LogFAQs > #938645152

Number of ways to draw 4 cards without getting a pair: (24*2)*(23*2)*(22*2)*(21*2)=4080384

Number of ways to draw 4 cards without any restrictions: 48*47*46*45=4669920

Therefore the odds that any particular person draws a hand without a pair is (4080384/4669920) or about 87.4%. (equivalent to xp's answer)

Note that player A obtaining a pair and player B obtaining a pair are not independent events since they're drawing uniquely from the same pool of cards, so you can't straightforwardly do (0.874)^12 to exactly get the probability of no player obtaining a pair. (tyder's reduced example is a clear demonstration of this.) Calculating this exactly would require more advanced combinatorics that I'm not capable of sorting through on the fly. If you want to use xp's method as an estimate, I don't have a great feel for how accurate it would be. My intuition says it would be a good estimate for the case of large numbers of players, but I don't know exactly how many you would need to meet that criteria, and I'm not 100% sure my intuition is correct here anyways.
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Oh woops. Putting Advokaiser in my sig like this until I think of something more clever