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For this example, I have 4 different events a ... d, each with a different probability of occurrence.

My sample space is the following: {a a b b b b c c c d d d}

p(a) = 2/12

p(b) = 4/12

p(c) = 3/12

p(d) = 3/12

Samples are drawn one at a time without replacement. As soon as a particular event is drawn, all other identical outcomes are removed from the sample space. For example, if I draw event b, then the other 3 b's are removed from the sample space leaving me with {a a c c c d d d}.

Is there an efficient way to calculate the probability of any particular event occurring in any particular position? In this example, I have 4 distinct events that can be drawn and 4 possible positions for each event. I'd like to be able to compute the probability of drawing event a in the 1st, 2nd, 3rd, and 4th positions, event b in the 1st, 2nd, 3rd, and 4th positions, and so forth.

Since the likelihood of a particular event occurring in particular position depends on the events selected before it, I am unsure how to come up with a formula that generalizes to larger datasets. I can compute this fairly straightforwardly with basic conditional probability in this example, but I'm wondering if there is a more efficient formulation out there when the numbers of possible events increase beyond a reasonable standard for manual calculation.

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    $\begingroup$ Your description seems unnecessarily complicated. If I follow correctly, you are describing sampling without replacement from $\{a,b,c,d\}$ with the given probabilities. With this simpler characterization in mind, can you identify anything that would cause the chance of any outcome to vary with position? You might consider drawing a probability tree... . BTW, stats.stackexchange.com/questions/20590/… looks like a strikingly similar question. $\endgroup$
    – whuber ♦
    Jan 28, 2014 at 3:53
  • $\begingroup$ A probability tree would help clarify the situation for small sample sizes, but I'm looking for an efficient way to calculate probabilities in each position when I have 100 different events to choose from. The reason the chance of any outcome varies with position is that the likelihood of event b occurring in position 2 is 4/10 if a is chosen first (2 outcomes are removed from the sample space) and 4/9 if either c or d is chosen first (3 outcomes are removed from the sample space in each case). $\endgroup$
    – Binder
    Jan 28, 2014 at 4:00
  • $\begingroup$ The point to the probability tree--and to asking the other questions I did--is that if there is any hope of obtaining a general formula, it will become apparent in looking at the situation with small numbers of events. $\endgroup$
    – whuber ♦
    Jan 28, 2014 at 4:11
  • $\begingroup$ Ah. Gotcha. I'll give it a shot. $\endgroup$
    – Binder
    Jan 28, 2014 at 4:14
  • $\begingroup$ A closely related question is at stats.stackexchange.com/questions/76662/… -- perhaps the answers there might suggest some useful approaches. $\endgroup$
    – whuber ♦
    Jan 28, 2014 at 13:38

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