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Imagine, for example, a 5x5 grid of cards, face down. The cards are either red or blue. Someone has given you the number of red cards for each specific row and specific column. Thus--focusing on a single row or column--if you picked a card at random from that row or column, you would know the likelihood of the card being red; if there's only 1 red card in a given row, the chance that you'll select a red card will be 1 in 5 or 0.2, and if there's no red cards in a given column, the chance that you'll select a red card will be 0 in 5 or 0.

How then, would someone calculate the probability of a specific card being red, being that the card is on the intersection of a row and column, each of which have a known probability?

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There's no way to calculate a probability since you didn't specify a distribution. I'll assume that what you meant is that the cards are independently chosen as red or blue with probability $\frac12$ each and then the row and column counts are provided, and you want the conditional probability of a given card being red or blue given the row and columns sums.

To determine this conditional probability, you need the counts of configurations that are compatible with the given constraints. The probability for the given card to be red is the number of compatible configurations in which that card is red divided by the total number of compatible configurations. I doubt that there's a simple expression for these numbers for arbitrary row and columns sums, but since there are only $2^{25}\approx3\cdot10^7$ different configuration, it's easy to count the compatible configurations on a computer.

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  • $\begingroup$ Wow, I had forgotten about this question. Since you're the only answer in two years I'll give an upvote, but as your answer boils down to brute-forcing the solution I'm not going to accept it on two points: A) brute force tends to be a solution for most topics of probability, albeit a poor one, and B) the question does specify calculating a probability. Not sure if there's an actual formula/calculation to my question, but thanks for this answer! $\endgroup$ – Ranger Apr 19 '16 at 13:59
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    $\begingroup$ @NexTerren: Yes, I wouldn't have given this sort of non- or half-answer if the question hadn't been sitting around without any answer for two years. If you're interested in the result, I'll be happy to write the code for it, but I agree that that's not an elegant way to solve a probability question. $\endgroup$ – joriki Apr 19 '16 at 14:02

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