Odds calculations in "Encyclopedia of Draw Poker", by Anno

[BadPenguin]

I picked up a copy of this book a little while ago. It has lots of odds calculations about various starting hands and draw results, including an analysis of the effects of keeping a kicker when drawing to a pair. When drawing three to a pair he gives the probability of improving to a given better hand as approximately:

two-pair = 1/6
trips = 1/9

These seem right to me, but he then looks at keeping a kicker and drawing two and says the chances in that case change to:

two-pair = 1/12
trips = 1/26

Neither of these seems right to me, so I though I'd give my reasoning and let some of the math guys (of which I used to be one in the distant past) tell me if I'm right or wrong.

Let's write the two possible situations (draw-3, draw-2) like this:

AAxxx
AABxx

In the two-pair case, it seems intuitive to me that your chances are no worse if you keep a kicker than if you draw three. My reasoning is that in the draw-three case, what does the first card you draw have to be to keep alive your chances of drawing exactly to 2-pair? All it has to do is not be an A, the chances of which are 45/47 (2 A's remaining, 47 cards left after your 2 A's and the three discards). Now you have AACxx, and from this point you're in exactly the same situation as if you'd kept the kicker, having to draw one more C in the last two cards. So it seems to me that the chances of getting to exactly two-pair by drawing two are very slightly greater than by drawing three, by a ratio of 47/45.

In the case of improving to exactly trips you need to draw exactly one A in each case. When you have a few events whose probabilities are very small you can get a decent approximation of the chance of hitting one by adding the probabilities up. Since this case fits that condition (and ignoring the very small probabilities of drawing to a full house of quads) the chance of getting trips when drawing two should be pretty close to 2/3 the chance of getting it by drawing three, not less than half as good as the book states that it is.

Can anyone confirm (or refute) my reasoning here?

[Al Mirpuri]

Dude you have to post this in the Probability forum. They love this sort of stuff.

Question: how good is the overall advice in Anno's book? How is the draw being played eg with/out a joker?

[BadPenguin]

[ QUOTE ]
Dude you have to post this in the Probability forum. They love this sort of stuff.

Question: how good is the overall advice in Anno's book? How is the draw being played eg with/out a joker?

[/ QUOTE ]

Thanks for the pointer to the Probability forum - I hadn't noticed it before. I will post this there.

The book seems pretty good, though I haven't gotten through the meatiest chapters on strategy toward the end. Lots of probability calculations (a couple of which I unsure of, as I mentioned). The draw he talks about is with antes, no joker (he makes some disparaging remarks about jokers in the beginning), no minimum openers. He talks mostly about no-limit, though since he always talks in terms of pot odds and probabilities rather than betting strategy per se you could adapt it to limit easily enough.

[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]