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There are quite a few questions regarding three prisoners riddle, but my question relating to a particular line of reasoning from Cassella and Berger, as highlighted in the image below.

Could you please explain how the authors derived the first equality? Was it based on some formula/theorem that I'm not aware of? Thank you.

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    $\begingroup$ $P(A) = \sum_iP(A,B_i)$. Let $A$ be "warden says B dies" and $B_i$ be who is pardoned. They are basically "integrating out" the variable containing information about who is pardoned. $\endgroup$
    – jbowman
    Commented Dec 10, 2019 at 3:20
  • $\begingroup$ Oh, is that all? Is the "intergrating out" a proper logic/method? I wonder why the authors didn't include the equality provided by you to make their explanation easier to follow? $\endgroup$
    – Nemo
    Commented Dec 10, 2019 at 3:31
  • $\begingroup$ Try drawing it out as a Venn diagram, with a circle divided into three equal pieces for A pardoned, B pardoned, and C pardoned. Then overlay an area representing "Warden says B dies" vs "Warden says C dies", more or less approximating the numbers above, and you'll see how they add up. $\endgroup$
    – jbowman
    Commented Dec 10, 2019 at 3:59
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    $\begingroup$ This is sometimes referred to as the Law of Total Probability $\endgroup$
    – knrumsey
    Commented Dec 10, 2019 at 5:23
  • $\begingroup$ Thanks, @jbowman. I thought Venn diagram was only used for sets, not for probabilty? Based on your instruction, I drew a circle with 3 equal parts labeled as A, B, C. To represent "B dies", I shaded parts A and C. For "C dies", I shaded parts A and B. Is that what you wanted me to draw? But I couldn't make out any insight from that visualisation! $\endgroup$
    – Nemo
    Commented Dec 10, 2019 at 5:29

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The wikipedia page for the problem helps pinpoint the reason for the 1/6 probability which I quote below. It also articulates the problem setting in a bit of a more verbose way which helps understand where that probability comes from so I recommend reading the problem statement on the wikipedia page as well.

The answer is that prisoner A did not gain any information about his own fate, since he already knew that the warden would give him the name of someone else. Prisoner A, prior to hearing from the warden, estimates his chances of being pardoned as 1/3, the same as both B and C. As the warden says B will be executed, it is either because C will be pardoned (1/3 chance), or A will be pardoned (1/3 chance) and the coin to decide whether to name B or C the warden flipped came up B (1 / 2 chance; for an overall 1/2 × 1/3 = 1/6 chance B was named because A will be pardoned)

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The question stipulates that exactly one of the three prisoners will be pardoned. This means that the sample space $\Omega$ can be partitioned into three disjoint events:

$$\mathscr{P}_A \equiv \{ \text{A pardoned} \},$$ $$\mathscr{P}_B \equiv \{ \text{B pardoned} \},$$ $$\mathscr{P}_C \equiv \{ \text{C pardoned} \}.$$

Letting $\mathscr{W}_B \equiv \{ \text{Warden says B dies} \}$ be the event of interest, the law of total probability then gives:

$$\mathbb{P}(\mathscr{W}_B) = \mathbb{P}(\mathscr{W}_B \cap \mathscr{P}_A) + \mathbb{P}(\mathscr{W}_B \cap \mathscr{P}_B) + \mathbb{P}(\mathscr{W}_B \cap \mathscr{P}_C).$$

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  • $\begingroup$ Thanks for your answer, @Reinstate Monica. (Beautiful symbols by the way). Could you please elaborate on how to calculate the first component on the right hand side of the equality? Intuitively, $P$(B dies $\cap$ A pardoned) = $P$(B dies $\cap$ C pardoned) = 1/3. However, the textbook says $P$(B dies $\cap$ A pardoned) = 1/6 as it takes into account the event of {C dies and A pardoned}. What is the logic for that method of calculation? $\endgroup$
    – Nemo
    Commented Dec 11, 2019 at 11:16

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