I assume the pool of 60 numbers consists of distinct numbers—that is, there are no duplicate numbers among the 60. I also assume that the selection is without replacement—that is, if you selected, say, 13, it is removed from the pool of numbers and so 13 is no longer available to be drawn. These are important factors and constraints to know in calculating the required probabilities.

In other words, you want that none of six specific numbers should be drawn, which means the first number for the second draw can be any one of 54 out of 60, the second number any one of 53 out of 59, …, and the sixth number any one of 49 out of 55. The probability that none of those numbers is drawn is thus P = (54/60)×(53/59)×(52/58)×(51/57)×(50/56)×(49/55) ≈ 0.5159.

Here it gets a bit more complicated. Suppose the first number matches one of the six previously drawn. The probability of this happening is six out of 60. The second number must not match any of those previously drawn, and so can be any one of 54 out of 59. Similarly for the third and subsequent ones. Thus, the probability of drawing a match first, followed by five non-matches is P = (6/60)×(54/59)×(53/58)×(52/57)×(51/56)×(50/55) ≈ 0.06317. However, we have only covered the case when the first number drawn is a match. Using a similar analysis and assuming that the second number drawn is the one that matches, the same probability P ≈ 0.06317 is found. In the cases where it’s the third, fourth, fifth or sixth number that matches, the same probability is again found in each case. Therefore, there are six different equiprobable ways of ending up with exactly one match, and so the overall probability is P = 6×(6/60)×(54/59)×(53/58)×(52/57)×(51/56)×(50/55) ≈ 0.3790.

(For two matches, P = (6|2)×(6/60)×(5/59)×(54/58)×(53/57)×(52/56)×(51/55) ≈ 0.09475 where “(6|2)” means “6 choose 2” = 6×5/2 = 15, which is the number of ways two items can be selected from six. Can you see how we get this answer for two matches?)

There is no difference in their probabilities. All four-number sequences have exactly the same probability of being drawn at random. The fact that the first one has an obvious pattern does not make it less likely.