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The goal is to (dis)prove a concept by comparing permutations of decks of cards.

A deck contains a number of blue cards(b) and red cards(r). These can be any permutation of that amount of b and r. If it's a 3 card deck with two blue cards and one red, the possibilities would be:

bbr, brb, rbb

For phase 1, the fact that "b1" and "b2" could be in either position is irrelevant. Those permutations are the same, by comparison.

Therefore,

  • A 9 card deck with 4 blue cards and 5 red cards would have outcome X.
  • A 10 card deck with 4 blue cards and 6 red cards would have outcome Y.
  • If you start with the above 4b/6r deck and randomly remove 1 red card, you will have outcome Z.

The possible permutations of outcomes X and Y are, of course, different. The outcomes of X and Z should be equivalent. (I believe.)

Phase 2 would now consider the probabilities of each permutation. So, for outcomes X and Z, we want to know if the odds of each permutation are the same between X and Z. (I believe they are, again.)

My bit of searching on my own has led me to believe I want to be using itertools, but I am having trouble putting that into the context that I need it for this exercise.

jonrsharpe
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Zovc
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    I believe that what you are after are not really permutations, but combinations – saiden Dec 06 '21 at 10:35
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    See [Combination](https://en.wikipedia.org/wiki/Combination) at Wikipedia. – trincot Dec 06 '21 at 10:48
  • I could be wrong, but my understanding is that the order does not matter with combinations, but it does with permutations. Based on that interpretation, it seems to me like I'm in the middle of those two things. – Zovc Dec 06 '21 at 11:59
  • Didn't mean to send that, was going for a line break, sorry. rbr, rrb, brr are the combinations, I suppose? r1 b r2, r2 b r1, etc... would be permutations, then? (Two different 'rbr's in that case.) – Zovc Dec 06 '21 at 12:00
  • So for some colours the order matter and for some others, it doesn't matter? – saiden Dec 06 '21 at 14:18
  • The first part of the proof, 'phase 1' is demonstrating that outcome X and Z have the same possible combinations. The second part of the proof, 'pase 2' is determing whether the probabilities of outcomes X and Z are the same. We already know how many blue and red cards are in the deck. We are concerned about where they are in the deck, and what the odds of drawing a red or blue card are based on the possible combinations. – Zovc Dec 06 '21 at 14:23
  • To rephrase, if it helps: The fact that with 2 red cards and one blue card, "rrb, rbr, brr" (3) are possibilities matters. The fact that "r1, r2, b" AND "r2, r1, b" (6?) are possibilities does not matter for the first part of the proof. I suppose it matters to some extent for the second part, if there actually is a difference between starting with 4b & 6r and removing a red randomly versus starting with 4b & 5r. – Zovc Dec 06 '21 at 14:45

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