In order for every number to have probability <3/7, there must be at least 8 numbers.

Consider the set of triples that the Picker preserves. (Each triple will be selected by the Picker to be the secret numbers with some probability, not necessarily equal.) Assume that the probability of every number being among the secret numbers is less than 3/7, and the set contains the triple {1,2,3}. Let p_{123} be the probability that the Picker's secret numbers are 1, 2, and 3, and for other combinations of 1, 2, and 3, replace the corresponding missing numbers with X's (so, for example, the probability that 1 is among the secret numbers but not 2 or 3 is p_{1XX}). Now, we have (A) p_{XXX}=0, since {1,2,3} has nonzero probability. We also have (B) p_{1XX}+p_{12X}+p_{1X3}+p_{123}<3/7, and (C) p_{X2X}+p_{12X}+p_{X23}+p_{123}<3/7, and (D) p_{XX3}+p_{1X3}+p_{X23}+p_{123}<3/7. Since (E) the sum of all 8 probabilities is 1, we can add A, B, C, and D, and subtract E, to get (F) p_{12X}+p_{1X3}+p_{X23}+2*p_{123}<2/7. Since all probabilities are nonnegative, this means (G) p_{123}<1/7. This is true for any possibility, so every triple occurs with probability less than 1/7. Keep this in mind.

If we take equation E and subtract A and B, we get (H) p_{X2X}+p_{XX3}+p_{X23}>4/7. Doing the same but replacing B with C or D gives us (I) p_{1XX}+p_{XX3}+p_{1X3}>4/7 and (J) p_{1XX}+p_{X2X}+p_{12X}>4/7. From these equations, and the fact that all probabilities are non(edit: what's wrong with me for forgetting the non here?)negative, we can deduce (K) p_{1XX}>1/7, (L) p_{X2X}>1/7, and (M) p_{XX3}>1/7. Since every triple occurs with probability less than 1/7, we can deduce that for any element of any triple, there are at least two triples that share only that element with the original triple. Now keep this in mind.

Now, suppose that along with {1,2,3} we have some triple that shares two numberswith it, letting it be {1,2,4}. Then there are at least two triples that contain 3 but not 1 or 2, and since they do not contain 1 or 2 they must contain 4 as well. Let these two triples be {3,4,5} and {3,4,6}. Then there are at least two triples that contain 5 but not 3 or 4, and since they do not contain 3 or 4 they must contain 6 as well. Furthermore, each must contain at least one of 1 or 2, so they must be {1,5,6} and {2,5,6}. Now consider a triple including 7. If it lacks 1 and 2 then it contains 3 and 4, but then it is mutually exclusive with {1,5,6}. Therefore it contains 1 or 2, and it also contains 3 or 4, but then it lacks 5 and 6, and then whichever of 1 or 2 it lacks, it is mutually exclusive with {1,5,6} or {2,5,6}. Therefore, there cannot be any triples involving any numbers except these six, but there must be at least 8 numbers. Therefore, no two triples share any numbers. This gives us (N) p_{12X}=p_{1X3}=p_{X23}=0.

Taking N into account with B, C, D, and G, we get (O) p_{1XX}>2/7, (P) p_{X2X}>2/7, and (Q) p_{XX3}>2/7. Therefore, there must be at least three triples that contain 1 but not 2 or 3, and furthermore none of them share any elements other than 1. Let these triples be {1,4,5}, {1,6,7}, and {1,8,9}. But there must also be a triple that contains 2 but not 1 or 3 (in fact there are at least three of them), so it must contain either 4 or 5, and either 6 or 7, and either 8 or 9, but that is impossible since it's only a triple. Therefore, this entire situation, where every number has probability <3/7, is impossible.