BadPenguin
04-25-2005, 12:54 PM
I originally posted this in the "Other Poker" forum because it deals with draw, but someone suggested posting it here as well, so here it is...
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?
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?