Question

You play a game where you first choose a positive integernand thenflip a fair coinntimes. You win a prize if you get exactly 2 heads. How should youchoosento maximize your chance of winning

Answer 1

Solution :

The probability of winning when you choose n is =  $^nC_2\left(\frac{1}{2}\right)^n$

$n\left(\frac{n-1}{2}\right)\times \left(\frac{1}{2}\right)^n = n(n-1)\left(\frac{1}{2}\right)^{n+1}$

Apply log on both the sides,

$f(n) = \log\left((n)(n-1)\left(\frac{1}{2}\right)^{n-1}\right) = \log n +\log (n-1)+(n+1) \ \log\left(\frac{1}{2}\right)$

Differentiation, f(x) is $f'=\frac{1}{x}+\frac{1}{(x-1)}+\log\left(\frac{1}{2}\right)$

Let us find x for which f' is positive and x for which f' is negative.

$\frac{1}{x}+\frac{1}{(x-1)} > 0.693$       , since $\log(1/2) = 0.693147$

For x ≤ 3, f' > 0 for $\frac{1}{x}+\frac{1}{x-1}+\log\left(\frac{1}{2}\right)>0$

$\frac{1}{x}+\frac{1}{x-1}-0.6931470$

That means f(x) is increasing function for n ≤ 3

$\frac{1}{x}+\frac{1}{x-1}< 0.693147$ for x > 4

f' < 0 for n ≥ 4, that means f(n) is decreasing function for n ≥ 4.

Probability of winning when you chose n = 3 is $3(3-1)\left(\frac{1}{2}\right)^{3+1}=0.375$

Probability of winning when you chose n = 4 is $4(4-1)\left(\frac{1}{2}\right)^{4+1}=0.375$

Therefore, we should chose either 3 or 4 to maximize chances of winning.

The probability of winning with an optimal choice is n = 0.375