# Dice probability with limited reroll

#### Dan K

##### New Member
Hello,

I'm having difficulty understanding this problem and I was wondering if someone had an answer.

Let
$$n = \text{Number of dice} = 3$$
$$p_s = \text{Single die success probability (when the face is 3 or more)} = \frac{1}{2}$$
$$k = \text{Number of successful die} \in 0, 1, 2, 3$$
$$P(n, k, p) = (^n_k) p^k(1-p)^{n-k}$$

The probability of getting 2 success $$(k=2)$$ if each die is rolled only once $$(p = p_s = \frac{1}{2})$$ is
$$P(3,\ 2,\ \frac{1}{2}) = (^3_2)\left(\frac{1}{2}\right)^2\left(1-\frac{1}{2}\right)^{3-2} = \frac{3}{8} = 0.375$$

If we have the option to reroll any failed die once,
then $$p = p_s + (1-p_s)p_s = \frac{3}{4}$$
and $$P(3,\ 2,\ \frac{3}{4}) = (^3_2)\left(\frac{3}{4}\right)^2\left(1-\frac{3}{4}\right)^{3-2} = \frac{27}{64} = 0.421875$$

What is the probability if we can reroll only one failed die (instead of all 3)?

#### Dan K

##### New Member
Out of the 3 dice, 2 $$(n_{1,2} = 2)$$ are rolled once $$(p_{1,2} = p_s = \frac{1}{2})$$ and 1 die $$(n_3 = 1)$$ is rolled twice $$(p_3 = p_s + (1 - p_s)p_s = \frac{3}{4})$$

The probability of rolling 2 successes when all three dice are rolled and then one is rerolled is:
$$P(\text{Rolling 2 successes on the 1-roll dices})\times P(\text{Rolling 0 successes on the 2-roll die})$$
+$$P(\text{Rolling 1 success on the 1-roll dices})\times P(\text{Rolling 1 success on the 2-roll die})$$

$$= P\left(2,\ 2,\ \frac{1}{2}\right)\cdot P\left(1,\ 0,\ \frac{3}{4}\right)$$ + $$P\left(2,\ 1,\ \frac{1}{2}\right)\cdot P\left(1,\ 1,\ \frac{3}{4}\right)$$

$$= (^2_2)\left(\frac{1}{2}\right)^2\left(1 - \frac{1}{2}\right)^{2 - 2}\cdot(^1_0)\left(\frac{3}{4}\right)^0\left(1 - \frac{3}{4}\right)^{1 - 0}$$
+$$(^2_1)\left(\frac{1}{2}\right)^1\left(1 - \frac{1}{2}\right)^{2 - 1}\cdot(^1_1)\left(\frac{3}{4}\right)^1\left(1 - \frac{3}{4}\right)^{1 - 1}$$
$$= 0.4375$$

However, this seems wrong because it's greater than if I were to roll 3 dice with rerolling $$P\left(3, 2, \frac{3}{4}\right) = 0.421875$$

What am I missing?

P.S. Sorry for the formatting. The tex on this site doesn't support \binom{x}{y} nor the '+' sign.