Suppose you're playing a three player game with wheat, barley, and corn. For one of the trading players to win with wheat, the two trading players must have nine wheat between them, which is the same as the shut out player not getting any wheat.
To get a hand, we choose 9 cards out of 27. There are C(27,9) ways of doing that, where C is the binomial coefficient. Once we add the restriction that the hand must have no wheat, we have 18 non-wheat cards to choose from, so the probability is (#of ways to get a hand with no wheat)/(total # of ways to get a hand) = C(18,9)/C(27,9), which is slightly more than 1%. An equivalent calculation for four players gets slightly more than 5%. We can then multiply the 1% by 3 (since there are three different commodities that the other players can use to win), and 5% by 4. This gets the approximation that with three players, the other players have a 3% chance of winning, and with four players, they have 20%.
This is a simplified version of the math, however. To get the exact number, we'll need something called the principle of inclusion/exclusion. This says that when we count all the ways of getting one commodity, we double count any configuration that has two commodities. For instance, any configuration for wheat and barley will appear on both the list for wheat, and the list for barley. So we need to subtract the number of ways of getting two. But then when we count the number of ways of getting two, we're overcounting the number of ways of getting three. And so on. So the total number is:
# of ways of getting 1 - (# of ways of getting 2 - (# of ways of getting 3 …)
This can be rewritten as
# of ways of getting 1 - # of ways of getting 2 + # of ways of getting 3 …
For three players, there's only one way the shut out player can be missing any two specific commodities: their hand must consist entirely of the third commodity. And there's no way they can miss all three commodities. So for three players, this doesn't change the answer much. As the number of players increases, however, this adjustment becomes more important (eventually, you'll be getting over 100% without the adjustment).
For four players, the adjustment still isn't very large. For one commodity we have 4 choices which commodity, then C(27,9) ways for the shut out player to be missing it. For two commodities, we have 6 choices what two commodities, and C(18,9) ways for the shut out player to be missing them. For 3 commodities, there are 4 choices which commodities, and 1 way for the shut out player to be missing them. So we have (4*C(27,9)-6*C(18,9)+4)/C(36,9), or 19.60%.
I also wrote some Python code to simulate it:
def simulation(commodity_count, iteration_count = 10**5):
commodities = range(commodity_count)
loss_count = 0
for i in range(iteration_count):
p = np.random.permutation([c for c in commodities for interation in range(9)])
loss = all ([c in p[:9] for c in commodities])
loss_count += int(loss)
This resulted in 0.96985 as the proportion of games lost in the three player case and 0.80308 for four, consistent with the calculated numbers.