The reciprocal rule is a special case of the quotient rule. You could even see all of these as a special case of the product rule.

" Suppose V1=U2 and y1=x2, V2=U3 and y2=x3,..., Vn-1=Un and yn-1=xn . Each U1, U2,…, Un contains one, and only one of the points from x1, x2,…, xn."

This is confusing and possibly nonsense - it's hard to tell. You don't need all those "y's". If you have two distinct points, x1,x2 in a Hausdorff space you can just say, "There exists disjoint open sets U1,U2 with x1 in U1 and x2 in U2 by the Hausdorff (or T2) property." No need to bring y's into it.

You've go the right idea proving it by induction - at least that's how I'd approach it. So set up the induction step carefully. First n=2 follows immediately from the Hausdorff property as above.

Induction Step for n>2.

Assume statement holds for n-1. Show it holds for n.

Holding for n-1 means there exist disjoint open sets U1,...,U(n-1) with xi in Ui for i=1,...,(n-1)

Now Show there exists disjoint open sets Vi, i=1,...,n with xi in Vi.

(Hint) The finite intersection of open sets is an open set.

PairTheBoard

Thanks PTB. I only introduced y because it's in our Hausdorff definition. I can make this happen now.

You could also do it in one fell swoop (without induction) with careful use of double indexing.

PairTheBoard

If we run 50 coinflips for let's say one buyin.
What is the probability that you are up 10 buyins after these?

You risk 1 buy in each time and try it 50.

0.0418591 for exactly 30 vs 20 or 10 up. ie 50!/30!/20!*(1/2)^50

and in case you are interested really in at least 10 then sum from 30 to 50

Sum[50!/n!/(50 - n)!*0.5^50, {n, 30, 50}]= 0.101319

0.0418% or 4.18%?

I ran 40 BI above EV in 150k hands.
Is that super unlikely?
Now I'm paranoid about running below EV and want to study the math behind it

So if you play like 50k hands, what would be the changes to run 10 BI above or below EV?
How likely? Like 10 to 1? 20 to 1?

my numbers are probability so multiply 100x if you want %.

On your 40 buy ins in 150k hands it depends on what your winrate and standard deviation are.

You define 1 buy in as 100bb?

We cannot answer what you ask without knowing your winrate and volatility. We can make guesses or estimate certain things if we know at least your volatility. Is it like 8bb/h for example?

Your avg is like 0.027bb/h that is small enough. If you were a breakeven player (0 win rate) and had a volatility of 8bb/h then the chance to have a result like this or better after 150k hands would have been about 10%. So likely you are a winning player. We can do further study if you want to assume what is the chance you are worse than some level in winrate but we need to know your volatility.

For 50k hands if we assume your avg result as your winrate 4000/150000 and sd 8bb/h i used above (but you could be 6 r 10 who knows plus depends on the tables and number of opponents on avg etc) then in 50000 hands the expected is 13.5 buy ins and the chance to have over 10 is 57.8%.

If you have excel study the function Normdist under help to see how to calculate these things if you know the winrate and sd. In N hands the expected will be winrate * N and the standard deviation of N hands will be like your sd*(N)^(1/2).

I will check standard deviation later. And yeah 1 BI = 100bb
Does standard deviation have more to do with luck, or with skill? Not really sure about that concept. We will have probabilities course soon so will learn in there.
If you mean how much change there is in my results, as 200nl player it can be really high. Long downswings are possible but upswings are less likely. Have lost 9 BI in a day few times.

It is very easy to see what your standard deviation is. Take a session and observe your stack at all hands played (what it is after each hand played). For example consider 300 hands in one session. You find the average of all those 300 hands say <X>. Then you create the sum (Xi-<X>)^2 over the 300 hands and divide it by 300 and take the square root of that and you have a decent estimate of your standard deviation for that session.

If you have 150000 hands you can in principle do the same using your balance, provided its the same levels not some mixed situation that is all over the place lol. Notice that your sd will depend on what game you play; heads up or 9 handed or 6 handed or 6 handed table but only 3 opponents for a while etc vs loose or very tight players, how deep they are (ie effective stacks) etc (all these things matter) . It will also depend on your game if you are LAG it will be higher than TAG etc and also on what they are so it may change from session to session. But if its 6 handed or 9 handed always full and usually 50bb-200bb it will tend to be the same and more like a reflection of your game.

6-max and small amount of HU.
I have been taking some loose lines. Like check/shoving missed AK, where you would normally check/fold. Also shoving some draws where you could just call or check behind.
Those lines will add variance, since sometimes they work, sometimes fail. It's definitely more tilting when those lines fail lol.
But yeah I used to play more nitty and there wasn't much variance.

Std dev 85 bb/100
And let's say I'm a 3bb/100 winner. Going to play 20000 hands. How to add these numbers in excel to NORMDIST?

I was playing tighter in January, and I was also running really good then consistently. Std dev 76bb/100
During downswing my std dev was also low (76-77). When I add winnings hands to the downswing sample, my std dev increases.
I was running a lot above EV during my downswing though.
Still not sure what is the correlation between standard deviation and luck.

If you have winrate 0.03bb/h and sd 8.5bb/h then in 20000 hands you will be approximated by a normal that has avg 0.03*20000=600bb and sd=8.5*(20000)^(1/2)=1202bb

So you use a normal that has avg 600 and sd 1202.

Basically you use the fact that the sum of n random variables that are normal is a normal with avg the sum of the averages and standard deviation the square root of the sum of squares of the individual standard deviations. If all are the same then it become m*n and sd*(n)^(1/2).

Thanks for the explanations. I still don't quite understand it, but I will when we go more in-depth into our probabilities course. I am putting the pieces together. Starts to make some sense.

So if I'm right, standard deviation does not include luck?
Like let's say I'm a 3 evbb/100 winner, at x % probability my results will be 0 EV AND I can also be below/above EV?
By using the variance calculator: With 3bb/100 ASSUMED (EV) winrate (no luck), std dev 85bb/100 and 20k sample. My chance to be losing would be 30.8842%
So this is just telling me that my results can change a lot over this sample?

Propability of running at or above observed win rate (3.37 bb/100) over 148200 hands.
Is this the luck factor? These are my 100nl results (pre-rb). With 0.65 evbb/100, 3.37 bb/100 and std dev 84.7

Conceptually, I'd say standard deviation is a way of measuring how much "luck" is involved in the game. It allows you to convert the following casual observation;


into precise statements of probability like;

But its telling you more than the following;

You knew that already. Standard Deviation provides the statistical measure which allows you to compute probabilities for how much your results are likely to differ from your EV. That's something you can use to manage your game size in relation to your bankroll to minimize chances of going busto.


PairTheBoard

to see brilliance in action even if I can't follow everything.

Standard deviation regarding the poker game results you have seen is a reflection of how volatile the game is naturally, how easy it is for your stack to change from one hand to the next if you play your natural game. Its both a characterization of the element of luck but also the particular game structure, situations and conditions played.

How can you avoid not going all in when KK vs AA, so how often that happens in a table is already thermalizing the game to be explosive then. But consider also how often you get 72o that leads to folds like over 95% of the time say so 72o is a hand that secures a fold or a loss of blinds usually, a pretty low volatility result. But the number of people matter too. A hand like A2s is probably a fold for most people utg 9 handed but becomes a raising or even pushing hand in other situations when short stacked in later positions or when raising in later positions to steal. So the structure and situation of the game itself is a source of volatility. That will be reflected in SD because 9 handed a ton of hands will at utg be folded and lead to no stack change. Other positions will be more volatile but they will still have a lot of folds in them if others have raised before from much earlier positions to give them credit. Now imagine how much more active a 3 handed table is.

Also your stack size affects SD. If you have 10bb you cannot expect to get as big a move when you get AA compared with what you get if you have 100bb effective.

So think of SD as some measure of how volatile the game is around the expected result, how likely a certain move away from average is to take place. It will reflect both the character of the game but also the conditions under which it is played and your personal style of playing together with the style of the opponents. A loose player may have sd 10bb/h and a tight one 6bb/h say 6 handed. Playing heads up will increase the SD even for a tight player as they are forced to raise and call and 3bet pretty wide now with hands they would have avoided even completely in 9 handed games from early positions.

AP Statistics Question (I'm a teacher)

When Hypothesis Testing for 2-tailed tests, do you (a) double your p-value, (b) cut the alpha-value in half, or (c) it doesn't really matter?

You mean you take into account both extremes (ends) so its double the one sided p-value.
Offer an example to see more if you mean that. Eg for coin flipping fair coin testing example;

http://en.wikipedia.org/wiki/P-value

"Before performing the test a threshold value is chosen, called the significance level of the test, traditionally 5% or 1% [1] and denoted as α. If the p-value is equal to or smaller than the significance level (α), it suggests that the observed data are inconsistent with the assumption that the null hypothesis is true, and thus that hypothesis must be rejected and the alternative hypothesis is accepted as true."


"Thus computing a p-value requires a null hypothesis, a test statistic (together with deciding whether the researcher is performing a one-tailed test or a two-tailed test), and data"

So continuing with this





I found that the partial derivatives were:

and

_______________________________

Now I'm being asked to show that there are no stationary points.

Which I do by setting the partial derivatives equal zero and showing (arguing) that there's no solution to neither of the equitions. I think this is right, but now it's gets more tricky.

I'm being asked to check if the function has any extrema on the set D.

It's my understanding that if a function doesn't have any stationary points, then the possible extrema will be boundary points. So I'm supposed to check the boundaries to find the extrema. But the problem with this set:

2x+y+2 /= 0

Rewriting to: 2x+y+2<0 and 2x+y+2>0

The set is not closed. Meaning that it doesn't contain its boundary points. Furthermore, the set is not "limited".
This suggests to me that the function doesn't have any extrema, because x and y could go to infinite numbers in each direction.
But I haven't found any source confirming this, and I'm not sure how to express this "solution" even if it's correct (I guess I could just write it out).
________________________

Am I on the right track here?

You don't need much to prove that your simple function doesn't have extrema.

1) f(x,y) is always strictly positive (I think this is obvious);

2) it can grow without bounds when (2x+y+2)->0;

3) it can be as close to 0 as you wish when either x or y approaches to infinity.

The second point tells you that there isn't a maximum and the first and third that there isn't a minimum.

Right, that makes a lot of sense. Thanks.

digital systems hw, i've spent > 6hrs trying to do this and still haven't got it to work.

the problem:


pg 150 in the book:


his lecture notes that he put on blackboard:


halp me plox