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I realize this is a rather trivial question when it comes to some that are asked on this exchange, but I was hoping for an intuitive (AND mathematical - I can't make sense of it if I don't understand the math) explanation of this problem:

This is actually an example problem in my book and I don't really understand the answer in non-mathematical or mathematical terms.

Roll two fair four-sided dice. Let $X_1$ and $X_2$ denote the number of dots that appear on die 1 and 2, respectively. Let A be the event $X_1 \geq 2$. What is P[A]? Let B denote the event $X_2 > X_1$. What is P[B]? What is P[A|B]?

So in my mind, the answer to what is P[A] = 3/4 - BECAUSE, die 1 can either be 2,3, or 4. BUT, according to their answer it is 12/16 = 3/4. Now, this is counter-intuitive to me because die 2 does not have anything to do with P[A] in this question..? (Note: They drew a diagram of 4x4 'grid' and said $X_2$ could be any of it's combinations.) This is somewhat bothersome to me, but the most bothersome is the second and third part of the question:

My response to P[B]?

If $X_1$ = 1, $X_2$ = 2,3,4 => 3/4

If $X_1$ = 2, $X_2$ = 3,4 => 2/4

If $X_1$ = 3, $X_2$ = 4 = 1/4

We then add 3/4 + 2/4 + 1/4 = 6/4 - which is of course wrong.

What exactly am I missing here? I understand that there are 16 combinations, but it doesn't seem to me that they matter at this point? The book uses a graph to explain this, but it is more or less equivalent to: 3/16 + 2/16 + 1/16 = 6/16 = P[B]. Why is it $a$/16 since we are only looking at die 2 with respect to die 1. That is, if a 1 is rolled on die 1, die 2 can either be a 2,3, or 4 (out of it's four possibilities)- which is why I have my error from above..

Then, we have P[A|B]. So, P[A|B] = $P[A\cap B]\div P[B]$. Let us assume that this problem was huge and I could not say (2,3),(2,4),(3,4) = 3/16 for P[AB] how then, would one easily determine P[AB]?

You are my hero if you are able to guide me in the probabilistic way of thinking on this!

Thanks in advance!

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2 Answers 2

up vote 1 down vote accepted

In the first question, you are not missing anything, just failing to see understand what your answer book is saying. The book says $(X_1,X_2)$ can be any of the $16$ equally likely outcomes $$\{(1,1), (1,2),\ldots, (1,4), (2,1), (2,2), \ldots (4,4)\}\tag{1}$$ and so $P(A) = P((X_1,X_2) = (2, d) ~\text{or}~(3,d)~\text{or}~(4,d)) = \frac{12}{16} = \frac{3}{4}.$ Here, $d$ means "don't care". You are ignoring $X_2$ entirely and saying that $X_1$ takes on values $1,2,3,4$ with equal probability $\frac{1}{4}$ and so $P(A) = \frac{3}{4}$. Notice that $((X_1,X_2) = (2,d))$ is the event $$((X_1,X_2) \in \{(2,1), (2,2), (2,3), (2,4)\})$$ which can be described more succinctly as $(X_1=2)$ and has probability $\frac{1}{4}$ regardless of how you describe it. Similarly for $(X_1=3)$ and $(X_1=4)$.

For the second question, just look at the $4\times 4$ array with entries $(X_1,X_2)$ in the set shown above in $(1)$ and count the number of entries that correspond to $B$. There are $6$ entries as you already found; each has probability $\frac{1}{16}$ and so $P(B) = \frac{6}{16}=\frac{3}{8}$.

I will leave the determination of $P(A\cap B)$ to you.

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The probabilities that you have for B are the conditional probabilities $P[B \mid X_1 = 1]$, $P[B \mid X_1 = 2]$ and $P[B \mid X_1 = 3]$. There you can look at the 4 equally likely possibilities for $X_2$ and compute the conditional probability.

For the general $P[B] = P[X_2 > X_1]$ you indeed have to look at the 16 equally likely possibilities of die 1 and 2 combined: both are relevant. If you really wanted, you could do $$P[B] = P[B \mid X_1 = 1] \cdot P[X_1 = 1] + P[B \mid X_1 = 2] \cdot P[X_1 = 2] + P[B \mid X_1 = 3] \cdot P[X_1 = 3] + P[B \mid X_1 = 4] \cdot P[X_1 = 4],$$ but that's quite a detour and in the end you're actually still considering all 16 possibilities for die 1 and die 2 together.

I'm guessing your text looks at all 16 possibilities to compute $P[A]$ as well just to make the intuition behind $P[A \mid B]$ more clear: you can then look at which of the 16 possibilities satisfy $B$ and among those figure out which satisfy $A$ as well.

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