Before I ask my question I will state the following "simple" problem.

Consider the experiment of drawing cards from a deck. Suppose that we select two cards at random from the desk. When we select the second card, we do not return the first card to the deck. We want to calculate the probability of the second card being spade (event B) given that the first card is spade (event A).

Solution :

Method 1 :

using the definition of conditional probability, the answer is = $\frac{\frac{^{13}C_2}{^{52}C_2}}{\frac{13}{52}} = \frac{4}{17}$

Method 2 :

Here we try to calculate the probability of event B on a reduced sample space (by removing the sample points corresponding to event A (removing the deterministic part from the randomized event) i.e. given that we know the first card selected was a spade, there are now 51 cards left in the deck, 12 of which are spades, thus the required probability = $\frac{12}{51} = \frac{4}{17}$.

Now I have seen that for some problems method 1 is not giving the same answer as method 2 (ex: https://mathoverflow.net/questions/117182/a-problem-on-dtmc). My question is when method 1 will give same answer as method 2 and when not. Hope my confusion is clear from above.

  • $\begingroup$ It is not fully clear what you mean? Both methods should yield the same answer. $\endgroup$
    – broccoli
    Dec 26, 2012 at 19:11
  • $\begingroup$ Somehow "red" in the statement of the problem became "spade" in the solutions. It matters because there are (in a standard deck, 26 red cards and only 13 spades.) $\endgroup$ Dec 26, 2012 at 20:08
  • $\begingroup$ Please go through the problem defined in mathoverflow.net/questions/117182/a-problem-on-dtmc or math.stackexchange.com/questions/264816/a-problem-on-dtmc. There method 1 and method 2 are not giving the same answer. Thta is my confusion. sometimes both are same, sometimes they are not. $\endgroup$
    – user21982
    Dec 27, 2012 at 0:30

1 Answer 1


The first method, using the definition, is of course always correct. By definition, the probability of event $A$ conditioned on (or "given") event $B$ is $$ P(A\;|\;B)=\frac{P(A \wedge B)}{P(B)} $$ whenever $P(B)\neq 0$. The question, I think, is when can you effectively reduce the sample space to one in which event $B$ is guaranteed and then calculate $P(A)$ within this reduced space? The result depends on how you force event $B$; it isn't always obvious how to do this in such a way as to preserve conditional probabilities.

The linked example is a good one; I'll summarize it here. Suppose a subject is standing in room $1$, and is asked to flip a (fair) coin once a minute until heads comes up. Once heads come up, the subject must move to room $2$, in which a hungry bear awaits. The bear will eat an uneaten subject with $99\%$ probability each minute. Now, given that the subject is still alive after $20$ minutes, with what probability is he still in room $1$? It seems plausible that restricting to the "uneaten" sample space by just ignoring the bear might give the right answer. This calculation yields probability $2^{-20}\approx 10^{-6}$ that the subject is still in room $1$. In fact, this isn't right: the longer the subject is in room $2$, the more likely he is to be eaten, and so ignoring the bear isn't neutral with respect to relative probabilities in the "uneaten" sample space. It biases the calculation toward cases where the subject is in room $2$ longer. Using the definition of conditional probability gives a much larger probability that the subject is still in room $1$. In any case where the "second method" gives a different result than the first, the second method is going to be wrong for an analogous reason.

  • $\begingroup$ Thanks for your reply. I did not understand the following statement : "It biases the calculation toward cases where the subject is in room 2 longer." Can you elaborate this statement ? What is wrong in the statement that the subject will be in the room longer if there is no bear. This is true always right. See, I want to find a method 2 for this problem solution. $\endgroup$
    – user21982
    Dec 29, 2012 at 1:45

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