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 ?
\(P = 1/4\)
For X and Y to be independent at the end, every ordered pair of face values (a, b) must finally occur with probability \(\frac{1}{13} \cdot \frac{1}{13} = \frac{1}{169}\), since each face value appears equally often in the deck and the marginals are uniform by symmetry.
In one random permutation, the probability that the top and bottom cards are a particular equal pair (a, a) is \(\frac{4 \times 3}{52 \times 51} = \frac{12}{2652}\), and for a particular unequal pair (a, b) it is \(\frac{4 \times 4}{52 \times 51} = \frac{16}{2652}\).
If X = Y the process always stops. If \(X \ne Y\) it stops only when the coin shows tails, with probability \(1 - P\); otherwise the deck is reshuffled and the trial repeats independently.
So the final probability of each pair is proportional to its stopping weight: \(12\) for an equal pair and \(16(1 - P)\) for an unequal pair.
All pairs become equally likely, and hence X and Y independent, exactly when \(12 = 16(1 - P)\), i.e. \(1 - P = \frac{3}{4}\), which gives \(P = \frac{1}{4}\).
Check: with \(P = \frac{1}{4}\) the unequal weight is \(16 \times \frac{3}{4} = 12\), equal to the equal-pair weight, so all 169 pairs share probability \(\frac{1}{169}\), and \(P(X = a, Y = b) = P(X = a) \, P(Y = b)\) holds.
Hence, \(P = 1/4\).
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