# A lottery with two cars

In this question on math the situation was given

A Tv Shows organize a lottery with two cars as prizes. Any watcher may send an SMS for participating independently of other watchers. If there are only one or two participants then they get a car. Otherwise, no one wins.

If you assume that all the other viewers will behave as you do, what should you do to maximize your expected winnings?

Added: Douglas Hofstadter had a piece "Dilemmas for Superrational Thinkers, Leading Up to a Luring Lottery" in the June 1983 Metamagical Themas (free download) like this with just one prize, which was a million dollars divided by the number of entries. It is an interesting essay.

• Can "not sending an SMS" lead to winning a car? Nov 10, 2017 at 17:39
• No, if you do not send an SMS you will certainly not win a car. Nov 10, 2017 at 17:44
• Does "what should you do" contain other actions besides sending / not sending an SMS? You know like rob a bank to distract other people from watching the tv show that broadcast the lottery or something like that... Nov 10, 2017 at 18:03
• Any watcher may send an SMS for participating independently of other watchers What does this mean? Nov 10, 2017 at 20:41
• Why probability is not an applicable tag here? Nov 11, 2017 at 15:06

If I understand correctly the assumptions we're supposed to be making, then essentially the only thing we can do is: send an SMS with probability $p$. (We get to choose $p$.) And I'll assume that we know the number $n$ of viewers exactly. Then

with probability $1-p$ we don't participate (hence don't win); conditional on our participating, with probability $(1-p)^{n-1}$ no one else participates and we win; with probability $(n-1)p(1-p)^{n-2}$ exactly one other person participates and we win. So our winning probability is $p(1-p)^{n-1}+(n-1)p^2(1-p)^{n-2}$. The ensuing calculation will be a bit nicer if we express it in terms of $q=1-p$; we get $(1-q)q^{n-1}+(n-1)(1-q)^2q^{n-2}$ or, expanding, $q^{n-1}-q^n+(n-1)q^{n-2}-2(n-1)q^{n-1}+(n-1)q^n$ or, simplifying, $(n-1)q^{n-2}-(2n-3)q^{n-1}+(n-2)q^n$. The derivative of this is $(n-1)(n-2)q^{n-3}-(n-1)(2n-3)q^{n-2}+n(n-2)q^{n-1}$ which is zero (unless $n$ is implausibly small) when $q=0$ (obviously an irrelevant solution since then there are always too many participants and no one wins) or when $n(n-2)q^2-(n-1)(2n-3)q+(n-1)(n-2)=0$. Solving the quadratic, this means $q=\frac{(n-1)(2n-3)\pm\sqrt{(n-1)^2(2n-3)^2-4n(n-1)(n-2)^2}}{2n(n-2)}$ or, simplifying the stuff inside the square root, $q=\frac{(n-1)(2n-3)\pm\sqrt{(n-1)(5n-9)}}{2n(n-2)}$. So $p=\frac{n+3\pm\sqrt{(n-1)(5n-9)}}{2n(n-2)}$ and clearly only the positive sign leads to an actually-possible probability, at least when $n$ is large.

So, in the unlikely event that I haven't made any mistakes in my algebra above, you should

pick up your phone and send the text with probability $\frac{n+3+\sqrt{(n-1)(5n-9)}}{2n(n-2)}$. For large $n$ this is approximately $\phi/n$ where $\phi=\frac{1+\sqrt5}2$ is the golden ratio.

• I didn't check every step, but the last step is wrong; the final expression behaves like $\sqrt{2}/n$, which is a plausible answer. Nov 10, 2017 at 18:21
• The expression$\frac12(n-1)p^2(1-p)^{n-2}$ should not have the $1/2$, because if there are two participants, then they both get cars. Nov 10, 2017 at 18:24
• Oo, an actual math answer while I was typing my silly one. The conclusion seems unlikely though: if a million viewers each participate with a one-in-a-thousand probability, there will be more than 2 SMSs. Also, looks like you didn’t account for there being two prizes.
– Bass
Nov 10, 2017 at 18:34
• D'oh! I misread the question as saying that there was one prize and you got it with probability 1/2 of there were two entrants. And yeah, it seems like the wrong order of magnitude. Unfortunately I'm on a small mobile device right now so it may be a while before I fix my boneheaded errors. Nov 10, 2017 at 20:08
• OK, might all be fixed now. Editing LaTeX on a phone is, um, error-prone so there may well be exciting new mistakes to replace the old ones. Nov 10, 2017 at 20:35

In order to maximize my chances, I (and everyone else, by extension) will adopt the strategy that I will send an SMS if and only if my StackExchange username is Nuclear Wang.

Only I send the text, and I will win 100% of the time.

• This raises the philosophical point of what is the same strategy. I would argue that the same strategy is for each person to say I will send an SMS if I can send it from my phone and do so. Now we get too many responses. Nov 10, 2017 at 17:56
• @RossMillikan If you're saying that "having the ability to send an SMS" and "Having the username Nuclear Wang on StackExchange" are essentially the same condition, I think I have to disagree... Nov 10, 2017 at 20:36
• Usernames on StackExchange are not unique. So I can get myself a car by registering as NuclearWang, and so can all my buddies...oh, wait. Nov 10, 2017 at 22:20

To maximize everyone’s chance of winning, while keeping the game fair,

everyone should look up the number of viewers for the show, double it, then find a dice with that many sides, roll it, and send the SMS iff they got a 13, a 42, or a 69.
That places the expected number of SMS’s at 1.5, which makes the no-win scenario as unlikely as possible.

(It’s probably possible to aim for a slightly higher EV to increase the chance of getting two cars instead of just one, but at some point the increased chance of ”too many” will balance that. Finding out whether there is a maximum somewhere else would require math, which I’ll gladly leave for someone else.)

Edit: that maximum was found in another answer, go vote that, since it’s beautiful. Now that this approach is fully investigated, here’s another one, which isn’t nearly as interesting, but it is fair to all participants, and will provide a better outcome on average. You’ll find yourself thinking, that can’t possibly be within the rules, and you are probably right. But you can plausibly interpret the question so that this would be allowed. :-)

Draw lots among all possible participants. Send the SMS if you got one of the two winning tickets.

• What's the significance of 13, 42 or 69? If the show only has 6 viewers, this strategy has a 0% chance of working. Nov 10, 2017 at 19:08
• I find it unlikely that a show with less than 35 viewers would thay on air, much less raffle off cars :) The significance is, of course, that they are, in a completely objective and absolutely undisputable manner, better numbers than 1, 2 and 3.
– Bass
Nov 10, 2017 at 19:23
• @NuclearWang The significance of the first is that it's unlucky. The second is the answer to the Ultimate Question. And the third I'll explain when you're older. Nov 12, 2017 at 8:11

If you don't send an SMS, then the chance of you winning a car is always going to be 0.

If you do send an SMS (assuming all other viewers do so), then if there is less than 2 viewers (you included), you'll win a car, if there is more the chance is 0.

So sending an SMS is the best option you got.

                  |# of viewers     |    1 |    2 |    3 |  ... |
|----------------------------------------------
|send SMS         |  win |  win | lose | lose |
|----------------------------------------------
|do not send SMS  | lose | lose | lose | lose |


Note: I used the term number of viewers and not number of participants because when assuming all other viewers will behave as you do then the two numbers are equal (if you do participate of course) and thus the terms are equivalent.

I noticed that there was a condition in the question: If you assume everyone else will respond as you do.

So you go down to the station, and agree if you are the first one or two standing outside, you will send the message, and pay off the others who participated in the event that you win, or accept the payout if you weren't the first.

But basing the entire answer on that premise is silly, so I prefer this answer:

Go to the station and steal the car(s).