# Flaw in a conditional probability argument

Imagine an experiment where you roll two fair, six-sided dice. Someone peeks at the dice, and (truthfully) tells you that "at least one of the dice is a 4". What is the probability that the total of the dice is 7?

It seems straightforward to calculate that the probability the total is 7 is 2/11.

However, the person who peeked at the dice could equally well have said "at least one of the dice is a 1" and you would come to the same conclusion - 2/11. Or they could have said "at least one of the dice is a 2" or "at least one of the dice is a 3", or indeed any number from 1 to 6, and you would still conclude that the probability that the total is 7 is 2/11.

Since you will always conclude that the probability that the total is 7 is 2/11, you could block your ears as they speak, and you'd still come up with 2/11. From there it's a short hop to conclude that even if they don't say anything the probability that the total is 7 is 2/11.

However, clearly if they don't say anything, the probability that the total is 7 is not 2/11, but rather 1/6.

Where is the flaw in the argument?

• You can simplify this question to asking about two flips of a coin, having the person say at least one of the flips is heads/tails and noting that the probability of the flips being different seems to rise from one half to two thirds. Nov 6, 2013 at 15:30
• One possible, although incomplete answer is that the conditional distribution over the total indeed changes in light of different evidence. The number 7 is the only magical number which probability will be the same given any evidence. Nov 6, 2013 at 17:47

The flaw in the argument is that the conditioning random variable is not well-defined.

The ambiguity lies in how our friend peeking at the dice decides to report the information back to us. If we let $X_1$ and $X_2$ denote the random variables associated with the values of each of the dice, then it is certainly true that for each $k \in \{1,2,\ldots,6\}$, $$\mathbb P(X_1 + X_2 = 7 \mid X_1 = k \cup X_2 = k) = \frac{2}{11} \>,$$ independently of $k$.

However, the events $\{X_1 = k \cup X_2 = k\}$ are clearly not mutually exclusive, and so clearly we cannot claim \begin{align} \mathbb P(X_1 + X_2 = 7) &\stackrel{?}{=} \sum_{k=1}^6 \mathbb P(X_1 + X_2 = 7 \mid X_1 = k \cup X_2 = k) \mathbb P( X_1 = k \cup X_2 = k ) \cr &\stackrel{?}{=} \frac{2}{11} \sum_{k=1}^6 \mathbb P( X_1 = k \cup X_2 = k ) \cr &\stackrel{?}{=} \frac{2}{11} \end{align}

Formally, we need to properly define a random variable, say $K$, that encodes the knowledge imparted by our peeking friend.

Our peeking friend could always report the value of the left-most die, or the right-most, or the larger of the two. She could flip a coin and then report based on the coin flip, or employ any number of more complicated machinations.

But, once this process is specified, the apparent paradox vanishes.

Indeed, suppose that $K = X_1$. Then, we have \begin{align} \mathbb P(X_1 + X_2 = 7) &= \sum_{k=1}^6 \mathbb P(X_1+X_2 = 7, K=k) \cr &= \sum_{k=1}^6 \mathbb P(X_1+X_2 = 7 \mid K=k) \mathbb P(K=k) \cr &= \sum_{k=1}^6 \frac{1}{36} = \frac{1}{6} \>. \end{align}

Similar arguments hold if we choose $K = X_2$ or $K = \max(X_1,X_2)$, etc.

• +1 Similar reasoning provides insight into the notorious Monty Hall problem.
– whuber
Nov 6, 2013 at 20:02

If $B$ is an event with the property that $P(B\mid D_i) = p$ for all events $\{D_1, D_2, \ldots\}$ in a countable partition of the sample space $\Omega$, (that is, $D_i \cap D_j = \emptyset$ for all $i \neq j$ and $\bigcup_i D_i = \Omega$), then the law of total probability tells us that $$P(B) = \sum_i P(B\mid D_i)P(D_i) = p\sum_i P(D_i) = p.$$ However, the law of total probability does not apply if the events $D_i$ are not mutually exclusive (even though their union is still $\Omega$), and we cannot conclude that $P(B)$ equals the common value of $P(B\mid D_i)$.

Let $A_i$ denote the event that at least one of the dice shows the number $i$ and $B$ the event that the sum of the two numbers on the die is $7$. We know that $P(B) = \frac{1}{6}$ and that $P(A_i) = \frac{11}{36}$. Also, $P(B\mid A_i) = \frac{2}{11}$. Now, $A_1\cup A_2\cup A_3 \cup A_4\cup A_5\cup A_6$ is the entire sample space $\Omega$ but we cannot use the fact that $P(B\mid A_i)$ is the same for all choices of $i$ to conclude that $P(B) = \frac{2}{11}$ because the $A_i$ are not mutually exclusive events. However, notice that regarded as a multiset, $A_1\cup A_2\cup A_3 \cup A_4\cup A_5\cup A_6$ contains each outcome $(i,j)$ exactly twice, once as a member of $A_i$ and again as a member of $A_j$. Therefore, $$\sum_{i=1}^6 P(B \mid A_i)P(A_i) = \sum_{i=1}^6 \frac{2}{11}\times\frac{11}{36} = \frac{1}{3}$$ which is exactly twice the value of $P(B)$.