How many such decks can be expected to exist? Well, there are 52 possble cuts, including "no cut", giving us 52 possible scenarios the dealer must win. Assumi

How many such decks can be expected to exist? Well, there are 52 possble cuts, including "no cut", giving us 52 possible scenarios the dealer must win. Assuming naively that winning one scenario does not affect the odds of winning another scenario, the probability of the dealer winning all scenarios is (1/n)^52, where n is the number of players. There are 51! possible decks (given that the deck is cyclic under cutting and we may hence assume the first card). Thus a naive expectation for the number of ice cold decks that exist for any n is 51!/n^52. This has the following values: 2 3.44e50 3 2.40e41 4 7.64e34 5 6.98e29 6 5.33e25 7 1.76e22 8 1.69e19 9 3.71e16 ... 10 1.55e14 11 1.09e12 12 1.18e10 13 1.84e8 14 3.90e6 15 1.08e5 16 3.77e4 17 1.61e2 18 8.25 19 0.49 20 3.44e-2 21 2.72e-3 22 2.42e-4 So the odds of such a deck existing are very strongly in our favor in all of the non-silly cases. Indeed, it appears likely that we can get almost all of the silly cases, too, especially if that assumption about the scenarios being independent was wrong. Finding them, now . . . that might be more work.