Probability of two dice against one

Question

If you roll two dice, what are the odds that at least one has a higher value than a third die you roll?" And for (B), it's the same as (A) except you roll three dice before checking against a fourth?

Answer 1

Let’s try asking “If I were to roll a number $k$ before rolling a $6$-sided dice, then what would be the chance the second die would have a lowet value than the other?”

Let us try $k=1$. We can roll the numbers $2,3,4,5,6$ to top that $1$. Since there are $6$ possibilities, and only $1$ does not meat the requirement, the probability of getting equal or less is $\frac16$. For $k=2$ there are $2$ possibilities=$\frac26$, for $k=3$ there are $3$ possibilities=$\frac36$ e.t.c.

Now what if we were to use $2$ dice? The second die has the same probability as the first as they are identical, so we square the chance for $1$ die of being under or equal to $1$ and the answer would be $\left(\frac16\right)^2=\frac1{36}$ for $k=1$.

But we’re not finding the probability of the dice being under or equal to $k$, are we? Thankfully the probability of getting over $k$ for $2$ dice would be $1-\frac1{36}=\frac{6^2-1^2}{6^2}$.

Finally, let’s factor in the various values of $k$. Remember, the deominater will not be $6^2$ but $6^3$ because that is the number of possibilities for $3$ dice (first and second dice, and the value of $k$) so the answer will be $$\frac{6^2-1^2+6^2-2^2+6^2-3^2+6^2-4^2+6^2-5^2+6^2-6^2}{6^3}$$, which is $\frac{125}{216}$

For the second answer, change the exponents by one to get $$\frac{6^3-1^3+6^3-2^3+6^3-3^3+6^3-4^3+6^3-5^3+6^3-6^3}{6^4}$$

Answer 2

Let $n$ denote the number of dice thrown prior to the last one (so you are asked about $n=2,3$).

it is easier to work with the complementary event...the probability that none of the first $n$ dice exceed the last one.

Suppose the final die shows $i$ (probability $\frac 16$ of course). Of course, the probability that a given die shows $≤i$ is $\frac i6$. Thus the probability that all $n$ of the prior dice show $≤i$ is $\left(\frac i6\right)^n$. It follows that the probability none of the first $n$ die exceed the final one is $$q_n=\frac 16\times \sum_{i=1}^6 \left(\frac i6\right)^n$$

It is easy to compute $$q_2=\frac {91}{216}\quad \& \quad q_3=\frac {49}{144}$$

So the answer you want, $p_n=1-q_n$ is given by $$p_2=\frac {125}{216}\quad \&\quad p_3=\frac {95}{144}$$

Sanity check: Note that, for large $n$, it is effectively certain that we'll get at least one $6$ in the initial $n$ tosses, so we expect $p_n\to \frac 56$ as $n$ gets large. Using a machine, it is easy to verify that $q_n\to \frac 16$ as $n$ gets large, as expected. Or, algebraically, it is clear that $\left(\frac i6\right)^n\to 0$ for $i<6$ so the only term in the sum that survives has $i=6$ which makes the limit apparent.