A three sided die?

Let the probabilities of getting 1,2,3 on the dice be $p,q,r$ , so $p+q+r=1$

Chance of getting out $C$ is $p^2+ q^2+ r^2$

Average score $A$ per round if not out is $\dfrac{4pq+6qr+6pr}{1-C}$

Let expected score be $X$ , if one round is over Alex has $C$ chance of being out or $1-C$ chance of scoring $A$ and playing on to score the expected score.

So $X=0*C+(1-C)(A+X) ⇒ CX= A(1-C) = 4pq+6qr+6pr$

$⇒X=\frac{4pq+6qr+6pr}{p^2+ q^2+ r^2}$

We know that at the maximum or minimum of $X$ , it’s partial derivative’s with respect to $p,q$ and $r$ will be all equal to zero

Taking the partial derivative’s and equating to zero we get the following three equations.

$2p(4pq+6qr+6pr)=(4q+6r)(p^2+ q^2+ r^2) ➝ 1$ $2q(4pq+6qr+6pr)=(4p+6r)(p^2+ q^2+ r^2) ➝ 2$ $2r(4pq+6qr+6pr)=(6q+6p)(p^2+ q^2+ r^2) ➝ 3$

Adding all these three equations results in:

$2(p+q+r)(4pq+6qr+6pr)=(10q+10p+12r) (p^2+ q^2+ r^2)$

Since $p+q+r=1$ and $X=\dfrac{4pq+6qr+6pr}{p^2+ q^2+ r^2}$

We get : $X=5p+5q+6r$ or $r=X-5$ and $p+q=6-X$

Rearranging equation 3 we get : $X × r = 3(p+q)$ Substituting for $X$ $X(X-5)=3(6-X)$ $⇒ X^{2}-2X-18=0$ $\therefore X=1+\sqrt{19}$ We are ignoring the negative root

$r=X-5=\sqrt{19}-4$

Since p and q are symmetric in equations $1$ and $2,p=q$

$p+q=6-X=5-\sqrt{19} ⇒ p=q=\dfrac{5-\sqrt{19}}2$

Give your ideas for a three faced die down here $\downarrow$