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We are assuming that every horse has an equally likely chance to win (or that you have no knowledge to the contrary.)
I. One method is this: the chance of selecting the first place finisher is 1/12; the chance of then selecting the second place finisher is 1/11 (as there are only 11 horses left to choose from since you selected first place correctly); and the probability of selecting the third place horse having already selected the first and second place horses is 1/10.
The probability of all three happening is 1/12*1/11*1/10=1/1320
II. Another method is to find the size of the sample space. Since order matters, we find the number of permutations of 12 items taken three at a time to be `_12P_3=1320` . As there is only one correct selection, the probability is 1/1320 as before.
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