Jtdylktuy

I have a boardgame call Enemy Coast Ahead: The Dambusters in which enemy FLaK poses a major problem.

Under some circumstances the game "AI" throws three instead of two dice and if the sum of any two of them is within a certain interval a hit is scored. Furthermore if doubles are rolled and that gives a hit the doubles are re-rolled. If that is a hit but not a re-roll the hit is score and then it stops. If that is a hit and a double hit is score and another re-roll follows.

For starters I'd just like to know the probability of the sum of the two dice being 2,3 or 4 when rolling three dice.

As I see it(and I have not idea whether I right): Dice 1,2 and 3, denoted A, B and C can roll a lethal combbination as AB, BC or AC.

At Flak Level 1 the lethal range is 2-4. That gives 1/36 + 2/36 + 3/36=6/36 =1/6 for a Hit.

Using three dice and combinations AB, DC and AC just one of them is enough for a hit so since OR goes into it we get 1/6 for each 1/6 + 1/6 + 1/6 = 1/2.

Best Regards
Peter

Let $A$ denote the event that $X_1+X_2in{2,3,4}$.

Let $B$ denote the event that $X_1+X_3in{2,3,4}$.

Let $C$ denote the event that $X_2+X_3in{2,3,4}$.

With inclusion/exclusion and symmetry we find:$$P(Acup Bcup C)=3P(A)-3P(Acap B)+P(Acap Bcap C)$$

Can you find these terms on RHS yourself?

If you calculate them correctly then your answer will be:$$3cdotfrac6{36}-3cdotfrac{14}{216}+frac{11}{216}=frac{77}{216}$$

You can't add the probability of non-mutually exclusive events like that. Suppose there were $10$ dice, so $binom {10}2=45$ pairs. Your method would give an answer greater than $1$.

Best here is brute force counting.

Counting the winning triples requires attention because there is a lot of overlap and because not all of them have the same number of permutations.

$6$ permutations: $${1,2,3}quad {1,2,4}quad{1,2,5}quad{1,2,6}$$
$${1,3,4}quad {1,3,5}quad{1,3,6}$$

$3$ permutations: $${1,1,2}quad {1,1,3}quad{1,1,4}quad{1,1,5}quad {1,1,6}$$
$${1,2,2}quad {2,2,3}quad {2,2,4}quad{2,2,5}quad{2,2,6}$$
$${1,3,3}$$

$1$ permutation: $${1,1,1}quad {2,2,2}$$

Thus there are $$6times 7+3times 11+ 1times 2=77$$ (ordered) winning triples. As there are $6^3=216$ unrestricted ordered triples, the answer is $$boxed {frac {77}{216}=.35648}$$

CAUTION: one problem with brute force enumeration is that it is quite error prone; it's hard to be sure you didn't miss a case or two. Check my list carefully. The first version of it left off two patterns and there might well be other errors that are still present.

Another way to brute force:
$$begin{array}{c|c|c|c}
D_1&D_2&D_3&Total\
hline
1&1-3&1-6&18\
1&4-6&1-3&9\
2&1-2&1-6&12\
2&3-6&1-2&8\
3&1&1-6&6\
3&2&1-2&2\
3&3-6&1&4\
4-6&1&1-3&9\
4-6&2&1-2&6\
4-6&3&1&3\
hline
&&&77
end{array}$$
Hence the required probability is $frac{77}{6^3}$.