Discussion
A deck of playing cards contains 52 cards, 4 each with face values in the set {A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K}. Let us define face values of the top and bottom cards as X and Y. The shuffling algorithm is defined as follows :
Sh1 : Permute the cards randomly due to that each arrangement occurs with probability 1/52 !
Sh2 : If \(X \ne Y\), flip the biased coin that comes up heads with probability P and go back to Sh1 if heads turns up otherwise stop.
Each coin flip and each permutation is assumed to be independent of all the other randomization. What value of P will make X and Y independent random variables once this process of shuffling will stops ?
Answer: A
The required value is P = 1/4.
After Sh1, a specific pair of face values for (top, bottom) has probability depending on whether the values match. The process keeps a matching pair at once, but keeps a non-matching pair only when the coin shows tails, with probability 1 - P. So the final probabilities are proportional to
\[ \begin{aligned} (x, x) &: \tfrac{4}{52} \cdot \tfrac{3}{51} = \tfrac{12}{2652} \\ (x, y) &: \tfrac{4}{52} \cdot \tfrac{4}{51}(1 - P) \\ &= \tfrac{16(1-P)}{2652} \end{aligned} \]
X and Y are independent, each uniform over 13 values, exactly when every one of the 169 pairs is equally likely:
\[ \begin{aligned} 12 &= 16(1 - P) \\ 1 - P &= \tfrac{3}{4} \\ P &= \tfrac{1}{4} \end{aligned} \]
With P = 1/3, 2/3 or 1/5 the matching pairs would be over- or under-weighted, so X and Y would stay dependent.
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