[BruceZ]

[ QUOTE ]
Hmm.... possibly the best way is just to count all the possible combinations including suits.

We know he holds the 2 of diamonds. The other potential cards he could hold: three 5s, four 4s, three 3s, two As.

A3 x6 - no
A4 x8 - no
A5 x6 - yes
34 x12 - no
35 x9 - yes
45 x12 - yes

That's 27 hands where the 5 pairs him, and 26 hands where it doesn't. 27/53 = there's a 50.94% chance he got paired on 4th street.

[/ QUOTE ]

Right except that there is only 1 ace unaccounted for. The possibilities then become:

A3 x3 - no
A4 x4 - no
A5 x3 - yes
34 x12 - no
35 x9 - yes
45 x12 - yes

This gives a probability of pairing the five of 24/43 =~ 55.8%.