TITLE: Game Theory - Unsure how to proceed with this question
QUESTION [14 upvotes]: A company has a competition to win a car. Each contestant needs to pick a positive
integer. If there’s at least one unique choice, the person who made the smallest unique
choice wins the car. If there are no unique choices, the company keeps the car and there’s
no repeat of the competition. It turns out that there are only three contestants, and
you’re one of them. Everyone knows before picking their numbers that there are only
three contestants. How should you make your choice?
Thoughts: Since there's no strategy you can adopt that the others cant, you need to aim for the situation where the other 2 pick the same number so you need to avoid them, so perversely aim high. However the other 2 can do the same as well, so do you aim lower of the high numbers, or does it make no difference? 
I understand the answer is not: pick 12, but once you've gone beyond the strategy of getting the other 2 to match, do you hedge your bets somewhat by keeping it low, or do you maximise your chances of your strategy working by going insanely high (utilising exponentials etc.).

REPLY [5 votes]: If we restrict our search to the geometric distributions with some parameter $p$:
$$S_k=(1-p)p^{k-1}$$
Then all the $Q_n$'s from Robert's answer will be equal to:
$$Q_n = \sum_{i=1}^{n-1} S_i^2 + (1 - \sum_{i=1}^n S_i)^2=\frac{1-p}{1+p}+\frac{p^{2n-2}}{1+p}(p^3+p^2+p-1)$$
So they will all be equal when $p$ is the only real root of:
$$x^3+x^2+x-1=0$$
Which is equal to $s \approx .543689012692076$. The first nine probabilities are therefore:  
$S_1 \approx .456310987307924$ 
$S_2 \approx .248091270169992$ 
$S_3 \approx .134884497736246$ 
$S_4 \approx .073335219401686$ 
$S_5 \approx .039871553032060$ 
$S_6 \approx .021677725302500$ 
$S_7 \approx .011785941067126$ 
$S_8 \approx .006407886662433$ 
$S_9 \approx .003483897572941$  
If all players play according to this strategy then the expected value of their payoff will be the common value of all $Q_n$'s i.e.:
$$\frac{1-s}{1+s} \approx 0.295597742522085$$
Which is also the unique real root of $x^3+x^2+3x-1=0$. It is less than $\frac{1}{3}$ because there is a:
$$\sum_{n=1}^{\infty}S_n^3=1-3\frac{1-s}{1+s}=\frac{4s-2}{s+1} \approx .113206772433746$$
Chance of all of them choosing the same number (it is the real root of $x^3-6x^2+36x-4=0$).