More Iterative Dice Rolling

Let $E_a$ be the total value of the dice rolled, if we start by rolling just two dice, while the third dice is already showing $a$ (for any $1 \le a \le 6$ ). Then, conditioning upon the outcomes of the two dice rolls, $\mathbb{E}[E_a] \; =\; \tfrac{1}{36}\sum_{u,v=1}^6\big(u + v + T_{a,u,v}\big) \; = \; 7 + \tfrac{1}{36}\sum_{u,v=1}^6 T_{a,u,v}$ where $T_{a,u,v} \; = \; \left\{ \begin{array}{lll} \mathbb{E}[X] & \hspace{1cm} & u = v = a \\ \mathbb{E}[E_a] & & u = v \neq a \\ \mathbb{E}[E_u] & & u \neq v = a \\ \mathbb{E}[E_v] & & v \neq u = a \\ 0 & & \mathrm{o.w.}\end{array} \right.$ and hence $\mathbb{E}[E_a] \; = \; 7 + \tfrac{1}{36}\left(\mathbb{E}[X] + 5\mathbb{E}[E_a] + 2\sum_{b \neq a}\mathbb{E}[E_b]\right) \; = \; 7 + \tfrac{1}{36}\left(\mathbb{E}[X] + 3\mathbb{E}[E_a] + 2\sum_{b=1}^6\mathbb{E}[E_b]\right)$ From this it is clear that $\mathbb{E}[E_1]=\mathbb{E}[E_2]=\mathbb{E}[E_3]=\mathbb{E}[E_4]=\mathbb{E}[E_5]=\mathbb{E}[E_6]= \mathcal{E}$ (say), where $\mathcal{E} \; = \; 7 + \tfrac{1}{36}(\mathbb{E}[X] + 15\mathcal{E})$ and so $21\mathcal{E} \; = \; 252 + \mathbb{E}[X]$ A similar argument gives us that $\mathbb{E}[X] \; = \; \tfrac{1}{216}\sum_{u,v,w=1}^6 \big(u + v + w + Q_{u,v,w}\big)$ where $Q_{u,v,w} = \mathbb{E}[X]$ if $u=v=w$ , $Q_{u,v,w} = \mathcal{E}$ if two if $u,v,w$ are equal with the other distinct, and $Q_{u,v,w} = 0$ if $u,v,w$ are all distinct. Hence $\mathbb{E}[X] \; = \; \tfrac{21}{2} + \tfrac{1}{36}\mathbb{E}[X] + \tfrac{5}{12}\mathcal{E}$ and hence $35\mathbb{E}[X] \; = \; 378 + 15\mathcal{E}$ Solving these equations simultaneously, we obtain $\mathbb{E}[X] = \tfrac{651}{40}$ and $\mathcal{E} = \tfrac{511}{40}$ . This makes the answer $651+40=\boxed{691}$ .

I solved this in a manner similar to how @Patrick Corn solved the Inspiration problem. This variation is just a tiny bit harder. With more opportunities to reroll, the expected value should be higher.

In fact, the first equation is identical:

$E=\frac{5}{9}(10.5)+\frac{5}{12}(10.5+E_{2})+\frac{1}{36}(10.5+E)$

What's different is if we see a single matched pair, there are more possibilities, so $E_{2}$ is different. There's a $\frac{5}{9}$ chance of rolling those two dice and seeing no matches, $\frac{5}{12}$ of seeing a single pair again, and $\frac{1}{36}$ of now having three that match. This last choice brings us back to $E$ . So

$E_{2}=\frac{5}{9}(7)+\frac{5}{12}(7+E_{2})+\frac{1}{36}(7+E)$

$E_{2}=12+\frac{1}{21}E$

Substitute this into the upper equation:

$E=\frac{5}{9}(10.5)+\frac{5}{12}(10.5+[12+\frac{1}{21}E])+\frac{1}{36}(10.5+E)$

$E=\frac{651}{40}=16.275$

$651+40=\boxed{691}$

If the expected value is $E$ , then we have $E=10.5+\frac 59(0)+\frac{5}{12}(E_2)+\frac{1}{36}(E)$ where $E_2$ is the expected value from rolling two dice. Notice that $E_2$ is exactly the same as $E$ except that we don't count the value of one of the three dice for $E_2$ . This means that $E_2=E-3.5$ and we can solve for $E$ to get $E=\boxed{\dfrac{651}{40}}$

We set up a Markov chain with 3 possible states:

By considering the probabilities when throwing the dice we get the transition probabilities [read (x->y) as P(new state=y|old state=x)]

Using the transition probabilities above, we set up the expected number of visits to each state (a, b and c):

From this, solve a=21/20, b=3/4, c=1 . When in A we throw 3 dice, when in B we throw 2 dice, when in C we throw none. The average points a die throw yields is 7/2 (no matter the state we're in, because the whole thing is symmetric with respect to the possible values 1..6), so we have for the expected total of thrown eyes:

E = (3a+2b+0c)×(7/2) = 651/40. The requested answer then is 651+40=691.

We can pair each game with a mirrored version, where we instead add the number facing downwards. The total amount from both of these games will be $7*$ (number of dice rolled). Hence, the average total for a game with $n$ rolls is $\frac{7n}{2}$ .

Let $E_d =$ expected number of rolled dice.

$E_d = 3 + \frac{1}{6^2} E_d + \frac{15}{6^2} E_2$ .

Where $E_2$ is the expected number from rolling 2 dice, leaving the 3rd face up.

$E_2 = E_d - 1$ .

Hence $E_d = \frac{93}{20}$ .

Therefore the expected total value is $\frac{93*7}{20*2} = \frac{651}{40}$ .

In this problem, three dices are rolled and we define:

X=Average score ( iterated) for 2-dice,

Y=Average score ( iterated) for 3-dice?

The data required is given:

Prob for 2-dice (same face)=(90/216),

Average score for 2-dice=(14/2),

Prob for 3-dice (same face)=(6/216),

Average score for 3-dice=(21/2),

The equations are:

Y=(21/2) + (90/216)X + (6/216)Y,

X=(14/2) + (90/216)X + (6/216)Y,

Therefore, subtracting the above two, we get,

Y - X = (7/2).

Solving, we get:

Y=(31/2) + (6/216)(1+5/7)Y,

Y=651/40 = 16.275

Answer = (651+40) = 691

I solved the system of equations with Gaussian Elimination. These people with actual solutions amaze me.