How much to spend on a high-variance option?
post by RolfAndreassen · 2013-01-03T18:38:18.801Z · LW · GW · Legacy · 37 commentsContents
37 comments
So the jackpot in the Ohio lottery is around 25 million, and the chance of winning it is one in roughly 14 million, with tickets at 1 dollar a piece. It appears to me that roughly a quarter million tickets are sold each drawing; so, supposing you win, the probability of someone else also winning is 1 - (1 - 1/14e6)^{250000}=2%, which does not significantly reduce the expectation value of a ticket. So, unless I'm making a silly mistake somewhere, buying lottery tickets has positive expected value. (I find this counterintuitive; where are all the economists who should be picking up this free money? But I digress.)
I pointed this out to my wife, and said that it might be worth putting a dollar into it; and she very cogently asked, "Then why not make it 100 dollars?" Why not, indeed! Is there any sensible way of deciding how much to put into an option that has a positive expected value, but very low chance of payoff?
37 comments
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comment by jimrandomh · 2013-01-03T20:35:25.858Z · LW(p) · GW(p)
Careful; lotteries frequently use tricks to deceive you about their EV, such as taking some of the winnings back as taxes, and reporting the total payout of an annuity that has lower cash value.
comment by JoshuaFox · 2013-01-03T19:09:44.181Z · LW(p) · GW(p)
Don't forget taxes and declining marginal value of money.
Replies from: faul_sname↑ comment by faul_sname · 2013-01-03T21:32:58.439Z · LW(p) · GW(p)
If you give the money to a charitable organization, the marginal value of money may decline much more slowly.
Which leads me to wonder if you can donate lottery tickets to a charity, and what the tax implications of those tickets winning would be.
comment by gwern · 2013-01-03T19:19:24.388Z · LW(p) · GW(p)
I find this counterintuitive; where are all the economists who should be picking up this free money?
Relevant reading:
- http://www.wired.com/magazine/2011/01/ff_lottery/all/1
- http://en.wikipedia.org/wiki/National_Lottery_%28Ireland%29#Lotto_6.2F36:_1988.E2.80.9392
- http://r6.ca/blog/20090522T015739Z.html
I believe the last link essentially answers your question: the Kelly Criterion, which is an optimal way to invest, advises investing less in a lottery than a single unit (ticket) costs.
Replies from: Vaniver, CronoDAS, RolfAndreassen, None, Metus, Vaniver↑ comment by CronoDAS · 2013-01-04T05:16:06.452Z · LW(p) · GW(p)
The Kelly Criterion describes how to invest if you have utility that goes up logarithmically with the amount of dollars you make. It's not a foolproof decision theory.
Replies from: jsteinhardt↑ comment by jsteinhardt · 2013-01-04T18:10:10.905Z · LW(p) · GW(p)
I was under the impression that for infinite repeated play, no matter what your actual utility function is (as long as it is increasing and the total number of dollars is bounded), it turns out that the optimal single-turn strategy "looks like" betting with a logarithmic utility function --- hence the Kelly Criterion.
I don't know much about this though so could be mistaken.
↑ comment by RolfAndreassen · 2013-01-03T22:18:24.116Z · LW(p) · GW(p)
I find the results of the Kelly criterion extremely counterintuitive, but it does answer my question. Thanks. I note that, presumably, we are discussing a lifetime strategy rather than a one-off, so the bankroll should not be my current cash reserves but the net present value of my expected lifetime income stream; but the Kelly fraction is so small that the optimal bet still works out to less than a dollar. Fascinating!
Replies from: shminux↑ comment by Shmi (shminux) · 2013-01-03T22:35:16.880Z · LW(p) · GW(p)
Note that if your thinking is something like "I should probably forgo buying a latte this morning and buy a lottery tickets instead", then the Kelly criterion does not apply (it does not affect your lifetime income). Instead you should consider how much of your revenue-neutral funds you can spare and weigh the emotional downside of forgoing one expense (a drink: mmm, feels good) against the actual and potential emotional upside of another (a lottery ticket: what if I win, what if I win! + potentially winning - bummer, I lost!).
Replies from: RolfAndreassen↑ comment by RolfAndreassen · 2013-01-03T22:44:14.119Z · LW(p) · GW(p)
Well, actually, I find that I cannot bring myself to alieve this counterintuitive result, even after watching a toy Monte Carlo simulation confirm it. So I guess I'll put in a dollar, and say "Kelly Criterion, obviously" if anyone asks me why not 100 dollars. :D
Replies from: gwern↑ comment by gwern · 2013-01-03T23:18:48.927Z · LW(p) · GW(p)
I don't blame you for not alieving. When I was doing Kelly on my prediction market trades, it was terrifying.
Replies from: RolfAndreassen, EvelynM↑ comment by RolfAndreassen · 2013-01-03T23:53:16.722Z · LW(p) · GW(p)
I did go a bit further towards alief by putting into my toy MC study, with the simple coin-toss game in your link, a bettor who puts in 50% of his bankroll every time - way, way beyond the Kelly fraction, and then having a think about how he managed to lose all his money. (Not literally, but enough that the remaining bankroll was 0 to my printout accuracy.)In ten thousand iterations my longest win and loss streaks are both of ten games. A loss streak of ten games will reduce this bettor's bankroll by a factor of 1024. But ten winning games will only increase it by about a factor 80. On the other hand, with the Kelly fraction of 4.5%, ten losses reduce your bankroll by about 40%, while ten wins increase it by 62%. The asymmetry in these specific examples is somehow more convincing than the final numbers from the toy MC run.
↑ comment by EvelynM · 2013-01-04T00:16:26.491Z · LW(p) · GW(p)
"It was terrifying" is evocative, but not informative.
Can you explain, preferably by including your evidence?
Replies from: gwern↑ comment by gwern · 2013-01-04T00:35:50.082Z · LW(p) · GW(p)
I mean pretty much exactly that: I plugged in the payoff numbers into the equation, thought hard about my past record of trades & predictions and how calibrated I was in each certainty range to determine my edge, looked at the result of the Kelly Criterion, and felt terror at the idea of committing that much of my Intrade bankroll to one trade. I discuss the KC in http://www.gwern.net/Prediction%20markets#how-much-to-bet
Replies from: EvelynM↑ comment by [deleted] · 2013-01-03T23:47:34.341Z · LW(p) · GW(p)
I don't think the straightforward Kelly Criterion quite answers the question. It would tell you how many identical tickets you should buy (in a lottery where the jackpot pays out for each winner rather than being shared). The question at hand is different, because by buying more tickets you increase the chance of winning, rather than the payoff for a win.
I'm sure there's a simple variation of the criterion you can use, but I'm too tired to work it out right now. (I still expect that for most people, the answer is "buy no tickets")
↑ comment by Metus · 2013-01-04T10:56:29.742Z · LW(p) · GW(p)
I am trying to understand the implications of the kelly criterion for a real world portfolio. What I get as a result is that if I have free choice on any bet at any odds and chances I should, in total, invest more than I have. (Result by integrating over all probabilities/odds that allow positive expected value) In fact, I should invest infinitely much money. The wikipedia page states that taking out credit to buy a bet would be formalized by the loss formula so the infinity result is not exactly interpretable as taking out a loan, if I can.
One obvious fix is to limit the odds and probabilites to realistic values but that seems quite arbitrary. Intuitively I would expect the Kelly criterion to give a finite sum for all bets with positive expected value, at least it does so for any given odds b in wikipedias terminology.
Replies from: gwern↑ comment by gwern · 2013-01-04T16:01:24.356Z · LW(p) · GW(p)
What I get as a result is that if I have free choice on any bet at any odds and chances I should, in total, invest more than I have.
If you invoke infinities or indefinite sets of bets, it shouldn't surprise you that regular results might not apply: if you decide to invest in a bet of n at 99% odds of doubling, wouldn't it be even better to invest n at 99% odds of tripling? Or even better than that, invest n at 99.9% odds of tripling? Or no, invest n+1 at 99.9% odds of tripling! I'm not sure why you'd expect anything useful from a KC or a variant with such arbitrary inputs.
One obvious fix is to limit the odds and probabilites to realistic values but that seems quite arbitrary.
It does?
Replies from: Metuscomment by Shmi (shminux) · 2013-01-03T19:23:51.705Z · LW(p) · GW(p)
Is there any sensible way of deciding how much to put into an option that has a positive expected value, but very low chance of payoff?
People here have been known to buy life insurance and sign it over to an enterprise promising just that.
Replies from: RolfAndreassen↑ comment by RolfAndreassen · 2013-01-03T21:35:57.477Z · LW(p) · GW(p)
Yes, but in that case you get a single take-it-or-leave-it offer. With a lottery you can buy up to the limits of your current cash reserves.
comment by Vaniver · 2013-01-03T19:59:52.002Z · LW(p) · GW(p)
I believe I've seen reports that discussed a spike in lottery ticket sales when it's positive EV; I think there are some people that do watch for that and behave accordingly. Putting your "actually risk neutral" investment money into it isn't a terrible idea, but whether or not you have that investment money depends on your situation.
[edit] That is, you should split your financial resources into "expected to be needed in the next 3 months," "expected to be needed in the next year," "expected to be needed in the next 5 years," "expected to be needed in the next 20 years," and "risk neutral." How much of your current income flow to put into each of those buckets depends on your projected income, projected consumption, and so on, and is a fairly complicated calculation if you want to do it carefully. Even so, eyeballing it is better than not running those numbers.
comment by buybuydandavis · 2013-01-04T08:04:27.426Z · LW(p) · GW(p)
so, supposing you win, the probability of someone else also winning is 1 - (1 - 1/14e6)^{250000}=2%
How'd you come up with only 250000 other tickets being purchased? When expected payoffs (by some estimates) exceed ticket cost, number of tickets purchased tends to skyrocket.
Replies from: RolfAndreassen↑ comment by RolfAndreassen · 2013-01-04T17:52:16.260Z · LW(p) · GW(p)
Admittedly this is the weakest part of the argument. I looked at the revenue for 2011, 42 million, and divided by the number of drawings, 3 per week for 52 weeks. Obviously this would miss a recent spike in sales. However, I tried the probability with some theoretical numbers, and to get a probability of someone else winning that significantly affects the expectation value, the number of tickets sold has to go way, way up from that baseline quarter million. A full order of magnitude increase in sales, to 2.5 million, only gets you a 17% probability of sharing the jackpot, conditioned on you winning.
Replies from: JRMayne↑ comment by JRMayne · 2013-01-05T00:25:52.772Z · LW(p) · GW(p)
I went wandering around ohiolottery.com (For instance, http://www.ohiolottery.com/Games/DrawGames/Classic-Lotto#4) and found this out:
- The cash payoff is half the stated prize.
- The odds to win the jackpot, as noted by the OP, are about 14 million-1.
- The amount of money being spent on individual draws is very low. The jackpot increase was $100K for the last drawing; I don't know exactly what their formula is, but I'd be shocked if they sold more than 400K tickets for the last drawing.
- Ohio is running a lot of lottery games; this is good for players who pick their spots.
There are also payoffs below the jackpot level, so I'm confident there's a positive EV per ticket.
The question as to how many tickets to buy, assuming you can effectively do so, is "All of them." Buy each individual ticket, take your 14 million tickets, and probably profit. (Remember, the jackpot kick will include some fraction of your 14 million, also. Plus, you'll have all the side prizes.) In practice, unfortunately, this requires a method to buy them effectively, some armored cars, and a staff of people to do it right. Failure to purchase all tickets results in some drama, for sure.
The execution expenses and risk are troubling; if those could be effectively mitigated, it's a great investment.
Assuming you're a few million short of that, though, it's harder. I buy CA lottery tickets when EV>1.20 per $1 invested. I have no strong justification for that number.
comment by RomeoStevens · 2013-01-03T18:54:04.938Z · LW(p) · GW(p)
An amount that does not affect your ability to take such deals in the future?
Replies from: Oscar_Cunningham↑ comment by Oscar_Cunningham · 2013-01-05T13:54:51.964Z · LW(p) · GW(p)
What's your justification for this? Also, doesn't any amount affect your ability to some extent?
Replies from: RomeoStevens↑ comment by RomeoStevens · 2013-01-05T22:33:08.760Z · LW(p) · GW(p)
It was my intuitive response, which I always hope someone will dismantle. If lottery is entertainment then using it as a substitute good for other entertainment that month from your entertainment budget shouldn't affect your future ability.
comment by [deleted] · 2013-01-03T19:10:20.818Z · LW(p) · GW(p)
Let's all pool bets. In the event we win, 1 mil goes to CFAR so EY can finish HPMOR.
comment by prase · 2013-01-03T19:27:22.549Z · LW(p) · GW(p)
Isn't there a standard reply containing utility functions? Assuming the usual diminishing marginal utility, if p(win) * U(jackpot) > U($1), then 2 p(win) * U(jackpot) > U($2) and so on. You should spend all your money! (Unless you are rich enough to buy so many tickets as to wander out of the domain of approximate linear dependece of p(win) on the number of tickets you own).
Replies from: prase, janos↑ comment by prase · 2013-01-04T18:41:43.818Z · LW(p) · GW(p)
Second attempt (thanks janos):
Isn't there a standard reply containing utility functions? You should buy N tickets where N is the solution to p(win | N tickets) * U(jackpot) + U(-N dollars) is maximal. Assuming the usual diminishing marginal utility of money and that U(jackpot) * p(win | 1 ticket) > -U($-1), a unique solution should exist for N > 0.
↑ comment by janos · 2013-01-04T02:33:55.512Z · LW(p) · GW(p)
I think you're making the wrong comparisons. If you buy $1 worth, you get p(win) U(jackpot) + (1-p(win)) U(-$1), which is more-or-less p(win)U(jackpot)+U(-$1); this is a good idea if p(win) U(jackpot) > -U(-$1). But under usual assumptions -U(-$2)>-2U(-$1). This adds up to normality; you shouldn't actually spend all your money. :)
Replies from: prase