Expected Value of Iterative Dice Rolling

Let $E$ be the expected value. There are three cases.

If we get three of a kind, the expected value is $\frac16(3 \cdot 1 + 3 \cdot 2 + \cdots + 3 \cdot 6) + E = 10.5 + E.$ This happens with probability $\frac1{36}.$

If we get two of a kind, the expected value is $10.5 + E_2,$ where $E_2$ is the expected value of the process starting with two dice. Note that the expected value of a two-of-a-kind set of three rolls is 10.5 by a symmetry argument: for every $xxy$ roll, there is a $(7-x)(7-x)(7-y)$ roll, and the sum of those two roll sums is $21.$

If we get three different numbers, the expected value is $10.5$ (by the same symmetry argument). This happens with probability $\frac{6 \cdot 5 \cdot 4}{6 \cdot 6 \cdot 6} = \frac59,$ so the probability of two of a kind is $1-\frac1{36} - \frac59 = \frac5{12}.$

So $\begin{aligned} E &= \frac1{36}(10.5 + E) + \frac5{12}(10.5 + E_2) + \frac59 (10.5) \\ E &= 10.5 + \frac1{36} E + \frac5{12} E_2 \\ \frac{35}{36} E &= 10.5 + \frac5{12} E_2 \end{aligned}$

To compute $E_2,$ there are two cases: two of a kind leads to an expected value of $7 + E_2$ and two different leads to $7,$ so we get the equation $E_2 = \frac16(7 + E_2) + \frac56(7) = 7 + \frac16 E_2,$ which gives $E_2 = 8.4.$ Plugging back in above gives $E = \frac{36}{35}\left(10.5 + \frac5{12} \cdot 8.4\right) = \fbox{14.4}.$

Let $E_3$ be the expected total for three dice, $E_2$ for two dice, and $E_1$ for one dice. It is straightforward to see that $E_1=3.5$ . For two dice, there is a $\dfrac 16$ chance of having a double, so we can write the equation $E_2=\dfrac 16 (7+E_2) + \dfrac 56 (7)$ The first term in this equation is for if we roll a double in which case we have an expected value of $7$ plus any more points we get from rolling the two dice again. The second term is for if we rolled two different numbers in which case we have expected value $7$ . Solving for $E_2$ gives $E_2=8.4$ . For rolling three dice, rolling a triple has a $\dfrac{1}{36}$ chance, rolling exactly a double has a $\dfrac{5}{12}$ chance, and rolling three different values has a $\dfrac 59$ chance. This leads to the equation $E_3=\dfrac{1}{36}(10.5+E_3)+\dfrac{5}{12}(10.5+E_2)+\dfrac 59 (10.5)$ by similar reasoning as before. Plugging in $E_2$ and solving for $E_3$ gives $E_3=\boxed{14.4}$

This problem can be generalized to the case of $n$ dice, where if multiple matches appear in one roll all sets are re-rolled separately. Let $\mu_{n}$ denote the expected value for a roll of $n$ dice while not considering re-rolling. All we need to consider now is the expected values of the additional re-rolls, which we will call $E_{n}$ , along with the probability for a group of $x$ dice to be the same, which we will denote $P_{n}(x)$ . The expected value for $n$ dice can then be represented as, $\begin{aligned} E_{n} &= \mu_{n} + \sum_{x=2}^{n} P_{n}(x)E_{x} \\ &= \mu_{n} + P_{n}(n)E_{n} + \sum_{x=2}^{n-1} P_{n}(x)E_{x} \end{aligned}$ $E_{n} = \frac{\mu_{n} + \sum_{x=2}^{n-1} P_{n}(x)E_{x}}{1 - P_{n}(n)}$

To find $\mu_{n}$ we must add together $6^{n}$ groups of $n$ numbers, each with a $\frac{1}{6^{n}}$ chance of occurring. Since each sum has the same probability, we can factor the probability out, and as we group the digits together their must be $n6^{n-1}$ of each number $1$ through $6$ , so

$\mu_{n} = \frac{1}{6^{n}}(n6^{n-1}(1 + 2 + 3 + 4 + 5 + 6))=\frac{21n}{6}$

We can write $P_{n}(x)$ explicitly be noting that you have $6$ options for the same set number, $5^{n - x}$ options for the remaining numbers, and the pair can be arranged $\binom{n}{x}$ ways, so

$P_{n}(x) = \frac{6 \times 5^{n - x} \binom{n}{x}}{6^{n}} = \frac{5^{n-x}\binom{n}{x}}{6^{n - 1}}$

Finally note that $P_{n}(n) = \frac{1}{6^{n - 1}}$ , so

$\begin{aligned} E_{n} &= \frac{\mu_{n} + \sum_{x=2}^{n-1} P_{n}(x)E_{x}}{1 - P_{n}(n)} \\ \\ &= \frac{\mu_{n} + \frac{1}{6^{n-1}}\sum_{x=2}^{n-1} 5^{n-x}\binom{n}{x}E_{x}}{\frac{6^{n-1} - 1}{6^{n-1}}} \\ \\ &= \dfrac{21n \times 6^{n-2} + \sum_{x=2}^{n-1} 5^{n-x}\binom{n}{x}E_{x}}{6^{n-1} - 1} = E_{n} \end{aligned}$

With this equation we can find that $E_{2} = 8.4$ and $E_{3} = \boxed{14.4}$

Different from the current given solutions, I made a program in Python:

import random

i = 1

n = 2000000

value = 0

while i <= n:

a = random.randint(1, 6)

b = random.randint(1, 6)

c = random.randint(1, 6)

value = value + a + b + c

if a == b == c:

    i -= 0

elif a == b or b == c or a == c:

    j = 1

    while j == 1:

        d = random.randint(1, 6)

        e = random.randint(1, 6)

        value = value + d + e

        if d != e:

            j = 0

            i += 1

        else:

            j = 1

elif a != b and b != c and a != c:

    i += 1

print(str(value/(n)))

This code, which basically describes the process of n experiments as mentioned in the problem, will approximate the expected value by executing the code n times. When n gets bigger, the result of the experiments devided by n will tend more and more towards the expected value (if n approaches infinity, the results of the individual experiments devided by n will equal the expected value). So when setting n equal to a number big enough, the result will approximate the expected value. I used n = 2000000 (two million), and did the experiment 3 times. This is what got printed:

14.401035

14.3966635

14.4023505

Taking the average of the three got me 14.4000163 which is approximately 14.4.

In rolling three dices, probability of three dices with same face is (1/36) and of two dices with same face is (15/36). The average sum of faces is 3×3.5=10.5

In rolling two dices, the probability of two dices with same face is (1/6). The average sum of faces is 2×3.5.

In the game defined in the problem, the iterated sum is given by:

(3×3.5){1+(1/36)+(1/36)^2+......}+ (15/36)(2×3.5){1+(1/6)+(1/6)^2+....}

= (3.5){3*(36/35)+1}=(3.5){143/35}

=14.3

Answer=14.3

My answer is 14.3 which is accepted by brilliant and posted here in the discussion. I didn't include the 2-dice iterated term in 3-dice iteration. If, I take three dice sum consistently, then, the expression can be rewritten to get answer Z as below:

2-dice term{iterated sum} = (15/36)(2×3.5){1+(1/6)+(1/6)^2+....} =3.5

3-dice term{non iterated sum}=10.5

Z= 10.5 + 3.5 +(10.5+3.5)*(1/36+(1/36)^2+(1/36)^3+......)

Z=14*(36/35)

Modified and Correct Answer=14.4

This isn't fully fleshed out, but it's how I convinced the OP his original answer was incorrect.

1st roll is $10.5$ no matter what, but $\frac{90}{216}$ of getting doubles which expects $8.4$ extra and $\frac{6}{216}$ of getting 3-of-a-kind which expects $14.4$ extra. $10.5+8.4\cdot \frac{90}{216}+14.4\cdot \frac{6}{216}=14.4$

To see the expected extra points for a matching pair consider the two dice version of the game. $7 + 7 \cdot \frac{1}{6} + 7 \cdot \frac{1}{6^{2}} + 7\cdot \frac{1}{6^{3}} ... = 7\cdot \frac{6}{5} = 8.4$

Similar reasoning for 3-of-a-kind.

Still working on it - I keep changing my mind! I am not sure 22.8 is the right answer but I can't seem to get rid of it