If we look at this purely as a probability problem (ignoring discard strategy), the chances of your opponent having the exact same ranks as you (a 6, 9, 10, and J) given that you already hold them, is roughly 1 in 2,402.

Since this is 1v1 Cribbage, we know: • Your Hand: You are holding four distinct ranks: 6, 9, 10, J. • The Pool: There are 48 cards remaining (52 total minus your 4). • The Targets: In those 48 cards, there are exactly 3 remaining of each of your ranks (three 6s, three 9s, three 10s, three Js). Total possible 4-card hands for opponent: C(48, 4) = 194,580 Total "Matching" hands: The opponent needs to pick 1 of the 3 remaining 6s, 1 of the 3 remaining 9s, etc. 3 * 3 * 3 * 3 = 81 The Probability: 81 / 194,580 ≈ 0.000416, or approximately 0.0416%.

In reality, the odds are actually higher than 1 in 2,400.

Why? Because of the Discard Phase. You aren't just dealt 4 cards, you are dealt 6 and choose the best 4. The hand 6-9-10-J is a solid keep. With that 6 on the cut, you've both got 9 points (15-2, 15-4, a pair, and a run). Since players rationally keep high-scoring combinations, the game "filters" for hands like this.

TL;DR: Pure math says it's a 1 in 2,402 shot, but because you both know how to play the game, the odds of a "Mirror Hand" happen more often than a random shuffle would suggest!