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:
$$\frac 58 \cdot \frac {31}{40} + \frac 38 \cdot \frac {23}{24} = \frac {31}{64}+ \frac {23}{64} = \frac{54}{72} = \frac{3}{4}.$$
Best Answer
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:
$$\frac 59 \frac 15 + \frac {5}{18} \frac {13}{40}+ \frac 16 \frac 38= \frac 19+ \frac {13}{144} + \frac {1}{16}= \frac{19}{72}.$$