Probability of drawing the unfair die

Question

I am trying to solve the following problem:

There are 10 dice. One die is unfair; it always gives 3. All 10 dice look identical. You draw one of the dice and roll it 5 times. The resulting number is 3 each time. Calculate the probability that you have drawn the unfair die.

If all dice were fair, no matter which you take, the probability to get 3 would be 1/6. Now we have one unfair die, which should increase this probability. How to calculate this increase? Bayes' theorem says that:

$$P(A|B) = \frac{P(A) P(B|A)}{P(A) P(B|A) + P(\bar A) P(B|\bar A)},$$

and $P(A)$ is the probability of picking the unfair die. How do I calculate the probability of rolling five 3s?

Answer 1

The question seeks to find the probability that the die drawn is unfair given that it was thrown $5$ times and all throws were $3$s. Hence, we seek to calculate $\mathrm{P}(\mathrm{Unfair}\,|\,\text{5 threes})$. According to Bayes' theorem, we have: $$ \mathrm{P}(\mathrm{Unfair}\,|\,\text{5 threes}) = \frac{\mathrm{P}(\text{5 threes}\,|\,\mathrm{Unfair})\cdot \mathrm{P}(\mathrm{Unfair})}{\mathrm{P}(\text{5 threes}\,|\,\mathrm{Fair})\cdot \mathrm{P}(\mathrm{Fair}) + \mathrm{P}(\text{5 threes}\,|\,\text{Unfair})\cdot \mathrm{P}(\text{Unfair})} $$ Now what's the probability to get $5$ $3$s when you picked the unfair die? Well it's $1$ (100%) so $\mathrm{P}(\text{5 threes}\,|\,\text{Unfair}) = 1$ because it always shows a $3$. What's the probability of getting $5$ $3$s when you picked a fair die? It's $(1/6)^5$, so $\mathrm{P}(\text{5 threes}\,|\,\mathrm{Fair}) = (1/6)^5$.

All that's missing are the probabilities of picking a fair or an unfair die. Figure those out and put all values in the formula to get the answer. Can you take it from here?

I find it interesting to plot how the posterior probability of having picked the unfair die depends on the number of $3$s we got. Here is the plot:

So if we throw the die once and get a $3$, the posterior probability is $0.4$ that the die is unfair. After 2 throws, both being $3$s, it's already $0.8$ and after three throws, all of them being $3$s, it's $0.96$.

Answer 2

According to Bayes’ theorem:

P(A | B) = ( P(A) * P(B | A)  )  /  (  P(A) * P(B | A) + P(not A) * P(B | not  A)  )

P(A) = P(unfair) = 1 / 10
P(not A) = P(fair) = 9 / 10

P(B | A) = P(5 threes | unfair) = 1
P(B | not A) = P(5 threes | fair) = 1 / (6^5)

P(A | B) = ( 1 / 10  * 1 ) / (  1 / 10  * 1  + 9 /10 * (1 / (6**5) ) )
P(A | B)  = 0.9988439306358382

P(B) = P(5 threes) =  P(A) * P(B | A) + P(not A) * P(B | not  A)
P(B) = 0.10011574074074074