100 numbers for $100 and ensure that I win. My expected payout would be $140 (I’d have $1.50 expected win for numbers 11-100, and a $0.50 expected win for numbers 1-10 since those are split 3 ways). $140 payout less $100 investment = $40 profit (ROI = 40%)

Alternatively, I could simply purchase numbers 11-100. This would cost $90 and yield an expected win of $135 (90 x $1.50 per ticket). $135 less $90 investment = $45 profit (ROI = 50%)

[cynyc]

I believe that is correct.

@joebuck20

Question - If you did end up spending $176 million on lottery tickets, could you write that off as a gambling loss on your taxes, and just end up paying on the amount you won above that (which in this case would be about $300 million)?

[cynyc]

@just_some_guy5

If it’s allright with you, I’ll slightly modify the parameters and try to explain my reasoning for avoiding popular numbers. As you said, let’s assume a $150 jackpot awarded for a random number drawn between 1 and 100. Let’s further assume that the “popular numbers” are 1 through 10, and they happen to be 3x as popular as numbers 11 through 100. To keep it as simple as possible, let’s assume that if we select a number between 1 and 10 we’ll split it with 2 other people, while a win with numbers 11-100 will not be split (this occurs b/c the numbers 1-10 are 3x as popular).

Given the above parameters, one option would be to simply buy all 100 numbers for $100 and ensure that I win. My expected payout would be $140 (I’d have $1.50 expected win for numbers 11-100, and a $0.50 expected win for numbers 1-10 since those are split 3 ways). $140 payout less $100 investment = $40 profit (ROI = 40%)

Alternatively, I could simply purchase numbers 11-100. This would cost $90 and yield an expected win of $135 (90 x $1.50 per ticket). $135 less $90 investment = $45 profit (ROI = 50%)

[brickbacon]

@just_some_guy5

If it’s allright with you, I’ll slightly modify the parameters and try to explain my reasoning for avoiding popular numbers. As you said, let’s assume a $150 jackpot awarded for a random number drawn between 1 and 100. Let’s further assume that the “popular numbers” are 1 through 10, and they happen to be 3x as popular as numbers 11 through 100. To keep it as simple as possible, let’s assume that if we select a number between 1 and 10 we’ll split it with 2 other people, while a win with numbers 11-100 will not be split (this occurs b/c the numbers 1-10 are 3x as popular).

But aren’t you missing some entrants? If we assume 101 participants, you and 100 other people. If numbers 1-10 are 3x as popular, we would still have some people betting in the 11-100 range, so you chances of splitting in that zone are greater than zero, right?

Your scenario should be that 30 tickets are bought for numbers 1-10, 70 tickets for numbers 11-80, and zero tickets for numbers 81-100. Then you return would be as follows:

1-10: You would be 1 of 4 winners for each number, so 150(.25)= 37.5
11-80. You would be 1of 2 winners for each number, so 150(.5)= 75
81-100: You would be the only winner for each number, so 150(1)= 150

So after taking the relative likelihood of those outcomes, we get:

37.5(.1)+75(.7)+150(.2)= $86.25

Which means a loss of $13.75, assuming my math is correct.

[just_some_guy5]

@brickbacon

But aren’t you missing some entrants? If we assume 101 participants, you and 100 other people. If numbers 1-10 are 3x as popular, we would still have some people betting in the 11-100 range, so you chances of splitting in that zone are greater than zero, right?

Your scenario should be that 30 tickets are bought for numbers 1-10, 70 tickets for numbers 11-80, and zero tickets for numbers 81-100. Then you return would be as follows:

1-10: You would be 1 of 4 winners for each number, so 150(.25)= 37.5
11-80. You would be 1of 2 winners for each number, so 150(.5)= 75
81-100: You would be the only winner for each number, so 150(1)= 150

So after taking the relative likelihood of those outcomes, we get:

37.5(.1)+75(.7)+150(.2)= $86.25

Which means a loss of $13.75, assuming my math is correct.

Your math is correct, however, my strategy of avoiding popular numbers in that case would avoid all numbers 80 and below. Numbers 1-10 aren’t appealing b/c the expected win for each would be 1% x $150 / 4 = $0.375. Numbers 11-80 are better but still a losing proposition given an expected win for each of 1% x $150 / 2 = $0.75. However, numbers 81-100 yield an expected win each of 1% x $150 / 1 = $1.50.

Based on the above, I’d want to buy all 20 numbers from 81-100 and would expect a probability weighted total win of $30. $30 payout less $20 for the tickets = $10 profit (ROI = 50%).

---

As an FYI, in my post above I assumed that 2 people had already purchased each of the numbers 1 through 10, and that 0 people had purchased numbers 11-100.

[just_some_guy5]

And just to be clear, this strategy clearly “works” better in a repeated-game setting, since you’re not going to receive exactly $30 on any one payout (but would expect to earn an average payout of $30).