[bigpooch]

It seems that the logic you present is correct. In the case
of two pairs, you would guess that the ratio is greater than
47/45 because of the bunching effect (if you hold AAB, there
are 3 Bs in the remaining 47 cards, whereas if you drew to
AA, discarding xyz, you may pick up an x/y/z as your first
card). In the case of trips, your point that the chances
are quite small is important so your logic is sound.

The calculations (ignoring other cards outstanding) are
rudimentary as a check:

(by the way, note that in practice, these odds underestimate
your real chances of improvement on average: do yo see why?)

Suppose you start with AAxyz.

I) Drawing three

For drawing three cards to AA, there are C(47,3)=
47x46x45/(3x2x1)= 16 215 combinations.

To make exactly trips, you need one of two aces and two
other cards. There are 9 non-(x,y,z) ranks and 3 ranks of
(x,y or z) so that there are C(9,2)x4x4+9x4x3x3+C(3,2)x3x3
possible combinations yielding altogether 927 combinations.
Thus, the chances of getting trips exactly is 2x927/16 215
or about 0.114339.

For two pairs, the calculation is easier. There are
9xC(4,2)x(44-3)+3xC(3,2)x(44-2)=2592 combinations and thus,
the chances of getting exactly two pair is about 0.159852.

II) Drawing two to AAx (discarding y and z)

There are C(47,2)=1081 combinations of draws.

Trips:
2x(44-2)=84 combinations of getting trips exactly yields an
approximate chance of 84/1081= 0.077706.

Two pairs:
3x(44-2)+9x6+2x3=186 combinations yields an approximate
chance of 0.172063.

Now comparing the ratios (drawing two/drawing three):

Two pairs:

(186/1081)/(2592/16215) = 155/144 = 1.07639
whereas 47/45 = 1.04444

Trips:

(84/1081)/(1854/16215)= 70/103 = 0.67961
whereas 2/3 = 0.66667

Overall, your reasoning will give a good approximation but
you need to consider the bunching effect in the two pair
case.

Unfortunately, if this was really in this book, I think
that quite a few 2+2ers could write a much better one! [img]/images/graemlins/smile.gif[/img]