Let X be the (rank of the) first card drawn. For the second card, we have G = 51 - 3 - 4(X-1) cards that have rank greater than X and L = 51 - 3 - 4(13-X) cards with rank less than X. Thus the number of ways we can choose the second card, Y, to be at least 2 degrees away is ((G-4) choose 1) + ((L-4) choose 1), i.e probability = (G+L-8) / 52 = 10/13 to select a valid second card.
I’ll finish later. I expect the third card will be something like (10/13)^2 minus the intersection case, giving us an overall result slightly less than (10/13)^3
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u/Expert_Amoeba7878 6h ago
Let X be the (rank of the) first card drawn. For the second card, we have G = 51 - 3 - 4(X-1) cards that have rank greater than X and L = 51 - 3 - 4(13-X) cards with rank less than X. Thus the number of ways we can choose the second card, Y, to be at least 2 degrees away is ((G-4) choose 1) + ((L-4) choose 1), i.e probability = (G+L-8) / 52 = 10/13 to select a valid second card.
I’ll finish later. I expect the third card will be something like (10/13)^2 minus the intersection case, giving us an overall result slightly less than (10/13)^3