mandelmonk

It's hype. Yes, like other junk food, it will keep you alive, but alive doesn't mean healthy. It lacks phytonutrients. It only has one vitamer of each vitamin. The nutrients it has won't be absorbed the same way. Bottom line is it's not a substitute for a healthy balanced diet. Nutrition is much more than an ingredients list with lots of vitamins (e.g. all the cereals with overly impressive "nutrition facts" are junk). Science is only scratching the surface of nutrition, and Soylent certainly isn't the state of the art regardless.

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

Not that I know of. Intuitively, given that there's life in our N=1 sample, one naturally assumes that the probably of a planet having life isn't something crazy like 1 / 10^50. However, our sample has total selection bias. As lifeforms, it's not like our sample is a random planet; our planet can only be one with life. So the fact that Earth has life means almost nothing.

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.

Edit #2 -- omg careless mistakes everywhere (1b, 1c, 1d). Fixed.
Edit -- Ok, fixed #1a and #2a now that I know that "badugi" means "rainbow".

For each question I'll break it down into parts a, b, c, d, e.

1a) Pr(rainbow) = 13^4 / C(52,4) ≈ 10.55%

1b) Pr(2:2) = C(4,2) * C(13,2)^2 / C(52,4) ≈ 13.48%
(edit: forgot to square the 78 before)

1c) Pr(2:1:1) = 4*3 * C(13,2) * 13^2 / C(52,4) ≈ 58.43%
(Edit -- 4 choices for which 3 suits are involved, but then also 3 choices for which suit is the double.)

1d) Pr(3:1) = 4*3 * C(13,3) * 13 / C(52,4) ≈ 16.48%
(Edit -- 4 permute 2 not C(4,2) because 2 choices for which suit is the triple.)

1e) Pr(monotone) = 4 * C(13,4) / C(52,4) ≈ 1.056%

2a) Pr(rainbow | no pairs) = Pr(both true) / Pr(no pairs)
N(both true) = 13*12*11*10 = 17160

Explanation: N(something) means "number of ways for something to happen". There are 13 possible ranks from an arbitrary "first" suit. Then since we don't want pairs, we only pick from 12 ranks when picking from the next suit. And so on.
It's not C(13,4) (picking 4 ranks with order not mattering), because order of suits matters -- it makes a difference which rank is which suit.

N(no pairs) = 52*48*44*40 / 4! = C(13,4) * 4^4 = 183040
answer = 17160 / 183040 = 3/32 or 9.375%

2b) Pr(2:2 | no pairs) = Pr(both true) / Pr(no pairs) = N(both true) / N(no pairs)

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%

My point about the 100% is that, of the 2/3 times you chose the goat and switched, you switch to the car 100% of the time because Monty is forced to reveal the other goat, leaving only one possibility (car) for the 3rd door. And like you said, it's still a 2/3 chance that you chose a goat because Monty will always reveal a goat no matter what you initially picked. Therefore P(switch to car) = P(chose goat initially) = 2/3

Re: "I'd be none the wiser". The whole point is, new information changes the probability. In an absolute sense, when you flip a coin, given the deterministic laws of physics, there's either a 100% chance it lands Tails or 0%, and there's really no such thing as probability. Probability only exists due to our human ignorance (and in that sense is subjective). However, randomness caused by ignorance is just as effective as hypothetical "true" randomness (the coin acting non-deterministically), therefore we have to resort to probability. And so, information and probability are closely related. If, when playing the game show, you ignore the new information then yes, in your brain it will be 50/50. Just like in poker, if you ignore the fact that say an Ace was exposed, then to you the chance of an Ace is still 4/52. But in both cases, you or someone could have used the information available to adjust the probability. So when we say "the" probability is 2/3, we mean the "most informed" probability (the one that factors in the new information), whereas 50/50 would be the "willfully ignorant" probability. All probability is ignorant, but we always discard the "optional ignorance" because the answers that result from it aren't as useful.

Hopefully by getting all philosophical, I've driven the point home. But in addition, monty hall is a problem that can be experimented in real life and simulated on the computer. Experiment/simulation will agree that the answer is 2/3, so this isn't just something in the abstract.

I see you mentioned binary choice again. Tom gave a good soccer example but I'll give another example and go a little deeper into it. Say you buy a lottery ticket. You'll either win the jackpot or you won't, that's 2 possible outcomes, but clearly you don't have a 50/50 chance of winning the jackpot. What it comes down to is, each outcome can be broken down into sub-outcomes (ways of happening) and for the lotto there are many more losing sub-outcomes (in this case, combinations) than winning ones. Same with any non-50/50 binary choice, one side has more
ways to happen (at least, in a deterministic universe).

Couple things you said I don't understand:

- What is the flaw in probability theory that you speak of?

- What do you mean by, "how can anything 50/50 have an outcome"?

Mr Pompos, the feeling you describe of seeing 2 lines of reasoning that seem to make equal sense (despite leading to different answers), I used to get that a lot and it's part of what fascinated me about probability (probability seemed to do that to me more than anything). My advice to you is, never can both lines of reasoning be correct simultaneously, so you just have to ponder it until you see the flaw in one of them. Think about it on the road, in the shower, when laying in bed. There is always a fallacy somewhere, otherwise both answers would be correct and all of mathematics and logic would fall apart (unless you're solving a quadratic in which case there are two answers!).

Here's another take on monty hall. 2/3 of the time, you picked a goat and by switching you'd have a 100% chance of getting the car (the key is, the host isn't revealing randomly, he's always revealing the goat). The other 1/3 of the time, you picked the car and by switching you'd have a 0% chance. So by switching, overall you have a 2/3 + 0 = 2/3 chance

What exactly is your thought process behind the answer of 50/50? (Assuming it's not the "binary choice" argument which Tom Coldwell already responded to.)

Now that I have a good amount of probability under my belt, what I enjoy the most when helping someone else is hearing a line of reasoning I haven't heard / thought of before. Even if it's wrong, sometimes I'm temporarily stumped as to why, and I love when that happens! That's the kind of stuff that expands the mind. Therefore I always like to know what someone was thinking.

He really was trash in that series last year, just a turnover machine. This year he's playing lights out.

Imo AV's are useless. A negative doesn't tell you something is clean (even if 40 different AV's all agree it's clean), and a positive can be false. Just have to take preventative measures like using open-source software whenever possible, and sandboxing your browser. For software I don't trust but want to install (i.e., Skype), I only do so in a virtual machine. (HEM is probably one I'd trust though. But I play on Bovada.)

Spurs gonna win it all. Like they should have last year.

Spurs/Heat prediction looking pretty good atm.

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.

The actual results approach the EV as the number of trials increases. If you plot deviation from EV vs time, the graph zigzags above and below the x-axis forever, but the zigs and zags begin tall and get flatter and flatter.
The number of trials required depends on how small you want the margin of error to be, and on the desired level of confidence for that interval. Khan Academy explains that stuff much better than I can. You can also get a feel for the time by playing around with the binomial distribution.

Turbulence, weather, white noise, human lungs, various plant life, lightning bolts. And then basically anything that isn't a perfect smooth shape (e.g. the aforementioned UK coast) can be approximated by fractals, which is why fractal image compression (pioneered by Michael Barnsley) is effective. Besides image compression / enhancement, you can generate all kinds of graphics with comparative ease using fractals.

Fractals even seem to occur in the stock market. Not to the extent that the movie Pi would have you believe, but there nonetheless.

I wouldn't say the monty hall problem is relevant to poker, but conditional probability in general / Bayes theorem can be.

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.

Thailand is under marshall law now. Maybe not the best place to live.