The point is that math whizzes who strive to be among the top one percent of poker players are at least ten times more likely to succeed then non math whizzes who strive just as hard.

Brushing aside the arbitrary figures I guess I can agree with this.

It's slightly misleading though in a significant way. Top 1% is really not a very significant designation in this context because over, perhaps well over, 90% of people who play poker with some frequency are losing players.

If you meant top 1% of legitimate pros, I don't agree with you. I think the skillset for being a really really good pro is a bit mysterious, but does not seem to involve being extremely good at math or even extremely intelligent in a more general sense.

If you had chosen chess I would be more inclined to agree with you (although I would just be guessing), because chess ability does seem to benefit more from pure analytical intelligence.

Poker seems more dependent on mundane skills like pattern recognition, concentration, emotional discipline, and very occasionally creativity. And emotional stability and competitiveness.


You may be assuming I am in the "poker is not about math, it's about reads!!11" school. I'm not. I'm just approaching this empirically. None of the really top players appear to be close to potential math whizzes, likewise none of the small number of legit math whizzes I know of who play poker seriously is close to being a really top player.

Social intelligence = +/- ability to do math (probability, prediction) on people.

I do agree on the level that if you can't do math on the cards, you will fail despite social intelligence.

I disagree that the math on the cards is beyond the potential of your average dumbass.

The problem is that social intelligence involves much more complex mathmatical relationships + perception of others + some other not other specified stuff.

there is a big demand for applied math, ie people who are good at mathematically modeling complex systems

A math whiz regarding poker is similar to a decathalon champ regarding football.

This is a much better analogy than I think you realize.

Elite football like poker requires a very specialized skill set (and build). The current US hopeful for Olympic lifting (in a certain sense the strongest person in the country) went into O-lifting because he was a mediocre collegiate-caliber football player.

So the Strongman didn't make it to the NFL. So what?

This is one of the more interesting threads in this forum, and we're all pretty lucky that jason1990 has been willing to devote so much time to it.

These recent comments *might* make for an interesting thread on their own, but they don't seem particularly on topic for this thread, and I'm a little worried that too many of them will cause jason1990 to lose interest in this thread.

That was just one example (and not of a strongman, that has a different connotation).

I don't know how to illustrate the point that football has a very specialized (innate) skillset apart from raw strength or power but I have heard this from people who are authorities on athletics.

So take my word for it or not. Whatever.

David's analogy was meant to imply that a math whiz would have a propensity to be good at many things including poker, whereas a poker whiz would only be good at poker. My point was that just as a decathlete would be mediocre at football, so would a typical math whiz be mediocre at poker.

The notation P(A | J) translates to "the probability of A given J". So no, the word "given" is not part of J.

The problem is that J is not a proposition. A proposition is a declarative sentence with a truth value.

I think the first great difficulty we encounter is not in answering your original question, but in asking it. The meaning of your original question may seem intuitively clear (to some people), but I think we will find that the attempt to translate this intuition into a precise and logically coherent question will be fraught with difficulty.

Ultimately, it is your question and your intuition, so you must be the one to provide the proposition J. But I will make a somewhat arbitrary suggestion to motivate your deliberations.

Mimicking the ideas described in this post (which were copied from this paper), we might suggest

J = the conjunction of all true propositions which either (a) describe the state of the world at 06-16-2010 10:25 PM GMT, or (b) describe a law of nature.

Of course, there are still difficult issues with this choice. It requires us, for example, to make sense of what we mean by "state of the world" and "law of nature". But let us imagine for now that all such issues are resolved, so that we may explore what might happen when we try to determine P(A | J). There are three possibilities.

(1) If J logically implies A, then P(A | J) = 1.

(2) If J does not logically imply A, then the problem of determining P(A | J) may be well-posed or not well-posed. (For more on well-posedness, see this thread.)

(2a) It is well-posed if the relationship between A and J exhibits a sufficient degree of symmetry. If it is well-posed, then it is in principle possible to determine an objective value for P(A | J).

(2b) If it is not well-posed, then P(A | J) cannot be objectively determined.

Personally, in the absence of evidence to the contrary, my default assumption would be that (2b) is the case.

There are many probability experts who do not even know what LHE is, so they are certainly not all fantastic players. However, I would guess that most probability experts would refrain from any kind of gambling, unless they were confident that it was profitable. So a probability expert who was a poor LHE player would probably choose not to play, in which case you would never see them.

Here is my suggestion to Vael, who asked a similar question. You may want to follow that recommendation, and combine your reading with active participation in discussions in the probability forum.

Other people can and have answered this question better than I could. So I will respond with a collection of links:

http://www.toroidalsnark.net/mathcareers.html
http://online.wsj.com/article/SB123119236117055127.html
http://www.maa.org/careers/
http://www.math.ucdavis.edu/~kouba/MathJobs.html
http://www.math.duke.edu/major/whyMajor.html
http://www.bls.gov/oco/ocos043.htm

I would also like to point out that "mathematician" is a career. According to the last link above,

It happens occasionally that sub-threads develop. It never bothers me or makes me lose interest. It does, however, make it more difficult for me to keep track of which questions still require a response.

I'm curious what your take on this problem is. Somebody posted about it a while ago in probability and I never figured out a good resolution to it.

I get your point on my point being obviously leading.

Not to make a fine point on it*, but do you have a lead attached to your point?

*TMBG lyrics ITT

Cool thread.

I'm a senior undergrad in Applied Math/Economics, probably entering into an MSc of Applied Math next year. Any general advice?

Thanks.

I assume you are referring to this thread. There, the problem is described somewhat differently. Mathematically, it is the same, but instead of prisoners living or dying, there are teams of players scoring points. I prefer this formulation, because I think it makes the counterintuitive elements of the problem more transparent.

I see there being two distinct counterintuitive elements here. One has to do with decision theory, the other with probability theory.

Regarding decision theory, the problem presents an example of a well-known situation where, if each player optimizes his or her individual behavior, then the team's performance will nonetheless be suboptimal. While this is interesting, it is not particularly novel. After all, the original prisoner's paradox makes exactly this point, but presents it in much simpler packaging.

Regarding probability theory, the problem presents a situation which seems to defy some basic intuition we have about the relation between probabilities of individual events and long run frequencies. Intuitively, the question is, how can each player on the team of probabilists have a 50% chance of earning a point, and yet the team, as a whole, only earns 48% (or 47.9%) of the possible points in the long run?

Let us begin by identifying the probabilistic principle that this situation appears to violate. Suppose we have a sequence of events A₁, A₂, A₃, ..., and suppose these events are exchangeable. (Exchangeability can be viewed as a generalization of independence.) Let S_n be the proportion of A₁, ..., A_n that occur. It can be proven that:

(This theorem can be viewed as a generalization of the law of large numbers.)

And yet, in this problem, we seem to have a situation that violates this. Let A_n the event that the n-th player selects the white stone. It seems that P(A_n) = 0.5 for all n. And yet, P(S_n → 0.48) = 0.5 and P(S_n → 0.479) = 0.5, so that E[lim S_n] = 0.4795. How can this be?

As with so many problems like this, I think the confusion arises because of our failure to explicitly account for the information on which the probabilities are based. Since all probabilities are conditional, the above theorem should technically read:

Now let Ω be the information I have after drawing 479 white stones and 520 black stones from the urn. If I am the first player, then it is true that P(A₁ | Ω) = 0.5. However, for every n > 1, we have P(A_n | Ω) = 0.4795.

All the other players are exchangeable with respect to my information. But there is an informational asymmetry between myself and them. I know that I first drew 479 white stones and 520 black stones. I do not know what the first 999 draws are for any other player.

If we look only at players who first drew 479 white stones and 520 black stones, then, based on my information, each one individually has a 50% chance of drawing the white stone. Moreover, for this subpopulation, P(S_n → 1 | Ω) = 0.5 and P(S_n → 0 | Ω) = 0.5, so that E[lim S_n | Ω] = 0.5, just as it ought to be.

I think what throws people off is that instead of thinking about P(A_n | Ω), they (implicitly) think about P(A_n | Ω_n), where Ω_n is the information the n-th player has after drawing 999 stones. In that case, it is certainly true that by symmetry, P(A_n | Ω_n) = 0.5 for all n. But this (as the puzzle demonstrates) does not tell us anything about the team's long term point distribution. If we neglect the role of information in probability, and think only in terms of P(A_n), then we leave the door open to this kind of ambiguity and confusion.

I am a mathematician. My point is lead.

Study probability.

In all seriousness, the role of probabilistic modeling in applied fields seems to be growing rapidly. I think it would be wise for anyone with an interest in applied mathematics to develop a working familiarity with the basics of measure-theoretic probability theory.

Have you read The Black Swan?

If so any particular thoughts on it?

So for example the second player, when he drew 479 whites and 520 blacks, he knows that the first player drew something similar (either the same or 480 white + 519 black), so he has that extra information. At first I was confused by how they were different because the other players don't know what others drew, however it is different because once they draw they know what people before drew (up to a difference of at most 1), letting them figure out how many white ones the first urn has (at least in expected value).

Wouldn't the expected number for the A_n's when n > 1 be a bit different than 479.5? I guess that is a good approximation though. So then the first one takes urn 1 and the rest take urn 2. Would the probabilitists have an edge with this technique? Since the first player has expected value .5 and the rest have .49, whereas the other team all of them have EV .49.

I have not. But I do know at least one seasoned, professional probabilist who has. If you would like to read his thoughts on it, they are here.

I think this is quite a bit different from something like the prisoner's paradox. Yes, in this case, if all the probabilists who see 479 or possibly 480 white balls out of their respective 999 ball peeks were to decide - on some basis - that all of them must be drawing from a bad urn and they could do better as a group if they all switch to the better urn containing 490 white balls, then they will do better as a group. Even though they each have an even better 50% chance of drawing a white ball from the bad urn conditioned only on what each individually knows.

But I think this is problematic when we look at a "basis" on which they might make that decision. Suppose, going into the experiment, they get together and agree to operate in such a way as to maximize thier results as a group. The puzzle is fooling our intuition into leading us to believe that the probabilists can make such a prior agreement so as to improve their group results by sometimes switching urns. Certainly in this case they could improve by switching. But I'm afraid this is fallacious thinking. What happens in general if they make a prior agreement to switch according to some criteria?

I think a natural criteria to look at is, "let's switch urns if we see so few white balls to conclude that we have an urn that's strictly worse than the 490 white ball urn." Certainly, with that criteria they will do better as a group in the puzzle's special case where the first probabilist sees 479 white balls. But how does that criteria work in general? The puzzle is driving our intuition to believe the probailists can improve their group results in general with such a criteria. But that's impossible! That would imply the expected value of the group's average result would be better - in general - than it would be if they all never switched, sticking to their individual 50% chances conditioned on their individual 999 ball peeks.

It's impossible because if they all peek and switch according to the same prior criteria then they are all identically distributed. By additive linearity of EV, that means the EV of their group average result equals the EV of their individual results (using the indicator that a white ball is chosen). But the EV of an individual who switches to a 490 ball urn based on a prior criteria for his 999 ball peek must be worse than 50%. In those cases where he switches he reduces his white ball chances from 50% to 49%. His individual EV is

(1) Pr(criteria not met)*(.5) + Pr(criteria met)*(.49) < .5

and so this must also be the EV for the group average result. It's impossible for them to agree to a prior criteria that improves their group average result in general.

So the question is, with a prior switching criteria where is the EV lost? Certainly not in the puzzle's special case where 1st peek sees 479 white balls - assuming the criteria has them all switching in that case. So where is the EV lost?

The natural criteria described above would have a probabilist switch if he sees 488 white balls or less. He could then conclude the urn has no more than 489 white balls and is a strictly worse urn than the 490 white ball urn. With this criteria, if the probabilists' urn has 490 or more white balls their group average results will be the same as if they never switch. If their urn contains 488 or fewer white balls then they will all switch to the 490 white ball urn and their group average will be better than if they never switched. So the only place where the EV can be lost with this switching criteria is if their urn contains exactly 489 white balls.

If we don't look closely at the 489 white ball case I think our intuition can fool us. There's a relatively small chance this case will occur and whatever EV might be lost here seems like it must be small compared to the gains in EV for all those cases where there are 488 or fewer white balls. Furthermore, the EV lost in the relatively unlikely 489 case must strictly outweigh the EV gained by switching in the much more likely 488 or fewer case. It must outweigh it by enough to validate the strict inequality in (1) above.

So what happens when the probabilist urn contains exactly 489 white balls? What happens is that 48.9% of the probabilists will see 488 white balls in their 999 ball peek and switch to the 490 white ball urn where they will pick a white ball 49% of the time. However, 51.1% of the probabilists will peek at 489 white balls and by the criteria will not switch urns. All of these 51.1% of the probabilists will chose the last black ball in the urn. In the 489 white ball urn case, the EV for the group average - and for prepeek individual probabilists - will be (0.489)(.49) = 0.23861 . That's a huge reduction from the .489 EV for never switching in this case. It's huge under this criteria, for example, compared to the EV improvement from 0.48 to 0.49 in the case of a 480 white ball probabilists' urn.

Unless I'm mistaken, the above discussion shows the following strict inequality to hold:


(2) Sum{k=1...488} [ C(1000,k)(.5)^1000 * (.49 - k/1000) ] <

< 0.489 - (0.489 * 0.49)

The left hand side calculates the improvement in EV under this prior criteria for urns with 488 or fewer white balls. The right hand side calculates the loss of EV in the 489 case where 48.9% of the probabilists switch under this criteria instead of getting their certain white ball had they not switched.

I don't see any easy way to show inequality (2) algebraicly. But I believe the probability discussion above proves it must hold. I'd be interested to see if anyone can show it algebraicly. I think a similair inequality could be easily shown algebraicly if the setup where a little different. Let the second urn be a fair one with 500 white balls. And let the probabilist prior criteria be to switch if his 999 ball peek shows 498 or fewer white balls.

So I don't think it's the case like in the prisoner's paradox that the probabilists are eschewing better group average results in favor of better individual decisions. Any prior criteria based on information in the 999 ball peeks that sometimes switches to the 490 white ball urn will in general produce the same EV for individuals as for the group average, and will be inferior to simply never switching. This even though it can improve both individual and group average EV for some probabilist urn composititions. By chance, the special case described in the puzzle happens to be one of them.

That is, it's one of them if 488 is the criteria cutoff number. If the probabilists switch on 479 or fewer white balls and the urn happens to have 480 white balls then probabilists like the one in the puzzle who see 479 white balls in their peek will drastically hurt their individual EV by following the criteria and at the same time will hurt the group average result. In this case the group average would be about half what it would be for probabilists who never switch and the drop in EV would be precisely due to probabilists like the one in the puzzle who peek at 479 white balls and switch "to help the group".


PairTheBoard

I don't know about live but that is way off for online players. All the large databases show over 30% of online players are net winners when you use some small minimum number of hands played as a filter (just a few thousand). That number gets higher with stakes too, where at the highest stakes as many as half are net winners, since they won a lot at lower stakes before moving up.

I was referring to live actually. Zero of the best online players are math whizzes.