I’m offering a deal on lottery tickets. On each ticket, you have a 75% chance of winning a great prize. The prize is such that you only want to win once: winning additional prizes brings no additional benefit. There’s two ways to get in on this deal:
The first ticket I sell will cost $1,000. If you lose on the first ticket, you can buy one additional ticket at a time for the discounted price of $300 per. No limit on how many tickets you can buy.
You can buy a set of six tickets for $1,750. If you lose on all tickets, you get a refund. If you win on the first ticket, you don’t get any rebate because you bought the tickets as a group.
Which offer would you take?
(I’m clear on what my answer would be, but somebody happens to disagree with my logic strongly.)
For the sake of argument, let’s say the prize has infinite value - so you’re going to keep buying tickets until you win or run out of money. For $1750 you can either have four chances or six chances - the six chances is the way to go.
Now say the prize is only worth $5000. Now the possibility of winning before we’ve spend $1750 matters. The first option gives us this possibility, and it may be worthwhile. It gets more and more worthwhile as your prizemoney goes down. I’m not going to calculate the break-even point here, but you could do it.
ETA: With infinite (or very large prizemoney) the ever-renewing six chances makes a difference too. With the six chances you never run out of money, because you keep getting your money back. With small prizemoney this doesn’t come into it. What matters is your chance of winning before you’ve spent the full $1750.
Agreed except for the fact that each ticket has a 75% chance of winning so, without actually doing the maths, you should win the prize in the first 2 tickets.
So $1300 vs $1750.
Option 2 is the sure fire way to give yourself the best shot, provided the prize is worth more than $1750, but the odds are in favour of getting it cheaper using option 1.
I’ve no idea - I like the break-even-on-failure clause of option 2, but if I had sufficient funds to play a few dozen times, it would be extremely unlikely that option 1 would never prevail - and if the prize is lots of money, I won’t care that I wasted some in winning it, once I eventually win.
The description of the prize is slightly suspect and makes me think there’s a ‘gotcha’ here - please can you confirm that "…a great prize … such that you only want to win once" does in fact refer to a phenomenally large sum of money and not, say, a life size marshmallow replica of the Arc de Triomphe? (I mean, I certainly don’t desire to win that twice).
For the first ticket, you have an expected utility of .75*prize_value - .25 * cost.
Prize_value isn’t given, but cost is stated as $1000.
For the set of 6 tickets we have to set it up with a binomial distribution P(n) = (6Cn).75[sup]n[/sup].25[sup]6-n[/sup], where n is the number of times you’ll win.
The expected utility of this is going to be
(sum from n=1 to n=6)[ n*P(n)*prize_value - (1-P(n)) * cost) ] + P(0) * cost
Where prize_value is the same value as above, and cost is $1750.
Then it depends on exactly what prize_value is. I think I’d generally go with option 2, because when I do the math, there appears to be a 99% chance of you winning at least one ticket.
Hmm, it occurs to me that since you can buy as many tickets as you want the first utility is actually
(sum from n=1 to m)(P(m,n) * n * prize_value) - m * cost
Where P(m,n) = (m C n) * .75[sup]n[/sup] * .25[sup]m-n[/sup]
Where “m” is the number of tickets you buy.
Also, in that original equation it’s not .25 * cost, you just subtract the cost since you always lose the cost money.
Similarly, the second equation should be
(sum from n=1 to 6)(P(n) * n * prize_value) - 6*cost + P(0)*cost
Since, again, the cost is constant unless you lose all of them.
One thing to note is that the cost of 2 is significantly less than the cost of 1 if m=6. That said, as m->infinity, the expected utility of 1 goes through the roof because the odds of winning are so high. However, if you can buy any number of option 2’s as you want, it will check out slightly better.
With option 1, on average you will spend $1100 to win the prize. This appears to be better than option 2, where (assuming you’re allowed to repeat until winning) you’ll spend $1750 exactly.
However, although you spend only $1100 on average you might spend much more, and might fail altogether if your capital is exhausted. For example, if your total capital is $1900 you have a 0.39% failure probability with option 1. With option 2 your failure probability would be either zero or 0.0244% (according as repeat buys are or aren’t allowed, respectively).
Thus, option 2 might be better if your capital is constrained. To answer exactly, it would be nice to know your “utility of money” function and whether repeat buys in option 2 are allowed.
But with option 2 you might win multiple times, which I think offsets the extra cost, if you win all 6 times your cost “per ticket” is only 1750/6, whereas if you win 6 times with repeat buys on option 1, you still spend 1k per ticket.
ETA: Oh, missed that winning multiple times brings “no additional benefit”
Maybe I’m missing something but isn’t the refund the big issue that outweighs all others? You only get the refund if you buy the six packs. So the smart thing to do is buy a sixpack of tickets and scratch them. If you win (and you probably will) you’ve won the “big prize” for an investment of $1750. But if the odds were against you, just collect the refund for your losing tickets and turn around and buy another sixpack. Keep doing this until you win. You’ll never have to invest more than your original $1750.
With the non-refund single tickets, there’s a small chance of having an open-ended run of bad luck with no potential end.
In this way, as long as the value of winning is greater than the cost, option 2 has no risk, and gives you a higher chance of return, but option 1 will give a greater return if you succeed on the first try. If you purchase additional tickets, you will always lose more than you would have taking option 2 since the cost compounds with each additional ticket on option 1.
Even if you can’t purchase additional runs of tickets with option 2, you have a >99% chance of succeeding. If the prize is so phenomenally mindblowing and offsets the cost of tickets immensely, it may be worth it to take option 1 to ensure you get it, but based on pure utility option 2 is far superior regardless of whether option 2 allows redos or not.
The only time option 1 might be preferred is if you have a very small reward and that extra guaranteed $750 loss is devastating (say the prize is $2000, you have to decide whether very good odds to get $1000 is better than almost perfect odds to get $250), but I doubt that is the case and I still feel you have far more to lose by risking having to buy a second ticket under option 1.
I hope OP comes back and tells us whether option 2 is repeatable or not, but even if he says Yes that option figures to cost you more. By buying up to three tickets in option 1, you win with probability 98.44% and save an average of $656 compared with option 2. In the limit, if you’re willing to buy an unlimited number of tickets in option 1, your savings will average $650 compared with option 2.
To solve the problem exactly, one needs the details. (You should know that by now!) Specifically we want to know your total capital and what its utility is to you.
(Utility = logarithm Total_Capital is a frequent, but by no means generally accepted, assumption.)
Can you check this? You’re not assuming that 2nd or subsequent prizes convey any benefit are you?
If you buy up to seven tickets in option 1, you have a higher success probability than with a single option-2 ticket, have spent (very slightly less than) $1100 on average and despite that there is no failure cost in option 2, are clearly better off if the prize is sufficiently humongous.
It has to cost more, if you win on your second attempt at option 1, since there’s no refund you’ve spent prize_value-$2000. There’s no way around that.
Edit:
We’ll call the utility of purchasing m option 1 tickets U[sub]1b[/sub]
U[sub]1b/sub = [if m < 4] (sum n=1->m)[P(n,m)] * prize_value - (sum k = 0->m-1)[$1000 - k * $300]
[if m > 4] = (sum n=1->m)[P(n,m)] * prize_value - $2200
(Assuming the cost can’t go negative, of course)
So you really get two tickets for less than the cost of option 2, but after the third ticket you get (1000+700+400)=$2100
ETA: Actually I have to admit that I’m not sure whether I should be using
(sum n=1->m)[P(n,m)] * prize_value
or just
P(m,m) * prize_value
The fact that you buy them one at a time is messing me up. The former, I think, would be more appropriate if you bought them in bulk since it’s more of “at least one”, the latter is closer to “the utility if you’re forced to buy m tickets”.
it all boils down to whether you can win the bet in 3 tries isn’t it? so the choice is clearly the first option since the odds are in your favour. however, as the size of your capital or gonads goes down, you may have no choice but to go with option 2.
What was I thinking? P(m,m) doesn’t even make sense, at best you want P(1,m), since P(m,m) is the probability of ALL your tickets winning. Herpdederpderp. I’m still not sure whether P(1,m)* or the sum is most appropriate, though.
I just realized I’m passing the arguments in the reverse of “standard” order, I wrote it as P(num successes, num trials) when usually it’s the other way… oops. I’ll leave it that way for consistency.
I’m not sure I can answer the “utility” question, mainly because the value of the “prize” is more of a personal experiential nature, rather than having a dollar value assigned to it. Think of the value as being something like throwing out the first pitch at a baseball game, or getting to have dinner with your childhood hero. There isn’t a gotcha with the prize.
I’m not sure how to give a specific answer to that, either. The sums are expressed as what I think would be a realistic reflection of the burden on a typical household income: $1,000 is a non-trivial amount of money, and you certainly do not want to waste it; on the other hand $1,750 is a substantial additional burden so that saving a few hundred bucks is an attractive option.
My view is that option #1 is substantially more attractive because the odds are quite high that the prize will be won within two or three tries. I view the increased cost of option #2 as being rather costly premium to very nearly guarantee a win. Given the odds, I think the refund policy is more about marketing than it is about adding a substantive factor to the decision-making. It is like lightning would have to strike to get the refund.
However, the compadre with whom I’m discussing this scenario fundamentally does not agree that the odds are quite high that one would probably (but not certainly) win after a few attempts at option 1. As in, simply doesn’t believe that there’s less than a 10% chance of “losing” a couple times in a row, because obviously there’s a 25% of losing on every ticket, and I’m committing a gambler’s fallacy. I’m not sure how to explain my disagreement with that conclusion.
The one other relevant point on this discussion is that the 75% odds of winning is kind of a WAG. To be strictly accurate, the probability of winning on any ticket is some undefined odds that would likely fall between 65% and 85%, as far as I can tell.
Assuming exactly 75% chance of winning with each ticket, you’ve got an expected cost of winning of $1,100 in option A, and $1,750 in option B, to win the same prize. So I’m going with Option A.
If you can’t explain to your friend that there’s only a 1/64 chance of spending more in option A than in option B, and less than 1 chance in 1000 that option A would be at least $750 worse than Option B, while you’ve got a 75% chance that option A will be $750 better than Option B, I’m not sure I can help you. Some people just don’t grasp compound and conditional probabilities.
Maybe he once saw or read the opening scene of Rosencranz and Guildenstern Are Dead, and it made a big impression on him.