Risk Problem

[stupidityismygam]

Though this does not follow the exact set up of Risk it is still an interesting problem based on Risk.

Consider 3 attackers and 3 defenders. Each attacker rolls a die (1-6) and each defender also rolls a die(1-6). The highest roll for the attackers goes against the highest roll for the defenders, second highest against second highest, and third highest against third highest. If the roll of the attacker is higher, the defender loses. If the roll of the defender is higher, the attacker loses. If there is a tie, the attacker loses.

What is the probability that the defenders win all three fights? (fraction please)

[AuSmith]

Has anybody tried this? Does Stupid have a solution? Should he just post it?

[stupidityismygam]

well...i know can # brute force it, however...i was hoping that there was a clever/nice solution out there

[AuSmith]

Here's what I got (during class of course).

You want to roll the 6 dice and order them into 3 pairs. Now, we will count how many triples of rolls are ordered correctly and are all three wins for the defender. I will call these "such" triples. But, wait. If both the defender and the attacker roll three 4's, isn't that much more unlikely than say if the defender rolls [unparseable or potentially dangerous latex formula] and the attacker rolls [unparseable or potentially dangerous latex formula]? Just counting all the triples treats them as equally likely. Hopefully, we can use a weight matrix to even things out, though this is the nasty part of the problem.

I was going to make a diagram in paint and then put it on photobucket... but I'm on a mac right now. Anyways, these ordered pairs [unparseable or potentially dangerous latex formula] represent the pairs in which the defender, rolling a [unparseable or potentially dangerous latex formula], wins against the attackers roll of [unparseable or potentially dangerous latex formula]. Furthermore, if the pair [unparseable or potentially dangerous latex formula] comes before the pair [unparseable or potentially dangerous latex formula] (following the arrows), then the second pair may be rolled after the first pair following our ordering.

(6,6)
--^
(6,5) -<- (5,5)
--^---------^
(6,4) -<- (5,4) -<- (4,4)
--^---------^----------^
(6,3) -<- (5,3) -<- (4,3) -<- (3,3)
--^---------^----------^----------^
(6,2) -<- (5,2) -<- (4,2) -<- (3,2) -<- (2,2)
--^---------^----------^----------^---------^
(6,1) -<- (5,1) -<- (4,1) -<- (3,1) -<- (2,1) -<- (1,1)

Therefore, a "such" triple is one that consists of rolls (not necessarily distinct) from the above directed graph (more specifically, the poset (partially ordered set)) that "follow the arrows". Probably the easiest way to track the middle roll. By this, I mean to consider each roll individually, then count how many such triples there are such that it is the middle roll.

Now, I want to make a matrix (but am having difficulty on the forum?) that counts triples that use that use both (-<-) and (-^-) arrows each move as worth 36, and triples that don't use any arrows on either move as worth 1. This is because there are [unparseable or potentially dangerous latex formula] equally likely rolls that produce the same ordered triple if both attacker and defender roll nothing the same (like [unparseable or potentially dangerous latex formula] but not [unparseable or potentially dangerous latex formula]) whereas there's only one roll that will produce the triple where both attacker and defender roll all the same number (like [unparseable or potentially dangerous latex formula]).

Okay, I'm going to leave now. I'll try to finish... later.