Vtyjkyuk

Suppose your skill (the ability to make a shot) follows a binomial distribution (with some unknown $p$). You need to make at least half of the shots. Would you rather take $4$ or $6$? More precisely, what should be the $p$ which is the threshold of your decision?

My Try: assuming $0 < p < 1$.
$$6choose3p^3(1-p)^3 =4choose2p^2(1-p)^2 \
implies p(1-p) = 6/20
$$

but apparently the answer is: $3/5$.

What's wrong with my calculation?

In the case of $B(6,p)$ we fail with probability ($0,1,2$ shots)

$$(1-p)^6+6p(1-p)^5+15p^2(1-p)^4.$$

In the case of $B(4,p)$ we fail with probability ($0,1$ shots)

$$(1-p)^4+4p(1-p)^3.$$

Now, we have

$$(1-p)^4+4p(1-p)^3-(1-p)^6-6p(1-p)^5-15p^2(1-p)^4=2p^2(1-p)^3(5p-3). $$ And thus we get $p=dfrac35.$

We can conclude that if $p<dfrac35$ it is better to choose $6$ shots and if $p>dfrac35$ $4$ shots.

Probability of making AT LEAST half shots out of 6 is

$$
binom63p^3(1-p)^3 +binom64p^4(1-p)^2 +binom65p^5(1-p) +binom66p^3
$$

and for making at least 2 out of 4 is

$$
binom42p^2(1-p)^2 +binom43p^3(1-p) +binom44p^4
$$

For threshold $p$, equating both, we get:
$$
10p^4-36p^3+48p^2-28p+6 = 0
$$
which reduces to
$$
2(x-1)^3(5x-3) = 0
$$
which gives $p = frac 35$ since $0 lt p lt 1$.