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Kleber's paper will certainly point you in the right direction if you can find it. (I have a printed out copy, and I don't remember where on the web I got it. Sorry.)

In it, he suggests thinking about a strategy as a pairing up of the $(n!+n-1){n-1}$ (n!+n-1)_{n-1}$messages with the${n!+n-1}\choose {{n!+n-1}\choose n} = (n!+n-1){n-1}$n!+n-1)_{n-1}$ hands the audience can give you. Of course, you can only pair a message with one of the n! hands that contain its cards, and you can only pair a hand with one of the n! messages you can make with it. So this is just like finding a perfect matching on an n!-regular bipartite graph, and by Hall's Marriage Theorem this is possible.

The bipartite graph you draw here is quite symmetric: aside from being n!-regular, it's also vertex transitive, so any of the n! edges you could choose from a particular vertex v are part of some perfect matching (since one of them is). This is where Kleber's claim that there are "at least n!" strategies come from.

We can get a much better lower bound if this paper is to be believed. Here it says there are at least $\left(\frac{(n!-1)^{n!-1}}{n!^{n!-2}}\right)^{{n!+n-1}\choose n}$ strategies, which for n=2 isn't very enlightening (it says there's at least one, when we know there are really two), but for n>2 puts our previous estimate to shame: with a three-card hand (and thus an eight-card deck), we get at least 2.54x10^21 strategies, a number so fantastically larger than our previous estimate of 6 that I'm still a little bit in shock. (It is, at least, many orders of magnitude lower than the naive upper bound of 6^56, where for each of the 3-subsets of 8 we choose one of the possible messages it could send without worrying about overlap with other hands.)

Anyway, I haven't read the linked paper, but the abstract suggests we might not get a much better lower bound than this. To improve, one might look for results on vertex transitive k-regular bipartite graphs, but I haven't found any.

1

Kleber's paper will certainly point you in the right direction if you can find it. (I have a printed out copy, and I don't remember where on the web I got it. Sorry.)

In it, he suggests thinking about a strategy as a pairing up of the $(n!+n-1){n-1}$ messages with the ${n!+n-1}\choose n = (n!+n-1){n-1}$ hands the audience can give you. Of course, you can only pair a message with one of the n! hands that contain its cards, and you can only pair a hand with one of the n! messages you can make with it. So this is just like finding a perfect matching on an n!-regular bipartite graph, and by Hall's Marriage Theorem this is possible.

The bipartite graph you draw here is quite symmetric: aside from being n!-regular, it's also vertex transitive, so any of the n! edges you could choose from a particular vertex v are part of some perfect matching (since one of them is). This is where Kleber's claim that there are "at least n!" strategies come from.

We can get a much better lower bound if this paper is to be believed. Here it says there are at least $\left(\frac{(n!-1)^{n!-1}}{n!^{n!-2}}\right)^{{n!+n-1}\choose n}$ strategies, which for n=2 isn't very enlightening (it says there's at least one, when we know there are really two), but for n>2 puts our previous estimate to shame: with a three-card hand (and thus an eight-card deck), we get at least 2.54x10^21 strategies, a number so fantastically larger than our previous estimate of 6 that I'm still a little bit in shock. (It is, at least, many orders of magnitude lower than the naive upper bound of 6^56, where for each of the 3-subsets of 8 we choose one of the possible messages it could send without worrying about overlap with other hands.)

Anyway, I haven't read the linked paper, but the abstract suggests we might not get a much better lower bound than this. To improve, one might look for results on vertex transitive k-regular bipartite graphs, but I haven't found any.