Product of 2 dice game

Question

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You are playing a game where you a paid the product of 2 dice

a) How much would you pay to have the opportunity to reroll any 1 of the dice?

b) How much would you pay to roll 3 dice and discard the lowest?

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I know the EV of product of 2 dice is 12.25

I calculated the Probability of the Product of 2 dice to be <= 12 as $ \frac {23} {36} $

Similarly the probability of the Product of 2 dice to be > 12 as $ \frac {13} {36} $

Now for Part a) New E.V. = $ \frac {23} {36} $ * (Increase in EV in this case) + $ \frac {13} {36} $ * (0 Increase)

However, I'm confused how to approach the Increase in EV in the case where we choose to reroll 1 of the dice with Probability of 23/36.

One approach I can think of for the above case is that we would only reroll a 1,2 or 3 and never a 4,5,6 as (4 * 4 = 16), (5 * 5 = 25), (6*6 = 36) is > than 12.25.

Then - $$ \frac {1} {3} * (3.5-1) + \frac {1} {3} * (3.5 -2) + \frac {1} {3} * (3.5-3) = 1.5 $$

Now, I'm not sure if it should be $ \frac {1} {3} or \frac {1}{6} $ for the probabilities above considering that it is conditioned on rerolling only if we get a 1,2 or 3

Now the 2nd die E[x] = 3.5 so we can say $$ 1.5 * 3.5 = 5.25 $$ which would be the increase in EV in the case where we choose to reroll 1 of the die

Hence increase in EV = $$ \frac {23} {36} * 5.25 = \frac {161} {48} $$

Not sure if this approach is correct

Part b) I'm not sure how to approach it

Answer 1

(a) Given the opportunity to reroll a singular dice, we note that this only makes sense to do if the number it is currently showing is $3$ or less, as the expected value of the reroll is $3.5$. Hence, the new expected value of a roll of $(m, n)$ is $3.5 \max(m, n)$ if any of $m$ or $n$ are $\leq 3$, and otherwise it remains $mn$. Splitting into $4$ cases, we have

Case 1: $m, n \geq 4$

In this case, we choose not to reroll, so the total sum of products is $(4+5+6)(4+5+6) = 15^2$.

Case 2: $m \geq 4$, $n \leq 3$

In this case, we choose to reroll the $n$-die, so the total average sum of products is $(4+5+6)(3\times 3.5) = 3 \times 3.5 \times 15$.

Case 3: $m \leq 3$, $n \geq 4$

This case is virtually the same as previous, and we get the same total.

Case 4: $m, n \leq 3$

In this case, we choose to reroll the smaller die, so the total average sum of products is $5 \times 3.5 \times 3 + 3 \times 3.5 \times 2 + 1 \times 3.5 \times 1$.

this gives an overall expected value of $$\frac{15^2}{36} + 2 \cdot \frac{3 \times 3.5 \times 15}{36} + \frac{5 \times 3.5 \times 3 + 3 \times 3.5 \times 2 + 1 \times 3.5 \times 1}{36} = \boxed{\frac{617}{36}}$$ Hence, you should be willing to pay up to $\frac{617}{36}$ to play this game (or $\frac{44}{9}$ more than the game without any reroll).

(b) We break this up into cases depending on what the smallest number is:

Case 1: Smallest number is $1$.

There are $91$ possible ways of this occurring, with total sum of product being $1 \times 1^2 + 3 \times 1 \times 20 + 3 \times 20^2 = 1261$.

Case 2: Smallest number is $2$.

There are $61$ possible ways of this occurring, with total sum of product being $1 \times 2^2 + 3 \times 2 \times 18 + 3 \times 18^2 = 1084$.

Case 3: Smallest number is $3$.

There are $37$ possible ways of this occurring, with total sum of product being $1 \times 3^2 + 3 \times 3 \times 15 + 3 \times 15^2 = 819$.

Case 4: Smallest number is $4$.

There are $19$ possible ways of this occurring, with total sum of product being $1 \times 4^2 + 3 \times 4 \times 11 + 3 \times 11^2 = 511$.

Case 5: Smallest number is $5$.

There are $7$ possible ways of this occurring, with total sum of product being $1 \times 5^2 + 3 \times 5 \times 6 + 3 \times 6^2 = 223$.

Case 6: Smallest number is $6$.

There is $1$ possible ways of this occurring, with total sum of product being $1 \times 6^2 = 36$.

Hence, the expected value of the game is $\boxed{\frac{1967}{108}}$, so this is the maximum you should consider paying.

Answer 2

The first roll is modeled by the random variables $X_1,X_2$. Let $X_{(1)}, X_{(2)}$ denote the order statistics.

Consider always replacing the smaller $X_{(1)}$ with a new roll $Y$. The expected value of the final payoff is $E[X_{(2)} Y] = 2E[X_1 Y 1_{X_2<X_1}] + E[X_1 Y 1_{X_2=X_1}] = E[Y](\frac 2{36} [\sum_{i,j}i 1_{j<i}] +\frac 1{36} \sum_i i ) = \frac {1127}{72}\approx 15.6$.

A strategy is then to reroll the smallest die if and only if $X_1X_2 < \frac {1127}{72}$.

As suggested in the comments section, a more sensible strategy is to reroll the smallest die with face $\leq 3$ (if any, otherwise no second roll). The expected payoff is then

$$\begin{aligned} &\phantom{=}E[X_1X_2 1_{\min(X_1,X_2)\geq 4} + X_{(2)}Y 1_{\min(X_1,X_2)\leq 3}] \\&= E[X_1X_2 1_{\min(X_1,X_2)\geq 4}] + 2E[X_1 Y 1_{X_2<X_1}1_{\min(X_1,X_2)\leq 3}] +E[X_1 Y 1_{X_2=X_1}1_{\min(X_1,X_2)\leq 3}] \\&= 25/4 + 371/36 + 7/12 = \frac{617}{36} \approx 17.1. \end{aligned}$$

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For the second question, you're looking for $E[X_{(2)} X_{(3)}]$, which can be decomposed as $$E[X_{(2)} X_{(3)}1_{\text{no ties}}] +E[X_{(2)} X_{(3)}1_{\text{a single tie}}] +E[X_{(2)} X_{(3)}1_{\text{all equal}}].$$

$E[X_{(2)} X_{(3)}1_{\text{no ties}}] = E[\sum_{\sigma\in S_3} X_{\sigma(2)} X_{\sigma(3)} 1_{X_{\sigma(1)}<X_{\sigma(2)}<X_{\sigma(3)}}] = 3! E[ X_{2} X_{3} 1_{X_{1}<X_{2}<X_{3}}] = \frac{21}2.$
I used Mathematica for the last computation.

$E[X_{(2)} X_{(3)}1_{\text{a single tie}}] = 3E[X_2^21_{X_1<X_2}1_{X_2=X_3}] + 3E[X_1 X_2 1_{X_1>X_2}1_{X_2=X_3}] = \frac{175}{24}.$

$E[X_{(2)} X_{(3)}1_{\text{all equal}}] = \sum_{i=1}^6E[i^21_{X_1=X_2=X_3=i}] = \frac{91}{216}.$

Finally, $$E[X_{(2)} X_{(3)}] = \frac{1967}{108}\approx 18.2.$$

This value is confirmed by the following numpy simulation:

import numpy as np
sample = np.random.randint(1,7,(10000,3))
sample_s = np.sort(sample)[:,1:]
print(np.mean(sample_s[:,0] * sample_s[:,1]))