Lots of things! They have non-integer dimension. They have infinite detail, a fractal is like an "inner universe" because you can keep zooming forever, always finding new details. Infinite perimeter despite finite length and width. Seemingly unpredictable complexity generated by simple deterministic rules. Accurate approximations of nature.

any out you have you can approximate by 2% and each street gives you one chance of hitting it

AQ x AK AIPF

3 queens help you = 3 x 2% = 6% x 5 streets = 30%

It is pretty rough but valuable in game playing. Implied odds have to take into account a lot of other stuff that I don't see a reason to start to calculate it in game. I rather just do it in my reviews.

Also a math guy here, we may chat on private ? I'm planning on doing some stuff on GT/Bayesian GT for RiO.

Thanks Raphael. In game, implied odds are useful for calulating whether you should call a draw on the turn. You look at your opponents bet size and also if you hit your nut draw then the average amount of money you will get on the river.

I'm pretty sure that it is important but a rough calculation can lead to a lot of mistakes. That's my point here. I think implied odds play better on a whole street game than one street, for example the turn. Your x/c range OTT is more 2nd pair/draw heavy and without a good read on tendencies calculating implied odds can be a waste of time and don't will improve your decision a lot, considering that you need to calculate something that is conditional to a river bet. In HU probably it is more relevant since you have a lot of more stats on river frequencies than usual cash/mtt games. 

Assuming that we don't donk that often OTR after x/c OTT, if we have a fd OTT and have to decide if to x/c or x/f. Adding 2 outs to flush draws and one to open enders are fine in terms of implied odds but again, it is too rough to change your decision or not. 

This guy made a while ago a video/spreadsheet on this but in game I don't like to use it.

https://www.youtube.com/watch?v=4jAlIsrfUJk

https://docs.google.com/previewtemplate?id=0AuG8Qqqo9hkydFJUMHVoMmQ4WlJpMkQ4YUg4RHA3RUE&mode=public

The higher the confidence you want, the more trials you need. The smaller the margin of error you want, the more trials you need. What you end up with is a confidence interval, e.g., "Our polls show, with 95% confidence, that 56 +/- 1 percent of voters favor Candidate A."

You might also be interested in hypothesis testing.

In poker, there's no practical use of knowing "how long the long run is". However, on a somewhat related noted, you might need to know your risk of having a downswing of a certain size, which is calculated using your winrate and variance. This archived post from some other forum (not by me) provides an excellent explanation of that.

As for binomial distribution. Maybe you flipped a coin 100 times and got 60 heads. Assuming the coin is fair, what's the probability of getting at least that many? That's a binomial distribution question. If you change the question to 600 heads out of 1000 flips, the answer will be substantially lower.

I'm only just learning the game theory side of poker myself, so I can't help ya there. Applying game theory to the Turn is indeed more complicated. It's talked about in Janda's book Applications of No-Limit Hold'em, and volume 2 of Tipton's book Expert Heads Up No-Limit Hold'em. I don't know if the answer to your question is found in those books because I still have to make time to read (and digest) them.

We don't know enough about biogenesis to know how probable or improbable it is on a random planet (it could be 1/1000 or 1/10^100 for all we know), and we lack a good sample of planets. If on a given planet it's 1/1000 then yes there are definitely aliens. But I don't think we have much clue what the probability of life on a planet is.

N(both true) = C(4,2) * C(13,2) * C(11,2) = 25740

Explanation: Similar to 2a. There are C(4,2) possibilities for the 2 suits involved. There are C(13,2) possible rank combos from the "first" suit and C(11,2) for the "second". The product of C(13,2)*C(11,2) partially counts order. It counts the 3 ways to put 4 ranks together into groups of 2 (the groups here being the suits), and it counts the 2 ways to order those groups. So it's 6 times as large as C(13,4), which can also be obtained by multiplying C(13,4) * C(4,2).

Answer = 25740 / 183040 = 9/64 or 14.0625%

By "badugi" I mean all 4 cards are different suits

2c) Pr(2:1:1 | no pairs)
N(both true) = 4*3 * C(13,2)*11*10 = 34320
(It doesn't matter where you put the "choose 2"; you can put it 2nd or last e.g. 13*12*C(11,2) equals the same thing.)
answer = 34320 / 183040 = 9/16 = 56.25%

For the rest of the problems, I'll just keep expanding this post.

2d) Pr(3:1 | no pairs) = 4*C(13,3)*30 / 183040 = 18.75%

2e) Pr(mono | no pair) = 4*C(13,4) / 183040 = 1.5625%

3a) Pr(rainbow | 2-pair) = 1/C(4,2) = 1/6

Consider one of the pairs. The other pair can be any combination of suits. The only valid combination is that of both other suits. That's 1 possibility out of C(4,2).

#'s 3b and 3c = 1/6 for same reason

#'s 3d and 3e = 0 because can't have a suited pair

I actually just found this out on my own, but if there is someone still around in this thread, lmk, I have some follow up questions regarding the numbers I found