This is probably very easy but I got no answers when I asked it before.
What's the formula to calculate in how many games, on average, someone would win the iPoker Jackpot given a certain win rate?
Game is sit n Go 6-handed
You have to win 6 consecutive to win jackpot.
Obviously if you win 6 consecutive, the streak is reset, so if you win 7 straight you don't win the jackpot twice.

How has your evaluation of decisions and decision making process changed (of day to day life) since you have started studying probability?

Under some idealized assumptions, the formula is

where p is the probability you win a given game.

I probably have very little day-to-day benefits from having studied advanced mathematical probability theory. The benefits, I think, come from having a job where I think about probability (even basic probability) on an everyday basis. They also come from having spent a lot of time thinking about the deeper meaning of probability in the real world.

One effect is that faulty probabilistic reasoning stands out like a sore thumb to me. I might, for instance, be in the kitchen, talking on the telephone, with the television on in the other room. If someone on TV makes some statement which contains flawed probabilistic thinking, I will probably notice it immediately.

I also think I am a better skeptic for having studied probability. I am more aware of the role of statistics in many branches of modern science, and I am also more aware of how horrible many scientists are at statistics. I am probably much more skeptical of scientific claims than the average layperson.

I also think I am pretty good at communicating probabilistic reasoning in ordinary spoken conversation. By that, I do not mean I am good at explaining math to laypeople. I mean I am good at explaining, in casual conversation, why something is or is not likely without using any math at all.

Are there any specific examples of this that you think show it extremely well?

jason-
when doing analysis for poker equities in a given game, say calculating the equity needed to shove over a 4bet (assume hell call 100%) yields a result in the form of:
E=(1-z)/(2+h) where z=3bet size as % of stack and h= dead money as % of stack).....

my question: is there any difference in how i should be treating a range (in probability terms 'set' if i remember correctly) compared to individual hands?

if i run it and comprise a set of hands w/ E vs a given range it still seems the player is paying a penalty for folding too much and could increase his EV by calling hands w < required equity...if theres not a difference in how to treat a set v the individual then i should create a range with the given equity found as such? correct?

What the probably that one meteor hit the earth and make all the volcano go into huge eruption?

What is the probably that i **** 4 girl in the same day?

This is an easy one.

The last example to really stick in my memory was briefly discussed in this thread. I eventually obtained a copy of the paper. To put it kindly, I remain thoroughly unconvinced.

I am not sure I understand your question. The equity needed to perform a certain action should, in general, not depend on what you know about your opponent's cards. The equity you actually have, on the other hand, will depend very much on that. I suspect you will have better luck generating a discussion of your question in Poker Theory.

Tough questions. My initial thoughts are summarized here.

no help there.

the basic questions is are ranges to be treated differently than individual hands?

for instance is there a difference between using that equation to interpret it as "this hand needs to have x% equity" vs "my range here needs to have this % equity"?

I am afraid I still do not properly understand your question. You might be able to generate more responses if you post a concrete example, such as a hand history, and begin by asking specific questions about that example. I would still recommend the Poker Theory forum, though. Good luck.

I probably put much more physics terminology into my question than was needed, sorry. In quantum mechanics, if you have a box full of electrons, you actually can't tell the electrons apart from one another and this turns out to greatly limit the type of mathematical functions we can use to describe their behavior. Classically, it is possible to sort of put labels 1 to N on each electron in the box, but the uncertainty principle does not allow this and this is actually a very important distinction in terms of physical properties.

I'm sure it doesn't impact your problem, but it is a very interesting and physically deep application of statistics to physics, based on some very basic math to somebody with your background.

This is interesting and sounds vaguely familiar, but I think I need help understanding. Please forgive and correct any stupid mistakes I make below. It has been a long time since I thought about the details of QM, and I will try to discuss this from memory.

Suppose we have two particles in a one-dimensional box. In this case, the wave function is some f(x,y) ∈ L²(R²). At this stage of the modeling, the particles are labeled, at least in the sense that the "first" particle is associated with the x-variable, and the "second" with the y-variable.

Now, if we measure the positions of the particles, the probability that the first particle is in [a,b] and the second is in [c,d] is

This need not be the same as the probability that the first particle is in [c,d] and the second is in [a,b]. So even at this level, the model is making a labeled distinction between the two particles.

Are you simply saying that if we take a snapshot of the two particles, then it is impossible to tell, with certainty, which of the two particles in the snapshot corresponds to the x-variable in the wave function? Or, to put it another way, if we take two consecutive snapshots, it is impossible to tell, with certainty, which particle in the second snapshot corresponds to which particle in the first.

I am probably misunderstanding what you are saying, but if this is the assertion, then why is this not also a problem in classical models of statistical mechanics? If two heavy particles are performing Brownian motion in an ambient medium and we take two consecutive snapshots, then mathematically (according to the probabilistic model) it is not possible to determine, with certainty, which particle in the second snapshot corresponds to which particle in the first.

The labeled distinctions are fine, the physical requirement is that the resulting states be symmetric (bosons) or anti-symmetric (fermions) under label exchange. Pauli exclusion for fermions is a consequence of this.

EDIT: As Max says, it is possible in principle to distinctly label the classical particles in some way, whereas with quantum ones it isn't. I am unclear whether this is indeed the logical consequence of other aspects of the theory (Max alluding to the uncertainty principle makes sense in that quantum particles won't have the well-defined trajectory that classical particles will, so that even if your classical particles aren't "labeled" you can watch their time evolution and keep track of which one is which, which Heisenberg prevents in the quantum case) or just an empirical axiom about the limits of our understanding in how to interact with matter.

EDIT AGAIN: Actually, it might be interesting if Max says a little bit about the spin-statistics theorem, which I don't know enough QFT to be able to explain, as that will likely say more about the reasons for this.

No problem at all. I have gotten alot out of your posts ITT, so I have no problem in trying to return the favor.


I think you actually get most of the issues with identical particles, but I think you may be confused about exactly what can be measured in quantum mechanics so the way you phrased things might not be the best way to think of it. In QM, you can't really measure the wave function squared, which is (correctly) interpreted as a probability distribution. All you can measure is the location (could be some other observable in QM but this is probably the best for an example) of the particle at some instant. The uncertainty in your measurement has no theoretical upper bound. You can measure its location down to basically a point if its an electron. Now you will get higher and higher uncertainty in momentum but that will only effect its position at times after t, which we don't really care about. Why the wave function squared is interpreted as a probability distribution is that if you take a bunch of identical systems and measure the location of each one individually and plot the results, you will get the wavefunction squared, so it sort of makes sense to think about the particle being stretched out in space and a measurement "collapses" its location to a particular point or small subset of points, but it is not exactly needed for our theory to work. All we really know is after you make a ton of measurements on a bunch of systems, we can predict what the shape will look like.

Now, I am not 100% on what I say next but it makes sense to me. Gumpzilla has a physics background also so if he disagrees I would like to hear. In order to make our probability interpretation of the wave function, we have all ready assumed that it is possible to make a bunch of identical systems which seems to imply that the particles are identical. I am not sure if we can formulate a reasonable interpretation of QM that looks anything like the experimentally confirmed idea I explained previously without assuming that all electrons are identical. And just to say again, identical particles is an assumption that I can only justify retroactively after the theory that we create assuming it gives correct experimental predictions. But to me, it seems like sort of a central idea that gets us from classical to quantum.


Yes, this is correct. it is a common problem in freshman QM to have 2 particles separated by some barrier and you can calculate the chances that the particles each tunnel and switch places next time you make a measurement.

Excellent question. There are a few things that sort of hide this effect and allow classical stat mech to actually exist. One is high energies (which in stat mech is temperature). Physically, you could guess this is true because the De Broglie wavelength goes down as momentum goes up, so basically when things are moving fast enough they are very close to point particles since their wavefunctions are localized. So also heaviness can hide this because heavy particles will have more momentum. It is actually easy to show that the high temperature limits of the quantum statistical functions is just the classical Maxwell Boltzmann distribution, which is a reassuring check.

I purposely didn't mention bosons and fermions in my previous post since i saw that you had mentioned it. It turns out ,somewhat surprisingly when you first see it, that there are actually 2 different ways in which particles can be identical to other particles of the same type. And this in turn has a profound impact on the physical behavior of these particles at any energy scale.

I think I sort of alluded to this previous and I am curious if you disagree or maybe can think of something I forgot about.

The spin statistics theorem is a fully rigorous statement about spin and even or oddness of the wavefunction given the particles are identical. It TECHNICALLY does not require quantum field theory but does require relativistic quantum mechanics so at that point most people just prove it with the language of QFT. What QFT does give is a very natural explanation of the assumption of identical particles in the first place. Since particles are just underlying excitations of the exact same quantum field, it makes sense for them to be identical.

I know that we cannot measure the wave function, but we can calculate it if we know its initial value, right? Speaking of this, perhaps you could answer a related question for me.

I have often read in books or articles something like this: "An experiment is set up in which such-and-such particle initially has such-and-such wave function..." Then they go on to apply the Schrodinger equation and calculate various things. My question is this. How do they set up an experiment like this? How do they know that the particle, after they have it set up, has the wave function they were trying to impose?

i think it is best to think of the wavefunction as sort of a tool that lets us calculate other things. We measure things like position, momentum and spin and we can predict what the EV of these measurements are by applying linear operators to the wavefunction and simplifying some expressions. So its not that we know the value of a wavefunction per say, its that we know a particle has some function that describes its wavefunction. With this, we can calculate things we can actually measure.

I am about as far as you can get from actual experiments in physics and I was a math major as an undergrad so i somewhat embarrassingly have never actually done a college level physics experiment, so Gumpzilla maybe can give better info here. Roughly, wavefunctions are solutions to the Schrodinger equation, so if you take an electron and apply some potential energy function to it, by an electric field for example, you can get calculate what the wavefunction will look like. So an experimentalist could say that these electrons are all in this particular state. From here you can change the potential or make it interact with another particle and you can use different techniques like perturbation theory to figure out how things changed.

My understanding was this. If the wave function at time t = 0 is f(x), then the wave function at time t, denoted by Ψ(x,t), can be found by solving the initial value problem

where H is the Hamiltonian operator. In other words, we need the initial wave function to obtain the wave function at a later time. The Schrodinger equation just tells us how the wave function evolves. In order to calculate something explicit, we need to know what it is evolving from. Is this right? If so, I am just wondering how the experimenters can set things up so as to produce a specific given initial wave function.

Oh I think we were talking a bit past each other as i did not really understand your question. What experimentalists typically do, or atleast what was done in the things you are reading, is solve the time independent Schrodinger equation. This happens when the wave function can be split into a part that depends on space only times a part that depends on time only, which is the standard technique of looking for separable solutions in diff EQ. A necessary condition for this to occur is that your potential function cannot be time dependent. So if you take an electron and put it in a non time dependent potential, it is possible to look for solutions of this form. So your solutions will only change phase, which cannot be measured, as you change time.

Once you have them in this state you can get the particles to interact or change the potential etc.

Maybe we are still talking past one another, or perhaps I am just not understanding. I might try to rephrase my question later when I have some time.

Is it the case that when you put the electron in the non time dependent potential the initial value of the wave function, f(x), doesn't matter because under those conditions the wave function evolves to a stationary solution (except for phase) which is the same for all initial values f(x)?


PairTheBoard