3 contestants choosing a smallest number to win a car Each of 3 contestants chooses a positive integer. The contestant who chooses the unique smallest positive integer number wins a car. If all of them chose the same number, then no body wins car. What is the optimal strategy?
I do not know how this is done, so I can only guess.
The players should play the same strategy, by the nature of the game. 
I started off assuming that it makes no sense choosing 4 or above (so I assumed one should choose 1,2 or 3).
After some algebra bashing, I arrived at the result.
maximise  $x(1-x)^2 + y(x^2+(1-x-y)^2)+(1-x-y)(x^2+y^2)$ subject to the condition $x+y\leq 1$
Somewhat surprisingly, this expression in symmetric in x and y. I used the Language's multiplier and arrive at $x=y=4/3-\sqrt{40}/6$, which is about 0.28. 
This suggest the contestants should choose 1 or 2 with probability 0.28 and choose 3 with probability 0.44. The chance of winning is about 0.29.
However, it is clear that this is not the optimal strategy. If players A and B uses such strategy, then C can choose 1 or 2 with probability 1/2, then he has about > 0.5 chance of winning. 
I suspected that the optimal strategy is all players choose 1 or 2 with probability 1/2. I seem to find is no way of deviating from this strategy which would improve each player's pay off. Is this right? how do we prove this?

If there is a Nash equilibrium, and the other two are playing it, then it doesn't matter what number the first player plays.  So if the others play with probability distribution p(1)=a,p(2)=b,p(3)=c,
First player wins playing 1 with probability $(1-a)^2$
First player wins playing 2 with probability $a^2+c^2$
First player wins playing 3 with probability $a^2+b^2$
So $(1-a)^2=a^2+c^2=a^2+b^2,a+b+c=1$
which has solution $a=2\sqrt{3}-3,b=2-\sqrt{3},c=2-\sqrt{3}$ and 
first player wins with probability $28-16\sqrt{3}=0.287$
If they use four numbers, then $(1-a)^2=a^2+(1-a-b)^2=a^2+b^2+d^2=a^2+b^2+c^2$
and Wolfram says the chance of a win is 0.294
When more numbers are used, it seems that $a/b,b/c,c/d,...$ all approach the same ratio.  That gives this solution for infinitely many:
$$(a_1,a_2,...)=a_1(1,x,x^2,x^3,x^4,...)\\a_1=1-x$$
Then, using $$(1-a_1)^2=a_1^2+(1-a_1-a_2)^2\\x^2=(1-x)^2+x^4\\x^3+x^2+x=1$$
we get the value of $x=0.5437$ for the Nash equilibrium, with a win probability of $(1-a_1)^2=0.2956$
