Six-sided and four-sided dice question contradiction

Question

Suppose we have two fair dice, with six sides (numbered 1 to 6) and with four sides (numbered 1 to 4). Suppose we pick a die at random and throw it, the result is announced to be 2 and the die is discarded. Then we pick the remaining die and throw it. What is the expectation of the second throw?

On the one hand, us getting 2 on the first throw, does not provide us with information on whether the first die was six-sided or four-sided, so we could argue that by total expectation the expectation of the second throw is just the average of expectations: $$ E(X)=(3.5)\frac{1}{2} + (2.5)\frac{1}{2}=3 $$

On the other hand, we know that we got 2 on the first throw, so the expectation of the second throw is: $$ E(X)=(1)\frac{2}{9} + (2)\frac{1}{9} + (3)\frac{2}{9} + (4)\frac{2}{9} + (5)\frac{1}{9} + (6)\frac{1}{9}=3.22 $$ It seems to me that there is a contradiction?

Answer 1

On the one hand, us getting 2 on the first throw, does not provide us with information on whether the first dice was six-sided or four-sided

This sentence is false.

Imagine that instead of a 6-sided die and a 4-sided die, you had a 1000000-sided die, numbered 1-1000000, and a 4-sided die, numbered 1-4. You select one of the two dice uniformly at random, and roll it, and the result is a 2. Do you really claim that you have no information about which die was selected? Intuitively, it is very probable that the 4-sided die was selected; if the 1000000-sided die had been selected, the result could have been anything between 1 and 1000000, and it would look like quite an unprobable coincidence that the result was a 2.

A great tool to formalise this intuition and turn it into precise numbers, and make it work with a 6-sided die instead of a 1000000-sided die, is Bayes' theorem.

Let $D_6$ be the event "we picked the 6-sided die first" and $D_4$ the event "we picked the 4-sided die first". Let $R_2$ be the event "we rolled a 2 on the first roll". Then, in the absence of information, the probability of picking the 6-sided die first is $P(D_6) = \frac{1}{2}$; but the probability of having picked the 6-sided die, knowing that we rolled a 2, is noted $P(D_6 \,|\, R_2)$ and can be calculated using Bayes' theorem:

\begin{align*} P(D_6 \,|\, R_2) & = \frac{P(R_2 \,|\, D_6)P(D_6)}{P(R_2)} \\ P(D_6 \,|\, R_2) & = \frac{P(R_2 \,|\, D_6)P(D_6)}{P(R_2 \,|\, D_6)P(D_6) + P(R_2 \,|\, D_4)P(D_4)} \\ \end{align*}

Filling in these values:

• $P(D_6) = P(D_4) = \frac{1}{2}$;
• $P(R_2 \,|\, D_6) = \frac{1}{6}$;
• $P(R_2 \,|\, D_4) = \frac{1}{4}$.

We get: \begin{align*} P(D_6 \,|\, R_2) & = \frac{\frac{1}{6}\times\frac{1}{2}}{\frac{1}{6}\times\frac{1}{2} + \frac{1}{4}\times\frac{1}{2}} \\ P(D_6 \,|\, R_2) & = \frac{2}{5} \\ \end{align*} Thus, after observing that we rolled a 2, the probability that we picked the 6-sided die is $\frac{2}{5}$, and the probability that we picked the 4-sided die is $\frac{3}{5}$.

(If we had used a 1000000-sided die and a 4-sided die, using the same reasoning we would have found that after observing a 2, the probability of having picked the 1000000-sided die is only $\frac{1}{250001}$, or $0.0004\%$.)

Now you can calculate the expectation of the second roll after having observed a 2 on the first roll. Calling $X$ the random variable equal to the result of the second roll:

\begin{align*} \mathbb{E}(X \,|\, R_2) & = \mathbb{E}(X \,|\, D_6)P(D_6 \,|\, R_2) + \mathbb{E}(X \,|\, D_4)P(D_4 \,|\, R_2) \\ \mathbb{E}(X \,|\, R_2) & = 2.5 \times \frac{2}{5} + 3.5 \times \frac{3}{5} \\ \mathbb{E}(X \,|\, R_2) & = 3.1 \\ \end{align*}

Thus the expectation of the second roll, having observed a 2 on the first roll, is 3.1. This is slightly higher than the expectation of the second roll we would have calculated if we hadn't observed the first roll: $$\mathbb{E}(X) = \mathbb{E}(X \,|\, D_6)P(D_6) + \mathbb{E}(X \,|\, D_4)P(D_4) = 2.5 \times \frac{1}{2} + 3.5 \times \frac{1}{2} = 3$$

Answer 2

Let $A$ represent the event we rolled the four-sided die in the first throw. Then $A^c$ represents the event we rolled a six-sided die in the first throw. Remember the second throw we are rolling the other die.

Let $B$ represent the event the outcome of the first roll is a $2$.

We are told $\Pr(A)=\Pr(A^c)=\frac{1}{2}$. We are told by the fact the dice are fair that $\Pr(B\mid A)=\frac{1}{4}$ and $\Pr(B\mid A^c)=\frac{1}{6}$.

Let $X$ be the random variable representing the outcome of the second die throw.

We are tasked with calculating $E[X\mid B]$.

For this, we look at $1\cdot \Pr(X=1\mid B)+2\cdot \Pr(X=2\mid B)+\dots+6\cdot \Pr(X=6\mid B)$

Let us examine $\Pr(X=1\mid B)$ more closely. We have $\Pr(X=1\mid B) = \dfrac{\Pr(X=1\cap B)}{\Pr(B)} = \dfrac{\Pr(X=1\cap A\cap B)+\Pr(X=1\cap A^c\cap B)}{\Pr(B\cap A)+\Pr(B\cap A^c)}$

$=\dfrac{\Pr(A)\Pr(X=1\cap B\mid A)+\Pr(A^c)\Pr(X=1\cap B\mid A^c)}{\Pr(A)\Pr(B\mid A)+\Pr(A^c)\Pr(B\mid A^c)}=\dfrac{\Pr(A)\Pr(B\mid A)\Pr(X=1\mid B\cap A)+\Pr(A^c)\Pr(B\mid A^c)\Pr(X=1\mid B\cap A^c)}{\Pr(A)\Pr(B\mid A)+\Pr(A^c)\Pr(B\mid A^c)}$

Recognize that $\Pr(X=1\mid A\cap B) = \Pr(X=1\mid A)$. Now... these we can actually find from the problem statement.

$=\dfrac{\frac{1}{2}\cdot\frac{1}{4}\cdot\frac{1}{6}+\frac{1}{2}\cdot\frac{1}{6}\cdot\frac{1}{4}}{\frac{1}{2}\cdot\frac{1}{4}+\frac{1}{2}\cdot\frac{1}{6}} =\dfrac{1}{5}$

Similarly we calculate for $X=2, X=3, X=4$. In the case of $X=5$ note that $\Pr(X=5\mid A^c)=0$ and not $\frac{1}{6}$ since there does not exist a face numbered $5$ on the four-sided die. That yields an answer of $\frac{1}{10}$ for those.

This gives a final calculation of our expectation as being:

$$1\cdot \frac{1}{5}+2\cdot \frac{1}{5}+3\cdot\frac{1}{5}+4\cdot\frac{1}{5}+5\cdot\frac{1}{10}+6\cdot\frac{1}{10}=3.1$$

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Note, a priori I would not have expected each face of the two dice to be equally likely results of the second throw given a particular result of the first throw and so would have rejected an answer that attempts to assume that fact. This analysis does show that this happens to be the case here. Remember that this is the exception, not the norm. Be very careful about trying to use $\Pr(A)=\frac{|A|}{|S|}$ as this is many times blatantly incorrect to do. There are two outcomes to a lottery, you win or you lose. You don't win with probability $\frac{1}{2}$ however.