顺序对

如果 Sam 再抽出两张与他刚刚抽出的牌相同的牌，则有 $2\choose2$ 方法可以做到这一点。

如果他抓到两张相同点数但与上一张不同的牌，则有 $4\choose2$ 方法可以实现此目的。队伍中，$12$和他上次拉的不一样。

假设牌组中剩余牌张为 $50$，且 Sam 将抽出更多 $2$，则总共存在 $50\choose2$ 可能的抽牌。

所以我们得到： $$\frac{1 + 12{4\choose2}}{50\choose2} = \boxed{\frac{73}{1225}}$$

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Original Explanation

If Sam pulls two more cards of the same rank that he just pulled, there are $2\choose2$ ways to do this.

If he draws two cards of the same rank but different from his last pair, there are $4\choose2$ ways of accomplishing this. Of the ranks, $12$ of them are different from his last pull.

Given that there are $50$ remaining cards in the deck and Sam will pull $2$ more, there exist a total of $50\choose2$ possible drawings.

So we get:

$$\frac{1 + 12{4\choose2}}{50\choose2} = \boxed{\frac{73}{1225}}$$