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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

Peter Lageri
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3 Answers3

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Let $A$ denote the event that $X_1+X_2\in\{2,3,4\}$.

Let $B$ denote the event that $X_1+X_3\in\{2,3,4\}$.

Let $C$ denote the event that $X_2+X_3\in\{2,3,4\}$.

With inclusion/exclusion and symmetry we find:$$P(A\cup B\cup C)=3P(A)-3P(A\cap B)+P(A\cap B\cap C)$$

Can you find these terms on RHS yourself?

If you calculate them correctly then your answer will be:$$3\cdot\frac6{36}-3\cdot\frac{14}{216}+\frac{11}{216}=\frac{77}{216}$$

drhab
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  • Argh! How do I read that P(A.... ? Is that "Set Theory"? – Peter Lageri Jan 03 '19 at 12:42
  • Is there something I should read to understand that? Its some 35 years since I went to school. – Peter Lageri Jan 03 '19 at 12:50
  • You should read $P(A)=P(X_1+X_2\text { equals }2, 3\text{ or }4)$. Here $A$ is indeed a set. A subset of the outcome space actually. It concerns the underlying theory for probability. If you want to endeepen then go looking for "probability space" on e.g. Wikipedia. – drhab Jan 03 '19 at 12:52
  • If this goes too deep then you better accept the answer of lulu. – drhab Jan 03 '19 at 12:59
  • Why 3P? Because there are three numbers that interest us? Was 6 in 6/36 found by brute force or calculated and if yes how? I would like to understand. – Peter Lageri Jan 03 '19 at 13:08
  • Actually $P(A)+P (B)+P (C) $ but these probabilities are equal so the summation equals $3P (A) $. That is the use of what I call "symmetry " in my answer. Also google on the principle of inclusion /exclusion to get more sight on it. – drhab Jan 03 '19 at 13:46
  • Aha, and the 6/36, the 14/216 and the 11/216 ? – Peter Lageri Jan 03 '19 at 14:22
  • $P(A)=\frac6{36}$ because in $6$ of the $36$ equiprobable cases we have $X_1+X_2\in{2,3,4}$. $P(A\cap B)=\frac{14}{216}$ because in $14$ of the $216$ equiprobable cases with have $X_1+X_2\in{2,3,4}$ and $X_1+X_3\in{2,3,4}$. $P(A\cap B\cap C)=\frac{11}{216}$ because in $11$ of the $216$ equiprobable cases we have $X_1+X_2\in{2,3,4}$, $X_1+X_3\in{2,3,4}$ and $X_2+X_3\in{2,3,4}$. – drhab Jan 03 '19 at 15:34
  • Yes, ok but did you arrive at 6, 14 and 11 by manuel brute force counting? – Peter Lageri Jan 03 '19 at 17:18
  • Yes, that's just a matter of counting. – drhab Jan 03 '19 at 17:39
  • Ok, thank you. I don't think I can come up with further questions. Here in Denmark apparently there is a principle called PIE. ( The principle of inklusion and exclusion) I'm trying to understand it. Thank you for your patience. – Peter Lageri Jan 03 '19 at 18:07
  • You are very welcome. PIE is indeed what is applied in my answer. – drhab Jan 03 '19 at 21:30
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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 $$6\times 7+3\times 11+ 1\times 2=77$$ (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.

lulu
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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 $\frac{77}{6^3}$.

farruhota
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