Walking back from class today, someone asked me if I’d like to take a 15 minute survey for a 1 in 20 chance of wining a Dell computer. It made me wonder, is there a mathematical way to determine if it would be worth my time?
Since 1/20 * $1000 (estimate of the price of the computer) = $50, does this mean that, in a way, I am being payed $50 for 15 minutes of my time? I’d take that check anyday.
Also, does it work the same way for the lottery? If the odds of winning x expected payoff exceeds the price of the lotto ticket, does that mean that it’s a good investment?
If my formula is flawed, what is the correct way to determine whether to participate in the game of chance?
Strictly speaking, you’re correct. The other factors to think about:
a) Is it really a 1/20 random chance that you’ll win something worth $1000? I’m doubtful that they want to spend $50 per interview, which is certainly a lot more than they’re paying for the interviewers time and the materials. Perhaps they’re lying about the 1/20 aspect, or perhaps they’re mis-representing the real prize which is a Dell handheld worth $150.
b) Even if they are paying it out, I wonder if it’s going out randomly. It may be true that of all the interviewees, 1/20th of them will win a prize. But perhaps they’re selling something, and only those that buy will win.
c) Value of the item to you. If it’s truly something you can quickly liquidate for $1,000, then all is good. Or if it’s something you really want and would pay $1,000 for today, then it’s good. But if it’s something that you want, but wouldn’t really buy it unless say it was priced at the steal price of $400, then that’s your value for the item.
d) Opportunity cost. If this happens to be a very important 15 minutes, as in the situation where you’re late for work or an important test at school, then the cost exceeds the reward.
Like Bill H. suggested, I think those odds (1 in 20) are seriously inflated on purpose.
Your expected value of $50 is correct. You’re not getting paid $50 for 15 minutes though, the correct interpretation is that, given many repetitions of that interview, on average you’d win $50 per interview, which sounds like too good a deal from my perspective.
Bill H’s “d) Opportunity cost”, is the main point to consider. The problem with your formula is that it overlooks the fact that as a single player, playing only once, there are only two possible outcomes : win the computer, or win nothing. There’s no outcome where you’ll be paid $50.
The same applies to a lottery, your average gains over a million trials aren’t relevant since you’ll never play a million times. The important thing is the opportunity cost - what do you forgo in exchange for a chance (however infinitesimal) of winning.
OTOH - I’m not sure that’d I’d call a lotto ticket an “investment” - the only people who are making a successful investment are the guys running the game.
I can’t endorse that particular survey company but keep in mind that they may get their “$1000” computer much cheaper. I do online market surveys for a reputable Australian market research company - they email me links to tests and I do them online while doing other things. I receive reward points that I can redeem for goods or store vouchers. Each point works out to be worth 10c. The last survey I did I earned $40. Several people I work with do market research for another company that pays $40 - 60 per session to taste test beers or sauces or look at marketting options. Usually sessions only last an hour.
I’m really not concerned with the technical details of this particular survey. My question is more general. If the average return (as computed by odds x reward) is worth more to me than my cost to participte (whether my cost is money, time, a missed test, etc.), does that mean I should play, even though I am only playing one time? In other words, even though I would either win $1000 or win nothing, does the fact that, if I played many times, my average return would be $50 mean that it is a prudent choice to play once?
Yes, that’s the point I was making. Averages are irrelevant in one-shot games or long-shot gambles.
In the case of a one-shot, there are only two possible outcomes - win or lose, so no ‘average’ gain.
In the case of long-shots like lotto the situation is similar. You will be playing a small number of times with a minute chance of a very large return - ‘averages’ would be irrelevant.
Average gain is only a useful measurement over a large population or a large number of trials.
If you’re tring to decide whether it’s worthwhile for you to buy a lotto ticket you need to consider the opportunity cost of your ticket price, and whether there is another more efficient way that you could achieve your goal of winning 10 million dollars.
Only you can decide that. Because only you can decide the values of the things I described, notably your value of the prize, and your value of the cost.
If the prize is worth $1000 to you, and you don’t mind losing the 15 minutes, then it’s a good deal.
If you multiply your odds of winning the lottery by the size of the jackpot, and it comes out to more than the cost of a ticket, then yes, buy, buy, buy! However, if you do the math and find that the quantities compare in this way, you’ve probably made a mistake. If the average lottery player wins money, rather than losing it, the lottery wouldn’t make a profit. We have a few lottery threads open now, if you go back two or three pages on GQ you can find everything you want to know about it.
They’re not irrelevant at all, since there are no true one-shots. Assuming that the OP’s description of the deal is accurate, that the prize is worth $1000 to him, and that his fifteen minutes are worth less to him than $50, then it’s a good gamble. Over the course of a person’s life, you’ll encounter many gambles, some good, some bad. If you take all of the good ones, then you’ll probably be better off than if you just consider each one a one-shot that isn’t worth anything.
**JFMichael ** –
As people have said, you’re right. The term used for it is ‘expected value’ often abbreviated EV. The idea is with a 1 in ten chance of winning $600, you expect to average a $60 win per try, so the expected value is $60.
To expand a little bit, if you have to pay $100 per try for that chance then your EV is -$100 (the cost) plus +$60 (expected winning) so the total EV is -$40. Not a good bet here. Gamblers (the smart ones who actually think, that is) sometimes call a good bet “+EV” (has a positive expected value) or a bad one “-EV”.
And since I’m writing a thesis here anyway, there are two other things to think about when deciding to take a bet, besides whether it’s positive EV. First is that every dollar doesn’t have the same value to you – who cares whether you have $100 million or $101 million in the bank? But the difference between $50 and $1 million is a lot. For small amounts, this probably doesn’t matter much.
But there’s also variance, which is how far the swings are from the expected value. For example, most people would take a job for $100,000 a year. But few would take a job that will give you a 1 in 10 chance to earn a million, but pay nothing the rest of the time. The EV is the same, but most people would like more certainty in how much they’ll be paid.
But in general, EV is a good place to start with deciding whether to take a bet.