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So, there is this Italian card game called "Scopa" (broom) I used to play when I was a kid (still do sometimes), and I've been wondering how many possible games can be played.

I'll try to explain how the game is structured but not the exact rules since they're not important for the calculation (if you're interestend in the game here's the Wikipedia article).

The deck is made of $40$ cards and the game is played in 2. At the beginning, of the 40 cards, 4 are dealt on ground, and then 3 cards are dealt to each player. At this point the players will each play one card at a time, until the cards in their hands end, and then they will be dealt 3 cards each again. This is repeat till the deck is empty.

What I've been thinking is:

  1. The number of ways the initial 4 cards can be dealt is $\binom{40}{4}$
  2. The number of ways the 3 cards can be dealt to the players is $\binom{36}{3}$ the first time cards are dealt, $\binom{33}{3}$ the second time, and so on...
  3. Each time a player receives the 3 cards, he has $3!$ ways of playing those cards.

Put all together, my estimates gives: $$\binom{40}{4}\cdot3!\cdot\prod_{i=1}^{12}\binom{i\cdot3}{3}\simeq9.37\cdot10^{37}$$

Is my reasoning correct?

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  • $\begingroup$ I used to play that game as a kid! If I recall correctly, you can drop a card on the ground without picking any up, or you can indeed take a card from the ground, depends on their number/type. Therefore, a previous configuration changes the available moves for the next player. Each move will make the other player some cards available (or not), so this isn't just as easy as counting which card to use first. That's if you distinguish using a card to lay it on the ground or to pick some other one, which I think you should because the scoring is different in each case. $\endgroup$ – Guido A. Oct 18 '17 at 14:52
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You made one mistake: The $(3!)$ should be raised to the $12$ power, since each time a player has to play $3$ cards you get that factor.

But generally, your reasoning and method was right.

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No, your $3!$ should be raised to the $12$th power because there were twelve times the order was invoked. A simpler way to calculate it is to choose the first four cards in $40 \choose 4$ ways as you have done, then recognize that you can just put the remaining $36$ cards in the order they will be played in $36!$ ways, making $${40 \choose 4} \cdot 36! \approx 3.4\cdot 10^{46}$$ ways. If you multiply your number by $6^{11}$ you get this one.

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  • $\begingroup$ Thanks for the quick response, same as @MarkFischler. I get now why i did that mistake: I first tought to use a sum instead of a product, so i multiplied the binomial coefficient sum by $3!$. Then when i switched to the product, i left the $3!$ outside. $\endgroup$ – pr0ves Oct 18 '17 at 15:06
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By far not a complete answer, but let me tell you a couple of facts that are explicative of the reason why the computation is quite hard.

First, if it were as simple as you put it, then you could just say that the number of games if $\binom{40}4\cdot 36!$. In fact, if a game (of people that do not use strategy) were completely determined by the initial cards on the field and the order in which each player plays his cards, then you could order the movers from first to thirty-sixth, notice that one player always plays the odd moves and the other one always plays the even moves, and claim that the number of games for each initial configuration is just $36!$. If you think about it, the fact that the moves are organised in rounds of six is irrelevant.

However, the fact is that the order in which the cards are played does not determine completely a game. Let's say the cards on the field are $\{2,3,4,5\}$ of some seeds, and you play a $7$. You can decide to capture either $\{2,5\}$, or $\{4,3\}$. An event like this duplicates the number of matches assigned to a specific order of moves.

"How can I know when a case like this happens?"

Well, that's why I'm outta here.

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  • $\begingroup$ You're right, i didn't think of that. I thought that, since you can play any card you have in your hand, the cards on the field don't affect the number of games. At least we can say that $\binom{40}4\cdot 36!$ is the lower bound for the number of games. $\endgroup$ – pr0ves Oct 18 '17 at 15:14
  • $\begingroup$ There are horrific matches where, starting from the empty field, people play $1,9,8,3,5,7,2$ and then a $10$. @pr0ves $\endgroup$ – user228113 Oct 18 '17 at 15:22

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