Alice and Bob sometimes lie; a die is thrown and they claim different results; what is the probability that Bob is being honest?

Question

Alice speaks the truth with probability $3/4$ and Bob speaks the truth with probability $2/3$. A die is thrown and both Alice and Bob observe the number. Afterwards, Alice asserts to Carl (who does not know the number) that the number is $3$ while Bob says (to Carl) the number is $1$. Find the probability that the number is actually $1$.

UPDATE: To clear ambiguity, note that if person decides to lie, he/she will choose a false answer randomly from all the possible false answer ($\{1, 2, \cdots, 6\}$ - $\{\text{The number that actually showed up}\}$). Also, a die is thrown, and then both Alice and Bob will see the number. Then, they will lie/say truth accordingly.

My attempt:

Case $1$: Number $1$ showed up.

Chance of all this happening = $\frac{1}{6} \cdot \frac{2}{3} \cdot \big(\frac{1}{4} \cdot \frac{1}{5}\big) = \frac{1}{180}$

Case $2$: Number $3$ showed up

Chance of all this happening = $\frac{1}{6} \cdot \frac{3}{4} \cdot \big(\frac{1}{3} \cdot \frac{1}{5}\big) = \frac{1}{120}$

Case $3$: Other number showed up

Chance of all this happening = $\frac{4}{6} \cdot (\frac{1}{4} \cdot \frac{1}{5}) \cdot (\frac{1}{3} \cdot \frac{1}{5}) = \frac{1}{450}$

So, total = $\boxed{\frac{\frac{1}{180}}{\frac{29}{1800}} = \frac{10}{29}}$

Is my attempt correct? If not, how to do this problem?

Answer 1

Nice problem! I think OP has the right answer using a correct method and I have nothing to add in terms of the calculation, but perhaps if we set up some careful notation and also be very precise about our independence assumptions, it will become more clear. For $k=1,2,3,4,5,6$, define the events

• $D_k$ that the die shows $k$;
• $A_k$ that Alice says it is $k$;
• $B_k$ that Bob says it is $k$.

Then the problem asks for $P(D_1\mid A_3\cap B_1)$. We'll assume the probabilities $$P(A_l\mid D_k)=\begin{cases}\frac34&\hbox{if $l=k$}\cr \frac14\cdot\frac15&\hbox{if $l\ne k$,}\end{cases}\qquad P(B_m\mid D_k)=\begin{cases}\frac23&\hbox{if $m=k$}\cr \frac13\cdot\frac15&\hbox{if $m\ne k$}\end{cases}$$ and also that Alice's and Bob's responses to any roll are independent in the sense that $$P(A_l\cap B_m\mid D_k)=P(A_l\mid D_k)P(B_m\mid D_k)$$ for all $k,l,m$.

Then we have $$\eqalign{P(D_1\mid A_3\cap B_1) &=\frac{P(D_1\cap A_3\cap B_1)}{P(A_3\cap B_1)}\qquad \hbox{(definition of conditional probability)}\cr &=\frac{P(D_1\cap A_3\cap B_1)}{\displaystyle \sum_{k=1}^6P(D_k\cap A_3\cap B_1)}\qquad\hbox{(total probability rule)}.\cr}$$ Now using the definition of conditional probability and the above assumptions, $$\eqalign{P(D_k\cap A_3\cap B_1) &=P(A_3\cap B_1\mid D_k)P(D_k)\cr &=P(A_3\mid D_k)P(B_1\mid D_k)P(D_k)\cr &=\begin{cases} \frac14\cdot\frac15\cdot\frac23\cdot\frac16&\hbox{if $k=1$}\cr \frac34\cdot\frac13\cdot\frac15\cdot\frac16&\hbox{if $k=3$}\cr \frac14\cdot\frac15\cdot\frac13\cdot\frac15\cdot\frac16&\hbox{if $k\ne1,3$,}\end{cases}\cr}$$ and substituting back above gives the answer $\frac{10}{29}$.

Answer 2

Since no one seems super confident in their answers, I figured I would contribute my own code that checks numerically. A pretty naive Python script that might take a minute or two on weaker hardware, but it works.

import random
pick a random number from the valid range.
def die():
    return random.randint(1,6)
keep picking a random number until it isn't the specified one
def rand_except(n):
    t = n
    while t == n:
        t = die()
    return t
def one_third_chance():
    return random.randint(1,3) == 3
def one_fourth_chance():
    return random.randint(1,4) == 4
assume bob and alice are both right by default, but assign them
new numbers if they lose a 1/3 roll or 1/4 roll respectively
def run():
    bob = alice = n = die()
if one_third_chance():
    bob = rand_except(n)

if one_fourth_chance():
    alice = rand_except(n)

return (n, bob, alice)


record 10 million results from our main function
data = [run() for _ in range(10_000_000)]
we're only considering cases where alice and bob disagree
cases = [r for r in data if r[1] != r[2]]
the amount of times bob is right
truthful = sum(1 for r in cases if r[0] == r[1])
chance = truthful / len(cases)
error = (chance - (10/29)) / (10/29)
print("Valid cases:", len(cases))
print("Bob was correct", truthful, "times")
print(f"Chance: {chance100}%")
print(f"10/29: {10/29  100}%")
print(f"Error: {error*100}%")

These were my results:

Valid cases: 4831063
Bob was correct 1667199 times
Chance: 34.509982585613147%
10/29: 34.48275862068966%
Error: 0.0789494982781197%

So out of 4.8 million cases the results were off by under 0.08% of what you predicted, meaning you're almost certainly correct as the problem is too simple to yield a fraction with enough precision to outperform your answer.

Answer 3

Edit: This is a tricky one! My initial solution made a silly error in calculating one of the ingredients for $P(A_3|\mathbb{1}^c)$. This updated solution agrees with the OP and the simulation study. I also made the problem a tiny bit more general by allowing the probabilities of lying to be arbitrary.

This is a Bayes Rule question, so there are only two ingredients we need: the prior odds and the likelihood ratio. The prior odds of rolling a one are $P(\mathbb{1})/[1 - P(\mathbb{1})] = 1/5$. Now we need the likelihood ratio: $$ \text{LR} = \frac{P(A_3,B_1|\mathbb{1})}{P(A_3, B_1|\mathbb{1}^c)} $$ Assume that, conditional on the true roll of the die, Alice and Bob make independent reports. This wasn't stated explicitly but it's reasonable in the context. Then the likelihood ratio simplifies to $$ \text{LR} = \frac{P(A_3|\mathbb{1})P(B_1|\mathbb{1})}{P(A_3|\mathbb{1}^c)P( B_1|\mathbb{1}^c)} $$ so we can calculate the likelihood ratio separately for Bob and Alice. Bob is easier, so we'll start with him. Let $p$ be the probability that Bob tells the truth. Then, $$ \frac{P(B_1|\mathbb{1})}{P(B_1|\mathbb{1}^c)} = \frac{p}{(1-p) \times 1/5} = 5 \left(\frac{p}{1-p}\right) $$ since Bob tells the truth with probability $p$, lies with probability $1-p$, and chooses uniformly at random from the 5 numbers that were not rolled when he lies.

Now we'll calculate the contribution from Alice's report. Let $q$ be the probability that she tells the truth. Then we have $$ P(A_3|\mathbb{1}) = P(\text{Alice Lies}\cap A_3|\mathbb{1}) = P(A_3|\mathbb{1}, \text{Alice Lies})P(\text{Alice Lies}|\mathbb{1}) = 1/5 \times(1-q). $$ The reasoning here is identical to the denominator for Bob, but I wanted to spell it out explicitly because the next step is more involved. For Alice's denominator, we have $$ \begin{aligned} P(A_3|\mathbb{1}^c) &= P(A_3|\mathbb{1}^c, \mathbb{3})P(\mathbb{3}|\mathbb{1}^c) + P(A_3|\mathbb{1}^c, \mathbb{3}^c)P(\mathbb{3}^c|\mathbb{1}^c)\\ &= P(\text{Alice Tells the Truth}) \times 1/5 + P(A_3|\mathbb{1}^c, \mathbb{3}^c) \times 4/5\\ &= q \times 1/5 + P(A_3|\mathbb{1}^c, \mathbb{3}^c) \times 4/5 \end{aligned} $$ so it remains to calculate $P(A_3|\mathbb{1}^c, \mathbb{3}^c)$. We can do this using the law of total probability: $$ \begin{aligned} P(A_3|\mathbb{1}^c, \mathbb{3}^c) &= P(\text{Alice Lies}\cap A_3|\mathbb{1}^c, \mathbb{3}^c) \\ &= \sum_{k=2, 4, 5, 6} P(\text{Alice Lies} \cap A_3| \text{True Roll} = k)P(\text{True Roll = k}|\mathbb{1}^c, \mathbb{3}^c)\\ &= \sum_{k=2, 4, 5, 6} (1 - q) P(A_3| \text{True Roll} = k) \times 1/4\\ &= 4 \times (1 - q) \times 1/5 \times 1/4\\ &= (1 - q) \times 1/5. \end{aligned} $$ Finally, we can compute the likelihood ratio contribution for Alice! It is given by $$ \frac{P(A_3|\mathbb{1})}{P(A_3|\mathbb{1}^c)} = \frac{(1 - q) \times 1/5}{q \times 1/5 + (1 - q) \times 1/5 \times 4/5} = \frac{(1 - q)}{q + (1 - q) \times 4/5}. $$ Now, combining the likelihood ratio contributions from Bob and Alice, we obtain $$ \text{LR} = 5\left( \frac{p}{1 - p}\right) \cdot \frac{(1 - q)}{q + (1 - q) \times 4/5} $$ and the posterior odds are simply the product of the likelihood ratio and the prior odds: $$ O = \frac{1}{5} \times \text{LR} = \frac{\displaystyle \left(\frac{p}{1-p}\right) }{\displaystyle \left(\frac{q}{1-q}\right) + 4/5}. $$ What's nice about writing things this way is that it shows that it's the odds of Alice and Bob telling the truth, respectively, that matter for the final solution. Using the values given by the OP: $p/(1-p) = 2$ and $q/(1-q) = 3$, so we obtain $O = \frac{2}{3 + 4/5} = 10/19$. Finally we can convert this to a probability: $O/(1 + O) = 10/29$.

Answer 4

If we didn’t have the observation then the chance that both lie is 1/12, the chance that both say the truth is 6/12, only Bob lies = 3/12, only Alice lies = 2/12. Our observation tells us they are not both saying the truth. The dice is 1 if and only if only Alice lies. Our observation had probability 6/12, only Alice lies had probability 2/12, so the ratio is 2/6 = 1/3. That’s the probability that the number is 1.

This looks suspiciously different to other answers. Let’s try again, and then try to figure out what’s wrong. (The reader might try that now, and finding a different result without finding the error only shows that some calculation is wrong, but not which one).

If we throw 1, the probability of this happening is 2/3 x 1/4 x 1/5 = 2/60 (2/3 that Bob says the truth, 1/4 that Alice lies, 1/5 that she picked 3 and not 2, 4, 5 or 6). If we throw 3, the probability is 3/4 x 1/3 x 1/5 = 3/60. If we throw 2, the probability is 1/3 x 1/4 x 1/5 x 1/5 = 1/60 x 1/5, added for 2, 4, 5, 6 gives 0.8/60. So what we observed happens with probability 5.8/60, it happens if 1 is thrown with probability 2/60, so the probability that 1 is thrown is 2/5.8 (and 3/5.8 for the number 3, and 0.8/5.8 that a different number is thrown, each number with probability 0.2/5.8. Or 10/29 for 1, 15/29 for 3, 1/29 for each other number.

So what’s wrong with the first argument? Here’s the mistake: We had a probability of 1/12 for “both lying”, but we can split that into 1/60 for “both lying in the same way” and 4/60 for “both lying in different ways”. And what we saw ruled out “both lie in the same way”, not just “both say the truth”. That creates a tiny difference in the result.