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After the top card is fixed, precisely three of the remaining fifty-one cards have the same rank. Thus a fresh uniformly shuffled deck has matching probabilityIn each independent round, the algorithm repeats with probability $r=(1-a)p=16p/17$ and stops with probability $1-r$. A match always stops. Summing the probabilities over the possible number of repeated rounds givesFor $0\le p\le 1$ the stopping probability per round is at least $1/17$, so termination is almost sure.

Permutation of the thirteen rank labels preserves both the uniform shuffle and the stopping rule. The two marginal distributions are consequently uniform:To establish independence, it is necessary to check the joint probability distribution, not just these marginals. For a specified rank $x$, the fresh-shuffle diagonal probability is $(1/13)(3/51)=1/221$. For different specified ranks $x,y$, it is $(1/13)(4/51)=4/663$. Reweight by the acceptance probability and sum over repeated rounds to obtainThe diagonal must equal $1/169$ for independence, forcing $17-16p=13$ and thereforeAt this value the off-diagonal expression is also $1/169$, proving sufficiency for every ordered pair. This is an example of match-biased reshuffling preserves uniform rank marginals: state-dependent rejection sampling changes dependence even though both individual rank distributions remain uniform.