Can a Pit Card Game be Won if two players refuse to trade with another particular player?

Question

My sister and I had an argument after I stated that I did not want to play Pit with them anymore because her husband and her son for some unknown reason refused to trade with me. Since she is always right, she said I was not telling the truth because the game is set up so that everyone has to trade in order for someone to win.

While I did not remind her that the game in question was not completed because I threw my cards down and quit, I still believe there is a possibility, however remote, that the game could be won if one player was frozen out by 2 or 3 others. Any mathematical statisticians out there that would like to come up with an answer for me? I really want to know the answer!!

Answer 1

This is actually deceptively simple. A full set includes every card of that type in the game. If you possess a card in a specific set, then nobody else can complete that set. If you possess cards from every set, then no set can be completed without trading with you.

The flip side is that if there is any set that you lack cards from, then that set can necessarily be completed without any trades to you.

Answer 2

(I'm mostly ignoring Bull and Bear rules here, but I think they can be added in with a little fiddling.)

If you just want to know whether it's possible, then it's not a question of probability but of showing that there is at least one arrangement of cards that would allow a player to win without trading. And it's pretty easy to see that this must exist, since it's how the game ends. So if you took player's hands at the end of a game, and then gave them out again in exactly the same order, the game starts with one player having a winning combination.

If you want a situation where the game is winnable if only 2 out of 3 players trade, then that's still possible - imagine stopping the game right before the winning trade is made. Then as long as the players who were going to make that trade aren't the ones being locked out, you know that they have everything they need to bring the game to a win.

What about a game where any one of the three players can be locked out and still guarantee a winnable state? Still pretty straightforward - give each player a hand consisting entirely of one commodity, then take one card from each player's hand and pass it to the player to their left. Now, whichever player is locked out, the other two players just need to make a trade of their single card, and one of them will get a winning hand.

Probabilities are a little trickier to work with, but do-able. If that's what you're really interested in then I can see what I can do.

Answer 3

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)
    return (loss_count/iteration_count)

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.