Order in joint probability
- Had a recent run through with the recurring issue of “order” in joint probability calculations. That is, my confusion regarding the perspective of certain situations where joint probabilities are used. The canonical example here is flipping two coins, and asking how we use the observed order of the coins to reason about a question like “What is the probability of seeing one head and one tail?”. Of course, the answer is trivial and the sample space is incredibly small, but how do we arrive there without looking into this sample space? We could say we allow the first observed coin to be anything, and then the second one must be the opposite. So what are the chances that the second coin is not one specific value on any given flip? 50%. And that’s the answer. In this case you don’t have to reason about other ways the coins might be arranged, since when the “second” coin is now the “first”, the outcomes (as in [H,T], [T,H]) will still look exactly the same. The ordering is just there for our convenience, and in this case not important at all.
- Then I was reasoning about a simple roll of three dice, and what the odds where that exactly two of the three were the same value. One way to think about it might be: let the first observed die be any value. Then the probability the second die matching that value is 1/6, and the probability the third die does not match that value is 5/6. So we have 5/61/6=5/36=30/216, where 216 is the cardinality of the sample space, for this particular form of dice roll. Note that the sample space is ordered: [1,2,3] is a different outcome than [3,2,1]. So our computation only accounts for a fraction of the total outcomes in our event of exactly two dice landing on the same value. Here we must consider the order, and include the possible observably different outcomes. In this case we need only consider the three possible positions of the die whose value does not match the other two; in each of those instances, swapping the other two die has no impact on the value of outcome (e.g. we’d have [1,1,2] and [1,1,2], the same outcome in our sample space). This we end up with 330/216=90/216 as our final result for this probability. This differs from the previous example because now the implied order
- Small note: I thought it be straightforward to extend this problem and work it out from the perspective of an unordered sample space, but I was unable to get the proper probability out of it. Here I was using the combination with repetition to get the size of the sample space, but my unordered outcomes didn’t line up. In this case I have 30 outcomes, and only 56 elements in the sample space. This is just confusion on my part about how to approach it, but I should look into this deeper to see what I’m doing wrong.