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Suppose the following solitaire with a standard deck. I turn four cards visible on the board and on each turn, I remove those suits that appears more than once in the board. Then I fill the board such that it has four cards and repeat removing. I win the game if I can remove all 52 cards from the board and lose otherwise, i.e. when all cards are from different suit. What is the probability to win this game?

I guess we need some kind of generating polynomial but I'm not sure how to solve that kind of problems.

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  • $\begingroup$ I think you can't win the game, removing all $52$ cards and clearing the table. The max you can remove is $51$ (so you'll have one card left on the table at the end). Simulation seems to bear this out, though I can't right now figure out a complete proof. $\endgroup$ – ShreevatsaR Jan 7 '14 at 17:42
  • $\begingroup$ It's certainly possible to win, for example with a deck configuration with 11 hearts followed by 1 spade, two hearts, 12 spades, and then similarly with clubs and diamonds. $\endgroup$ – Jonas Granholm Apr 9 '14 at 16:18
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I don't see any way to do this mathematically. Here's code that computes the winning probability using dynamic programming and checks the result with a simulation. You didn't specify what happens when the deck runs out and you can't fill the board; I assumed that in this case you nevertheless remove cards of identical suits in the partially filled board and win if you empty it. The result for the winning probability is

$$ \frac{4611922675628644134029}{721031421579150441387600}\approx0.0063963\;. $$

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