Winning probability to choose a chip of a box.

Question

The Player got 10 blue and 10 red chips, and he was offered a game where he should distribute those chips into 2 similar boxes. (All chips should be placed in one of the boxes.) Then, another person randomly chooses a box and then takes out one random chip from this box (unless the box is empty). The player wins if a red chip is drawn and lose otherwise. What's the maximum probability of the player's winning outcome assuming that he distributes chips the right way?

I have been trying this problem for the past 1 week and not able to proceed further. The best I have reached is:

P(getting R) = P(getting Red | in chosen Box 1)*P(choosing Box 1) + P(getting Red | in chosen Box 2)*P(choosing Box 2)

But not able to think of a method to maximise the probability. Can anyone please help me with this?

Answer 1

It is more or less clear, that the distribution $$ \color{red}{1} + \color{blue}{0} \qquad\text{ and }\qquad \color{red}{9} + \color{blue}{10} $$ in the first box, and respectively in the second box is optimal. (Or switch boxes.)

The probability of winning is then $$ \frac 12\cdot 1+\frac 12\cdot \frac 9{19}=\frac{14}{19}\ . $$ This is the maximal value among all possible ways to split the $20$ chips, $$ \color{red}{p} + \color{blue}{q} \qquad\text{ and }\qquad \color{red}{(10-p)} + \color{blue}{(10-q)} $$ where the probability is $$ \frac 12\cdot \frac {p}{p+q} + \frac 12\cdot \frac {10-p}{20-p-q}\ , \qquad p+q\not\in\{0, 20\}\ . $$

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For the sake of completeness, i'll add a...

Proof: Here is a quick estimation. If $p=0$, we obtain a probability $\le 1/2$, this is not optimal. So we may and do assume by symmetry that $1\le p\le 5$.

The difference $\displaystyle \frac 12\left(\frac {p}{p+q} + \frac {10-p}{20-p-q}\right)-\frac {14}{19} $ has positive denominator, and the numerator is $$ -5p^2 + 5p + 9pq + 14q^2 - 185 q \ . $$ As a function in $q$, the part $9pq + 14q^2 - 185 q$ is zero in $q=0$, then it is monotone decreasing on $J:=[0, (185-9p)/2]\supseteq[0, (185-45)/14/2]=[0,5]$, then it increases on an interval of the same length as $J$, "following" $J$, so the value in $10$ is zero or less. This shows that an optimal value is obtained for $q=0$. It remains to maximize $-5p^2+5p$ on $[1,5]$. Of course, $p=1$ gives the maximum.

$\square$

Answer 2

Another derivation.

Without loss of generality we may assume that the left box contains $n\le10$ chips with $p$ chips from them being red ($0\le p\le n$).

Then the probability to win $$ \frac12\left(\frac pn+\frac{10-p}{20-n}\right)=\frac{5n+p\,(10-n)}{n\,(20-n)} $$ for $n<10$ is obviously maximized by the choice of the largest possible $p$, i.e. $p=n$, so that the expression to maximize is: $$ \frac{5n+n\,(10-n)}{n\,(20-n)}=\frac{n(15-n)}{n(20-n)}, $$ which is clearly minimized by the choice of the smallest possible $n$ except for $0$, i.e. $n=1$.

That means that one red chip should be placed in one box, and the other 19 chips in the other box.