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gorillapaws's avatar

Is there ever an odds to payout ratio that makes purchasing a lottery ticket a mathematically rational choice?

Asked by gorillapaws (16531 points ) April 4th, 2012

I realize that you have to factor in multiple winners splitting the prize, the odds that will happen etc. Let’s have the most basic case, where you buy a ticket for $1 with 1 number between 1 and 10 and the winner gets $100 trillion. In such cases, the mathematician would buy as many of each number as he could possibly afford. Of course everyone else would be doing the same. Even still, I suspect that such a lottery would still net a nice return even split several billion ways.

So how would a rational mathematician calculate when the odds are actually in her favor to win, given that there could be many other winners?

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14 Answers

anartist's avatar

I bought tickets the day after the mega-million win. I don’t need a mega-million. 100,000 will do nicely. And the competition is way down.

PhiNotPi's avatar

Actually, I think that the answer would be no, probably always no, due to the fact that the size of most jackpots is determined by the number of tickets sold, which also influences the number of expected winners.

I’m going to use your example.

If you assume that half of all money spent on tickets goes into the pool, then there would be 200,000,000,000,000 tickets, so about 20,000,000,000,000 winning tickets since there are ten numbers. If you win, you will get a return of five dollars (100 trillion divided by 20 trillion). There is a 10% chance of making $4, and a 90% chance of losing $1, so it would not be worth it.

gorillapaws's avatar

@PhiNotPi That’s a very good point. I didn’t realize that the jackpot was determined by the number of tickets sold (although it makes perfect sense—I’m not sure why I didn’t figure that out). Let’s say the jackpot has accumulated to that size from many previous months of loosing, and therefore the current round’s jackpot is independent of previous weeks’ sales.

PhiNotPi's avatar

You would have to obtain good estimates as to the number of active tickets, and then the calcutions would be similar.
If O is the chance of winning, J is the jackpot, T is the number of tickets, and C is the cost of a ticket, your expected return R would be:
R = J/T – C
That’s it. The neat thing is that by simplifying the equation, I managed to remove all reference to the odds of winning. This formula works perfectly for the above example, giving the correct result of expecting to lose 50 cents on every ticket.

Edit: I haven’t had enough time to test this enough, but I will tell the results as soon as I can.

PhiNotPi's avatar

OK, so I checked over the work, and it appears that my formula is correct as long as the expected number of winners (O*T) is above one. If it is below one, it does not take into account the fact that the jackpot will go unawarded. I’ll try to find a better method later.

jerv's avatar

If the lotteries are set up like the ones I’ve seen, then not really… except that it’s worth a couple of bucks to keep the dream alive. How much is a bit of happy fantasizing worth to you?

funkdaddy's avatar

So in the simplest sense the jackpot times your odds of winning should be larger than your wager

So if there are 200 tickets sold for a dollar and one winner will be drawn you could figure your break even point as

Jackpot * (Your Tickets / Total Tickets) = Your Wager

J * (1 / 200) = 1

The only thing you’re adding with a real world lotto is the fact that more than one can win because you’re not concerned about your chances of not winning (and the pool continuing) only your chances of winning and having to split the pot (which reduces your expected gain).

I’m guessing the best you could do without analyzing the numbers people tend to play is assume an even distribution and try to figure the number of probable winners that way.

So you’d have to correct your expected winnings for the number of probable winners. This gets fuzzy because you won’t have a partial winner, but I believe for figuring your odds, a partial winner would be fine.

So you’d be looking at

Probable Winners = Total Tickets / Possible Scenarios (a specific set of numbers)

(Jackpot / Probable Winners) * Odds = Cost

So if your scenario is 200 tickets, sold for a dollar, numbered from 1 to 100, and all money is awarded your equation would be

(200/2) * 1/100 = 1

That figures your break even point, so really you’d want the left side to be larger than the right.

To fill in some blanks with the mega million jackpot

Jackpot was 640,000,000
Odds are 1 in 175,711,536
Jackpot before the big one was $363 million, so we can assume at least 300 million tickets
Tickets are $1

So we’re looking at

Probably winners = ~300,000,000/175,711,536 – make this 2 for ease

(640,000,000/2) * (1/175,711,536) = 1.82

Which is bigger than the $1 cost of a ticket and doesn’t figure in smaller prizes.

zenvelo's avatar

I look at the odds on winning (in California, roughly 1 in 13 million for the Super Lotto) and compare that to the projected jackpot. So for my own justification if the jackpot is over 14 million, I figure it is paying off at “close to true” rate. And since the jackpot is a rollover, I figure the number of tickets sold will be roughly equal to the 13 million.

So, that’s my calculation story and I am sticking to it.

ragingloli's avatar

No. Though my math teacher played lotto, even though he admitted that it is insane to do so.

whitenoise's avatar

You would give up a small amount of money in exchange for a chance to a fortune you could not acquire through any other way.

That can still be rational, even though the overall benefit would be negative for all participants.

It is akin to taking an insurance. Through an insurance you spend an amount you are sur you can miss to prevent an expenditure that you cannot handle, even though the chances are small of it happening and the insurance company making money overall.

SmashTheState's avatar

Yes. In instances where no one has won previous lotteries and the pot is folded into the current one, it’s statistically possible that each lottery ticket possesses a slightly greater likelihood of return than 1:1. In these cases, a few investment groups usually spring up so that people can invest their money into a pool in order to minimize the effect of chance. They’ll buy hundreds of thousands of tickets, then split the total winnings among all members, to provide a small, statistically guaranteed return on investment.

LostInParadise's avatar

To state the obvious, any lottery that consistently pays out more than it takes in cannot be expected to be around for very long.

PhiNotPi's avatar

The expected payout can go over the cost of winning the ticket, but only through the use of rollovers.

I’m going to reanalyze the Mega Millions expected payout.

First, you must calculate the expected number of other tickets. The jackpot went up by about 300 million, and according to CBS, only sixty percent goes into the pool, so there are actually 500 million tickets.

So, there are ~500/175 = ~2.85 winning tickets, and in real life there where three.

So part of the jackpot that any particular winner would get is about 640 million / 2.85 =~ 224 million.

So, there is about a 1/175,000,000 chance that you would get this number, but no matter what you have lost that particular dollar that you spent. This makes sense because if you get a return of five dollars, you will only make a profit of four.

The expected return is
R = ~224,000,000/175,000,000 – 1
R = ~1.28 – 1
R = ~0.28

It would actually be worth it, with an expected profit, although not as much as @funkdaddy expected.

gambitking's avatar

all this math is giving me a headache. the simple answer is ‘no’.

reason: by the time you invest enough to “beat” the odds, you’re breaking even. so you dont really “win”

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