I always seem to struggle with even the most basic probability problems. Last night, I thought up the following simple scenario:
Alice and Bob are rolling standard (6-sided) dice. Due to gender
discrimination, Bob gets two dice to Alice's one. What's the probability
that the sum of Bob's two dice will exceed Alice's?
I have attempted to solve this (see below). However, I am wondering if my method is (i) correct and (ii) efficient. For such a simple problem, this really took me a long time.
After some reading, I figured I could start with this:
$$\begin{eqnarray}
P(B>A)&=&\sum_{k=1}^6 P(B>A \cap A=k) \\
&=&\sum_{k=1}^6 P(B>A | A=k)P(A=k) \\
&=&\frac{1}{6}\sum_{k=1}^6 P(B>A | A=k) \\
&=&\frac{1}{6}\sum_{k=1}^6 P(B>k)
\end{eqnarray} $$
where A is Alice's total and B is Bob's. However, even those $P(B>k)$ look like some work to calculate. If I consider Bob's two dice values separately (so $B = B_1+B_2$):
$$\begin{eqnarray}
P(B>k)&=&\sum_{n=1}^6 P(B>k \cap B_1=n) \\
&=&\sum_{n=1}^6 P(B>k | B_1=n)P(B_1=n) \\
&=&\frac{1}{6}\sum_{n=1}^6 P(B>k | B_1=n) \\
&=&\frac{1}{6}\sum_{n=1}^6 P(B_2>k-n) \\
\end{eqnarray} $$
Working these out by hand, I get:
$$\begin{eqnarray}
P(B>1)&=&1 \\
P(B>2)&=&\frac{35}{36} \\
P(B>3)&=&\frac{33}{36} \\
P(B>4)&=&\frac{30}{36} \\
P(B>5)&=&\frac{26}{36} \\
P(B>6)&=&\frac{21}{36}
\end{eqnarray} $$
Adding them up and dividing by 6, I get:
$$P(B>A)=\frac{181}{216}\approx 0.83796$$
Now, I have no idea how to check if this is correct. I did a few more sums and it seems that Bob is more than 9 times more likely to win than Alice, which I find quite surprising!
Please note: A similar (but more general) question was asked here, but no answer was given for multiple dice. I followed the advice of one of the answers and wrote a computer program, which gave me the same result as above.
The question here looks very similar, but I think it's actually very different.
Best Answer
Your answer is correct. We can confirm it by complementary counting.
Suppose Bob rolls a blue die and a green die. For $k \leq 7$, the number of ways Bob could obtain a sum of exactly $k$ is the number of solutions of the equation $$b + g = k$$ in the positive integers, where $b$ is the number displayed by the blue die and $g$ is the number displayed by the green die. Each solution corresponds to the placement of a single addition single addition sign in the $k - 1$ spaces between successive ones in a row of $k$ ones. To illustrate with $k = 7$, $$1 \square 1 \square 1 \square 1 \square 1 \square 1 \square 1$$ the outcome $$1 1 1 + 1 1 1 1$$ corresponds to $b = 3, g = 4$, while $$1 + 1 1 1 1 1 1$$ corresponds to $b = 1, g = 6$.
The number of solutions of $b + g = k$ is $k - 1$ since there are exactly $k - 1$ places to put the addition sign.
Hence, the number of ways that Bob rolls a sum of at most $n$ is $$\sum_{k = 1}^{n} (k - 1) = \frac{n(n - 1)}{2}$$ Since there are $6^2 = 36$ possible outcomes, the probability that Bob rolls a sum of at most $n$ is $$\Pr(B \leq n) = \frac{1}{36} \cdot \frac{n(n - 1)}{2}$$
If we denote the value Alice obtains by $A$ and the sum that Bob obtains by $B$, the probability that Bob's sum is at most the value Alice obtains is \begin{align*} \Pr(B \leq A) & = \sum_{n = 1}^{6} \Pr(A = n)\Pr(B \leq A \mid A = n)\\ & = \frac{1}{6}\sum_{n = 1}^{6} \Pr(B \leq A \mid A = n)\\ & = \frac{1}{6}\left[\frac{0}{36} + \frac{1}{36} + \frac{3}{36} + \frac{6}{36} + \frac{10}{36} + \frac{15}{36}\right]\\ & = \left(\frac{1}{6}\right)\left(\frac{35}{36}\right)\\ & = \frac{35}{216} \end{align*} Hence, the probability that Bob's score exceeds Alice's score is $$\Pr(B > A) = 1 - \Pr(B \leq A) = 1 - \frac{35}{216} = \frac{181}{216}$$ as you found.