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So I had an assignment due yesterday, regarding the probability of the attackers second highest dice roll, being strictly bigger than the defenders second highest dice roll, in the game Risk.

In the game, the attacker has three dices and the defender has two. The highest dice of the attacker is compared with the highest dice of the defender, and the second-highest dice of the attacker is compared with the second highest dice of the defender. For the attacker to win, their dice must be strictly larger than the defender. The dice of the defender are $C_1$ and $C_2$, ordered as $C_{(1)} \leq C_{(2)}$, and the dice of the attacker are $B_1$, $B_2$ and $B_3$, ordered as $B_{(1)} \leq B_{(2)} \leq B_{(3)}$

I, with the help of this board wrote my answer as

P(A>D) = $B_{(2)}>C_{(1)}$:

$$\sum_{k=2}^{6} \left(\sum_{n=1}^{k-1}\frac{13-2n}{36}\right) \cdot \frac{-6k^2+42k-20}{216} =\frac{1181}{1944} =0.6075 $$

Yet my TA had another answer. We both had the same probability for $B_{(2)}$, ie. $P(B_{(2)}=k) = \frac{-6^2+42k-20}{216}$ but he used the distribution function for $B_{(2)}, P(B_{(2)}$≤k)$ = \frac{9k^2-k^3}{108}$, instead of the probability function for $C_{(1)}$ which I used.

He then multiplied the possibilities and took the sum, and got close to the same answer, but not the exact.

$$\sum_{k=1}^{6} \frac{21k-3k^2-10}{108} \cdot \frac{9k^2-k^3}{108} = 0.598 $$

My question is, which of these are the correct way to find the probability of the attackers second dice being strictly larger than the defenders second dice roll?

Greg Martin
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Markus Andeasen
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1 Answers1

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I find that it helps in cases like this to use clear symbology in the setup of the formula that shows (as much as possible) exactly why you chose the terms you chose, rather than going directly to polynomials.

The probability you set up was the probability that $B_{(2)} > C_{(1)},$ taking each value of $B_{(2)}$ as a disjoint event to be summed:

\begin{align} P(B_{(2)} > C_{(1)}) &= \sum_{k = 1}^6 P((B_{(2)} = k) \cap (C_{(1)} < k)) \\ &= \sum_{k = 1}^6 \left( P(B_{(2)} = k) \cdot P(C_{(1)} < k) \right) \\ &= \sum_{k=1}^6 \left( P(B_{(2)} = k) \cdot \sum_{n=1}^{k-1} P(C_{(1)} = n) \right)\\ &= \sum_{k=1}^6 \left( \frac{-6k^2+42k-20}{216} \cdot \sum_{n=1}^{k-1} \frac{13-2n}{36} \right). \end{align}

That's your formula (up to harmless changes in the order of multiplication) and it makes perfect sense.

The best sense I can make out of your TA's formula is:

\begin{align} \sum_{k=1}^6 \left(\frac{21k-3k^2-10}{108} \cdot \frac{9k^2-k^3}{108}\right) &= \sum_{k=1}^6 \left(\frac{21k-3k^2-10}{108} \cdot \sum_{n=1}^k \frac{21n-3n^2-10}{108} \right) \\ &= \sum_{k=1}^6 \left( P(B_{(2)} = k) \cdot \sum_{n=1}^k P(A_{(2)} = n) \right) \\ &= \sum_{k=1}^6 \left( P(B_{(2)} = k) \cdot P(A_{(2)} \leq k) \right) \\ &= \sum_{k=1}^6 P((B_{(2)} = k) \cap (A_{(2)} \leq k)) \\ &= P(B_{(2)} \geq A_{(2)}), \end{align}

where $A_{(2)}$ is the second-lowest value on another set of three dice rolled at the same time as the three dice that are rolled for $B_{(2)}.$ That is, your TA has computed the probability that the second lowest die from one set of three dice will beat or tie the second lowest die from another set of three dice.

I cannot find any evidence in your TA's formula that it considers the defender's two dice at all. Any event depending only on the defender's two dice must have a probability that is a multiple of $\frac1{36}.$ We have $P(A_{(2)} \leq 3) = \frac12$ and $P(A_{(2)} \leq 6) = 1,$ which are multiples of $\frac1{36},$ but the other values of $P(A_{(2)} \leq k)$ are not multiples of $\frac1{36}.$

In short, your TA seems to have used $P(A_{(2)} \leq k)$ (the probability of an event on the roll of three dice) where you correctly used $P(C_{(1)} < k)$ (the probability of an event on the roll of two dice). The mistake isn't just $\leq$ versus $<$; your TA's formula isn't even looking at the same number of dice. The fact that the result is as close as it is to the correct result is due to the two errors partially canceling out; giving the defender three dice reduces the chance that the attacker's second-best die wins, but making ties a win for the attacker rather than the defender increases the attacker's chance of a win.

David K
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