Dice probability problem

[GameGod]

2 defending dice do battle with a attacking dice, where a can be any positive integer. The highest scoring defender die is paired with the highest scoring attacking die; the second-highest scoring defending die is paired with the second-highest scoring attacking die. If, in each pair, the defending die has a greater or equal score to the attacking die, the defending die wins (conversely, if the attacking die has a greater score than the defending die, the attacker wins that pair).

In terms of a, how do we calculate the probability that i) the defending dice will win both pairs, ii) the attacking dice will win both pairs, iii) 1 pair will be won by a defending die and the other by an attacking die?

I see I'm required to show work, but I really don't know how to start! I solved the problem for when there is only 1 defending die and it's paired up with the highest attacking die. I landed up realizing that P(Defending die wins pair)=(1/6)*Sigma(i=0 to 6)[(i/6)^a]. (Please excuse the lack of LaTeX - my abilities do not reach far enough so far to write sigma functions in LaTeX) but how do I find a function for this problem now there are 2 defending die?

 

[BGM]

I think a very simple way is that first you condition on the result of the two defending dice.

Let \( (D_1, D_2), D_1 \geq D_2 \) be the result of the two defending dice.

So by independence and counting the possible permutations, you know that

\( \Pr\{D_1 = d_1, D_2 = d_2\}
= \left\{\begin{matrix} \displaystyle \frac {1} {36} & \text{if} & d_1 = d_2 \\[1em]
\displaystyle \frac {2} {36} & \text{if}& d_1 > d_2 \end{matrix}\right. \)

and \( d_1 \in \{1, \ldots, 6\}, d_2 \in \{1, \ldots, d_1\}\)


Let \( (C_1, C_2), C_1 \geq C_2 \) be the maximum 2 dices out of \( a \geq 2\) dices. By the same principle we can generalize the joint pmf of the maximum 2 dices:

\( \Pr\{C_1 = c_1, C_2 = c_2\}
= \left\{\begin{matrix} \displaystyle \frac {a(a-1)c_2^{a-2}} {2 \times 6^a} & \text{if} & c_1 = c_2 \\[1em]
\displaystyle \frac {a(a-1)c_2^{a-2}} {6^a} & \text{if}& c_1 > c_2 \end{matrix}\right. \)

Note that when \( a = 2 \) you just reduced to the same joint pmf as \( (D_1, D_2) \).

Given the result of the two defending dice is \( (d_1, d_2) \),

- for \( a = 1 \) you can calculate the probability easily as the probability will be \( \frac {d_1} {6} \)

- for \( a \geq 2\) attacking dice, both defensive dice will be winning when
\( C_1 = c_1 \leq d_1, C_2 = c_2 \leq d_2 \)

So at last you just sum all those probabilities:

\( \sum_{d=1}^6 \sum_{d_2 = 1}^{d_1} \sum_{c_1 \leq d_1, c_2 \leq d_2} \Pr\{C_1 = c_1, C_2 = c_2\} \Pr\{D_1 = d_1, D_2 = d_2\} \)

 

[GameGod]

I think a very simple way is that first you condition on the result of the two defending dice.

Let \( (D_1, D_2), D_1 \geq D_2 \) be the result of the two defending dice.

So by independence and counting the possible permutations, you know that

\( \Pr\{D_1 = d_1, D_2 = d_2\}
= \left\{\begin{matrix} \displaystyle \frac {1} {36} & \text{if} & d_1 = d_2 \\[1em]
\displaystyle \frac {2} {36} & \text{if}& d_1 > d_2 \end{matrix}\right. \)

and \( d_1 \in \{1, \ldots, 6\}, d_2 \in \{1, \ldots, d_1\}\)


Let \( (C_1, C_2), C_1 \geq C_2 \) be the maximum 2 dices out of \( a \geq 2\) dices. By the same principle we can generalize the joint pmf of the maximum 2 dices:

\( \Pr\{C_1 = c_1, C_2 = c_2\}
= \left\{\begin{matrix} \displaystyle \frac {a(a-1)c_2^{a-2}} {2 \times 6^a} & \text{if} & c_1 = c_2 \\[1em]
\displaystyle \frac {a(a-1)c_2^{a-2}} {6^a} & \text{if}& c_1 > c_2 \end{matrix}\right. \)

Note that when \( a = 2 \) you just reduced to the same joint pmf as \( (D_1, D_2) \).

Given the result of the two defending dice is \( (d_1, d_2) \),

- for \( a = 1 \) you can calculate the probability easily as the probability will be \( \frac {d_1} {6} \)

- for \( a \geq 2\) attacking dice, both defensive dice will be winning when
\( C_1 = c_1 \leq d_1, C_2 = c_2 \leq d_2 \)

So at last you just sum all those probabilities:

\( \sum_{d=1}^6 \sum_{d_2 = 1}^{d_1} \sum_{c_1 \leq d_1, c_2 \leq d_2} \Pr\{C_1 = c_1, C_2 = c_2\} \Pr\{D_1 = d_1, D_2 = d_2\} \)

Thanks for the response.

Your last function calculates the probability of the defending dice winning both pairs, right? So what would be the probability of the attacking dice winning both pairs?

My other issue is, I don't understand what your final summation is meant to represent. If you mean to vary c1 from 1 to d1, and c2 from 1 to d2, why not simply write this as 2 further sigma operators?

 

[BGM]

For both attacking pairs, you just need to change the summation bound to

\( \sum_{d_1 = 1}^5 \sum_{d_2 = 1}^{d_1} \sum_{c_1 = d_1 + 1}^6 \sum_{c_2 = d_2 + 1}^{c_1} \)

For you issue yes the interior summation is just a double summation:

\( \sum_{d_1 = 1}^6 \sum_{d_2=1}^{d_1} \sum_{c_1 = 1}^{d_1} \sum_{c_2 = 1}^{\min\{c_1, d_2\}} \)

 

[GameGod]

Thanks again. Do the same functions, with same symbols, for Pr{C1=c1,C2=c2} and Pr{D1=d1,D2=d2) continue to hold when used with the summation bound you just suggested for the attacking dice winning both? (As you originally wrote them for the defending dice winning both)

Is there a way of writing the whole of either of these as a closed-form function of a, for case a≥2, just using the sigma operators (and the elementaries)? For instance, perhaps by separating out the sections of the function (i.e. operators) for c_1=c_2 and c_1>c_2, and d_1=d_2 and d_1>d_2; or just in general another way of writing this which could be computed directly.

I would understand if such a solution doesn't exist.

 

[GameGod]

I used the attached closed-form solution, for the probability of the attacking dice winning both pairs. Can you check and tell me if it is correct?

What does this evaluate to, though? For a=3, the correct answer should come to 1445/3888. Instead, for me, it comes to 85/216 using your formula. Have I solved it incorrectly, or evaluated it incorrectly? Or are there adjustments needing to be made to the formulae?

 

[GameGod]

Please help... anyone who understands BGM's solution properly, I would be thankful for you to point out the problem. It could be in 1 of three stages:

1) BGM's solution itself
2) My attempt to rewrite it, for the case of the attacking dice winning both pairs (see the PDF)
3) My evaluation of my rewriting of the solution

But I have checked the figures 3 times now and never arrived at the correct answer, so I don't think it is 3). If anyone else evaluates it and finds 1445/3888 when a=3 then I will certainly check my numbers again.

 

[BGM]

Sorry previously I use the Inclusion-Exclusion method to count which is not good and missing the intersection case. Now I used another method to count and this one should be correct (at least checked from \( a = 2 \) to \( a = 10 \) which the joint pmf sum to 1):

\( \Pr\{C_1 = c_1, C_2 = c_2\}
= \left\{\begin{matrix} \displaystyle \frac {c_2^a - (c_2 - 1)^a - a(c_2 - 1)^{a-1}} {6^a} & \text{if} & c_1 = c_2 \\[1em]
\displaystyle \frac {a[c_2^{a-1}-(c_2-1)^{a-1}]} {6^a} & \text{if}& c_1 > c_2 \end{matrix}\right. \)

 

[GameGod]

Thank you very much, this function works perfectly for all \(a \geq 2\).

What is the function for \(a = 1\)? Here we have total 2 outcomes, attacking die wins against highest defending die or attacking die loses against highest defending die. How do we calculate either of these? (The other could simply be 1-the first.)

 

[BGM]

For \( a = 1 \) it becomes very easy.

First you can calculate the marginal pmf for \( D_1 \):

\( \Pr\{D_1 = d_1\} = \frac {2d_1 - 1} {36} \)

Since the defender dice win whenever the attacking dice is less than or equal to \( D_1 \), therefore given \( D_1 = d_1 \) the probability will just be \( \frac {d_1} {6} \).

As a result by the law of total probability, the winning probability is

\( \sum_{d_1 = 1}^6 \frac {d_1} {6} \times \frac {2d_1 - 1} {36} \)

 

[GameGod]

Thanks very much.

I have another thread which follows on from this one, on a slightly different route: http://www.talkstats.com/showthread.php/44025-Game-probability-problem. Your help would be greatly appreciated.