# When is it appropriate to count outcomes to solve a combinatorial probability problem

I have been looking at problems on this site such as selecting cards from a deck of cards or marbles from a bag of marbles. One thing I have been struggling with is when I can solve the problem simply by counting outcomes, or, equivalently, when I can assume that outcomes have equal probability. I have two recent, motivating examples.

1. Probability of 2 Cards being adjacent. The problem here is to do with the probability of picking cards and finding a seven next to a king. In theory we can solve the problem by considering of 3 groups with identical items in each group: 4 sevens, 4 kings, and 44 others. We then count the number of outcomes of interest, and divide by the total number of outcomes (the total is given by the multinomial theorem).

2. probability of occcuring alternative colors. Here there is one bag, with $n_1$ balls of one color and $n_2$ balls of another, $n_1 \neq n_2$, and we are interested in the probabilities of outcomes when we draw $k < n_1 + n_2$ balls from the bag (with replacement). This is a case where the probabilities of outcomes are not equal.

To quote from the answer to the second question

counting outcomes is not a good way to approach this problem.

I can see in each case, and other cases, when counting is ok -- when outcomes are equally probable, using common sense and intuition. But this took me quite a long time in the first case. It is also error-prone. It seems like more skilled mathematicians understand when counting is a good way to approach a problem, so my question is:-

Are there any rules or principles that I can apply to decide whether or not counting outcomes is reasonable?

(related, more general, question about how to approach combinatorial problems: Combinatorics: When To Use Different Counting Techniques)

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In the problem 2 that you mention, there is only one bag. The problem can be perfectly well solved by counting how many of the $7^4$ equally likely possibilities satisfy the alternating colour criterion. The details are essentially the same as the ones we get when we use the "probabilistic" approach. That said, I prefer direct calculation of probabilities. For one thing it is more general, with equal ease it can deal with fair dice and loaded dice. – André Nicolas Aug 1 '13 at 16:25
@AndréNicolas thanks I have fixed the error in my question. Re counting $7^4$ equally likely possibilities, I suppose the key point is that some of these equally likely possibilities refer to the same outcome (this is related to multiplying by $^n\rm{C}_x$ to get binomial probabilities for an outcome -- I was wondering whether that fitted in). That helps, thanks. – TooTone Aug 1 '13 at 16:37
For $7^4$ we need to treat the balls as having Student Numbers. To count the alternating strings count the WBWB and BWBW. Let us count the WBWB. First slot can be filled in $4$ ways, for each the second can be filled in $3$, and so on. Or else the two W slots can be filled in $4^2$ ways and for each the two B slots can be filled in $3^2$ ways. – André Nicolas Aug 1 '13 at 16:51