What's the probability that when we pick another die and roll it, we'll get a number smaller than N

Question

There are 3 dice in the bag: 6-sided, 12-sided and 20-sided. We pick one die at random, roll it, and it gives us a number N. What's the probability that when we pick another die and roll it, we'll get a number smaller than N, if: a) $N = 12$; b) $N = 4$?

My attempt:
The probability to pick a certain die is 1/3.
b)Let N = 4. I decided to break this problem into several cases:
Case 1: We first pick a 6-sided die and then a 12-sided, then the favorable outcomes are: (4,1),(4,2),(4,3) - 3 favorable outcomes out of 12 (since the number on the first die has already been decided)
Case 2: 6-sided die and followed by a 20-sided. We have the same favorable outcomes, but this time it's 3 out of 20.
Case 3: 12-sided die and followed by a 6-sided $\Rightarrow$ 3 out of 6.
Case 4: 12-sided die and followed by a 20-sided $\Rightarrow$ 3 out of 20.
Case 5: 20-sided die and followed by a 6-sided $\Rightarrow$ 3 out of 6.
Case 3: 20-sided die and followed by a 12-sided $\Rightarrow$ 3 out of 12.

Therefore, the total probability is $$\frac{1}{3}\cdot\frac{1}{3}\cdot\frac{3}{12}+\frac{1}{3}\cdot\frac{1}{3}\cdot\frac{3}{20}+\frac{1}{3}\cdot\frac{1}{3}\cdot\frac{3}{6}+\frac{1}{3}\cdot\frac{1}{3}\cdot\frac{3}{20}+\frac{1}{3}\cdot\frac{1}{3}\cdot\frac{3}{6}+\frac{1}{3}\cdot\frac{1}{3}\cdot\frac{3}{12} = \frac{1}{5}$$

As for part a) there are less cases, since we can't pick the 6-sided die first, so the probability is $$\frac{1}{3}\cdot\frac{1}{3}\cdot 1+\frac{1}{3}\cdot\frac{1}{3}\cdot\frac{11}{20}+\frac{1}{3}\cdot\frac{1}{3}\cdot 1 +\frac{1}{3}\cdot\frac{1}{3}\cdot\frac{11}{12} = \frac{52}{135}$$

I'm not entirely sure that my solution is correct. Can someone take a look at it please? Thank you.

EDIT: This is my second attempt after reading all the comments. I'll use Robert's approach.

Clearly, if we know that we rolled a $12$, then the first pick can't be a 6-sided die. Furthermore, if we pick a 12-sided and a 20-sided dice $60$ times, we'll roll a $12$ on the $12$-sided die $5$ times, and we'll roll a $12$ on the $20$-sided die $3$ times. Thus, if we know that we rolled a $12$, that means the probability that the first pick was the $12,$ and $20$-sided die is $\frac {\frac{5}{60}}{\frac{8}{60}}=\frac{5}{8}, \frac {\frac{3}{60}}{\frac{8}{60}}=\frac{3}{8}$ respectively.

If the first pick was the $12$-sided die, then the probability the next roll is smaller is $\frac 12 (1 + \frac {11}{20}) = \frac{31}{40}$.

If your first pick was the $20$-sided die, then the probability your next roll is smaller is $\frac 12 (1 + \frac{11}{12})=\frac{23}{24}$.

So the total probability is:

$$\frac 58 \cdot \frac {31}{40} + \frac 38 \cdot \frac {23}{24} = \frac {31}{64}+ \frac {23}{64} = \frac{54}{72} = \frac{3}{4}.$$

Answer 1

I'll do (b). Using this technique, you should be able to do (a) on your own.

Let's perform this trial $180$ times. Then I'll pick each die $60$ times. I'll roll a $4$ on the $6$-sided die $10$ times, I'll roll a $4$ on the $12$-sided die $5$ times, and I'll roll a $4$ on the $20$-sided die $3$ times. Thus, if I know that I rolled a $4$, that means the probability that my first pick was the $6, 12,$ and $20$-sided die is $\frac 59, \frac {5}{18}.$ and $\frac 16$, respectively.

Your proposed solution is incorrect because you assume that each die was equally likely to have been picked as your first die, but once you know that your first roll was a $4$, that's no longer true.

If your first pick was the $6$-sided die, then the probability your next roll is smaller is $0.5(\frac{3}{12}+\frac{3}{20})=0.2$.

If your first pick was the $12$-sided die, then the probability your next roll is smaller is $0.5(\frac 36 + \frac {3}{20}) = 0.325$.

If your first pick was the $20$-sided die, then the probability your next roll is smaller is $0.5(\frac 36 + \frac{3}{12})=0.375$.

So the correct probability is:

$$\frac 59 \frac 15 + \frac {5}{18} \frac {13}{40}+ \frac 16 \frac 38= \frac 19+ \frac {13}{144} + \frac {1}{16}= \frac{19}{72}.$$

Answer 2

Let $N$ be the result of the first roll. Let $X_i$ be the number of sides of the $i^{\rm th}$ die selected; i.e., $X_i \in \Omega = \{6, 12, 20\}$, for $i = 1, 2$. Let $M$ be the result of the second roll.

We want to compute $$\Pr[M < N \mid N],$$ the probability that the second roll does not exceed $N$, given $N$. To this end, we note $$\Pr[M < N \mid N] = \sum_{x \in \Omega} \Pr[M \le N-1 \mid X_1 = x]\Pr[X_1 = x \mid N].$$ The conditional probability $$\Pr[M \le N - 1 \mid X_1 = x]$$ represents the probability of rolling below $N$ given that the first die had $x$ sides. This can be computed as follows. With probability $1/2$ each, $X_2$ occurs among those values in $\Omega$ that are not equal to $x$, and given that the second die has $X_2$ sides, the probability of not exceeding $N$ on the second roll is $\min\{\frac{N-1}{X_2}, 1\}$. Thus $$\begin{align} \Pr[M < N \mid N] &= \frac{1}{2} \Bigl( \left(\min \left\{ \tfrac{N-1}{12}, 1 \right\} + \min \left\{ \tfrac{N-1}{20}, 1 \right\} \right)\Pr[X_1 = 6 \mid N] \\ &\quad+ \left(\min \left\{ \tfrac{N-1}{20}, 1 \right\} + \min \left\{ \tfrac{N-1}{6}, 1 \right\} \right) \Pr[X_1 = 12 \mid N] \\ &\quad+ \left( \min \left\{ \tfrac{N-1}{6}, 1 \right\} + \min \left\{ \tfrac{N-1}{12}, 1 \right\} \right)\Pr[X_1 = 20 \mid N] \Bigr). \end{align}$$ All that is left is to compute the conditional probabilities of $X_1$ given $N$. By Bayes' theorem, we have $$\Pr[X_1 = x \mid N = n] = \frac{\Pr[N = n \mid X_1 = x]\Pr[X_1 = x]}{\Pr[N = n]}.$$ The denominator is, by the law of total probability, $$\Pr[N = n] = \sum_{x \in \Omega} \Pr[N = n \mid X_1 = x]\Pr[X_1 = x] = \frac{1}{3} \sum_{x \in \Omega} \Pr[N = n \mid X_1 = x],$$ since $\Pr[X_1 = x] = 1/3$ for all $x \in \Omega$. Then we have $$\Pr[N = n \mid X_1 = x] = \frac{1}{x} \mathbb 1(n \le x).$$ Therefore, $$\Pr[X_1 = x \mid N = n] = \frac{\frac{1}{x} \mathbb 1 (n \le x)}{\frac{1}{6} \mathbb 1 (n \le 6) + \frac{1}{12} \mathbb 1 (n \le 12) + \frac{1}{20} \mathbb 1 (n \le 20)}.$$ From here, it is best to construct a table for $N \in \{1, \ldots, 20\}$ with the desired probabilities: $$\begin{array}{c|cccc} N & \Pr[X_1 \mid N] & \min\{\frac{N-1}{X_1}, 1\} & \Pr[M \le N \mid N] \\ \hline 1 & (\frac{5}{9}, \frac{5}{18}, \frac{1}{6}) & (0, 0, 0) & 0 \\ 2 & (\frac{5}{9}, \frac{5}{18}, \frac{1}{6}) & (\frac{1}{6}, \frac{1}{12}, \frac{1}{20}) & \frac{19}{216} \\ 3 & (\frac{5}{9}, \frac{5}{18}, \frac{1}{6}) & (\frac{1}{3}, \frac{1}{6}, \frac{1}{10}) & \frac{19}{108} \\ 4 & (\frac{5}{9}, \frac{5}{18}, \frac{1}{6}) & (\frac{1}{2}, \frac{1}{4}, \frac{3}{20}) & \frac{19}{72} \\ 5 & (\frac{5}{9}, \frac{5}{18}, \frac{1}{6}) & (\frac{2}{3}, \frac{1}{3}, \frac{1}{5}) & \frac{19}{54} \\ 6 & (\frac{5}{9}, \frac{5}{18}, \frac{1}{6}) & (\frac{5}{6}, \frac{5}{12}, \frac{1}{4}) & \frac{95}{216} \\ 7 & (0, \frac{5}{8}, \frac{3}{8}) & (1, \frac{1}{2}, \frac{3}{10}) & \frac{11}{16} \\ 8 & (0, \frac{5}{8}, \frac{3}{8}) & (1, \frac{7}{12}, \frac{7}{20}) & \frac{23}{32} \\ 9 & (0, \frac{5}{8}, \frac{3}{8}) & (1, \frac{2}{3}, \frac{2}{5}) & \frac{3}{4} \\ 10 & (0, \frac{5}{8}, \frac{3}{8}) & (1, \frac{3}{4}, \frac{9}{20}) & \frac{25}{32} \\ 11 & (0, \frac{5}{8}, \frac{3}{8}) & (1, \frac{5}{6}, \frac{1}{2}) & \frac{13}{16} \\ 12 & (0, \frac{5}{8}, \frac{3}{8}) & (1, \frac{11}{12}, \frac{11}{20}) & \frac{27}{32} \\ 13 & (0, 0, 1) & (1, 1, \frac{3}{5}) & 1 \\ 14 & (0, 0, 1) & (1, 1, \frac{13}{20}) & 1 \\ 15 & (0, 0, 1) & (1, 1, \frac{7}{10}) & 1 \\ 16 & (0, 0, 1) & (1, 1, \frac{3}{4}) & 1 \\ 17 & (0, 0, 1) & (1, 1, \frac{4}{5}) & 1 \\ 18 & (0, 0, 1) & (1, 1, \frac{17}{20}) & 1 \\ 19 & (0, 0, 1) & (1, 1, \frac{9}{10}) & 1 \\ 20 & (0, 0, 1) & (1, 1, \frac{19}{20}) & 1 \\ \end{array}$$