# Solved – Probability of drawing the unfair die

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?

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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 $$mathrm{P}(text{5 threes},|,mathrm{Fair}) = (1/6)^5$$.
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 $$0.96$$.