# 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.

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,⋯,6} - {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 = $\frac{1}{6}\cdot \frac{2}{3}\cdot \left(\frac{1}{4}\cdot \frac{1}{5}\right)=\frac{1}{180}$
Case 2: Number 3 showed up
Chance of all this happening = $\frac{1}{6}\cdot \frac{3}{4}\cdot \left(\frac{1}{3}\cdot \frac{1}{5}\right)=\frac{1}{120}$
Case 3: Other number showed up
Chance of all this happening = $\frac{4}{6}\cdot \left(\frac{1}{4}\cdot \frac{1}{5}\right)\cdot \left(\frac{1}{3}\cdot \frac{1}{5}\right)=\frac{1}{450}$
So, total = $\overline{)\frac{\frac{1}{180}}{\frac{29}{1800}}=\frac{10}{29}}$
Is my attempt correct? If not, how to do this problem?
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Bridget Acevedo
Step 1
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 $P\left({D}_{1}\mid {A}_{3}\cap {B}_{1}\right)$. We'll assume the probabilities

and also that Alice's and Bob's responses to any roll are independent in the sense that
$P\left({A}_{l}\cap {B}_{m}\mid {D}_{k}\right)=P\left({A}_{l}\mid {D}_{k}\right)P\left({B}_{m}\mid {D}_{k}\right)$
for all k,l,m.
Then we have
$\begin{array}{rl}P\left({D}_{1}\mid {A}_{3}\cap {B}_{1}\right)& =\frac{P\left({D}_{1}\cap {A}_{3}\cap {B}_{1}\right)}{P\left({A}_{3}\cap {B}_{1}\right)}\phantom{\rule{2em}{0ex}}\text{(definition of conditional probability)}\\ & =\frac{P\left({D}_{1}\cap {A}_{3}\cap {B}_{1}\right)}{\sum _{k=1}^{6}P\left({D}_{k}\cap {A}_{3}\cap {B}_{1}\right)}\phantom{\rule{2em}{0ex}}\text{(total probability rule)}.\end{array}$
Step 2
Now using the definition of conditional probability and the above assumptions,
###### Did you like this example?
limfne2c
Step 1
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 $P\left(\mathbb{1}\right)/\left[1-P\left(\mathbb{1}\right)\right]=1/5$. Now we need the likelihood ratio:
$\text{LR}=\frac{P\left({A}_{3},{B}_{1}|\mathbb{1}\right)}{P\left({A}_{3},{B}_{1}|{\mathbb{1}}^{c}\right)}$
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
$\text{LR}=\frac{P\left({A}_{3}|\mathbb{1}\right)P\left({B}_{1}|\mathbb{1}\right)}{P\left({A}_{3}|{\mathbb{1}}^{c}\right)P\left({B}_{1}|{\mathbb{1}}^{c}\right)}$
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,
$\frac{P\left({B}_{1}|\mathbb{1}\right)}{P\left({B}_{1}|{\mathbb{1}}^{c}\right)}=\frac{p}{\left(1-p\right)×1/5}=5\left(\frac{p}{1-p}\right)$
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
$P\left({A}_{3}|\mathbb{1}\right)=P\left(\text{Alice Lies}\cap {A}_{3}|\mathbb{1}\right)=P\left({A}_{3}|\mathbb{1},\text{Alice Lies}\right)P\left(\text{Alice Lies}|\mathbb{1}\right)=1/5×\left(1-q\right).$
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
$\begin{array}{rl}P\left({A}_{3}|{\mathbb{1}}^{c}\right)& =P\left({A}_{3}|{\mathbb{1}}^{c},\mathbb{3}\right)P\left(\mathbb{3}|{\mathbb{1}}^{c}\right)+P\left({A}_{3}|{\mathbb{1}}^{c},{\mathbb{3}}^{c}\right)P\left({\mathbb{3}}^{c}|{\mathbb{1}}^{c}\right)\\ & =P\left(\text{Alice Tells the Truth}\right)×1/5+P\left({A}_{3}|{\mathbb{1}}^{c},{\mathbb{3}}^{c}\right)×4/5\\ & =q×1/5+P\left({A}_{3}|{\mathbb{1}}^{c},{\mathbb{3}}^{c}\right)×4/5\end{array}$
so it remains to calculate $P\left({A}_{3}|{\mathbb{1}}^{c},{\mathbb{3}}^{c}\right)$. We can do this using the law of total probability:
$\begin{array}{rl}P\left({A}_{3}|{\mathbb{1}}^{c},{\mathbb{3}}^{c}\right)& =P\left(\text{Alice Lies}\cap {A}_{3}|{\mathbb{1}}^{c},{\mathbb{3}}^{c}\right)\\ & =\sum _{k=2,4,5,6}P\left(\text{Alice Lies}\cap {A}_{3}|\text{True Roll}=k\right)P\left(\text{True Roll = k}|{\mathbb{1}}^{c},{\mathbb{3}}^{c}\right)\\ & =\sum _{k=2,4,5,6}\left(1-q\right)P\left({A}_{3}|\text{True Roll}=k\right)×1/4\\ & =4×\left(1-q\right)×1/5×1/4\\ & =\left(1-q\right)×1/5.\end{array}$
Step 2
Finally, we can compute the likelihood ratio contribution for Alice! It is given by
$\frac{P\left({A}_{3}|\mathbb{1}\right)}{P\left({A}_{3}|{\mathbb{1}}^{c}\right)}=\frac{\left(1-q\right)×1/5}{q×1/5+\left(1-q\right)×1/5×4/5}=\frac{\left(1-q\right)}{q+\left(1-q\right)×4/5}.$
Now, combining the likelihood ratio contributions from Bob and Alice, we obtain
$\text{LR}=5\left(\frac{p}{1-p}\right)\cdot \frac{\left(1-q\right)}{q+\left(1-q\right)×4/5}$
and the posterior odds are simply the product of the likelihood ratio and the prior odds:
$O=\frac{1}{5}×\text{LR}=\frac{\left(\frac{p}{1-p}\right)}{\left(\frac{q}{1-q}\right)+4/5}.$
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: $p/\left(1-p\right)=2$ and $q/\left(1-q\right)=3$, so we obtain $O=\frac{2}{3+4/5}=10/19$. Finally we can convert this to a probability: $O/\left(1+O\right)=10/29$