Hard statistical question: Erik beating Rama on OH

[Swoncen]

With that equation Erik has a chance of ~30,7% to win a single duell against Rama. This is due to the small amount of Eriks Results an the top 100 compared to Rama. Consider 99 Results of Rama and 1 of Erik.. if Erik's result is the 70th-best he would have a chance of 30%.

 

[AvGalen]

Lots of people seem to be just guessing and this question is just way to hard for that.

TomZ: You are making a big assumption with the SD and you changed the question for Mats by taking different data. I guess your calculation does give an approximation though.

Swoncen:

Int_-inf_+inf(D1(x)*D2(x)*dx). D1 and D2 are the distributions. Now it should be clear for all of us.

Yes, completely clear I also don't understand how you could do "point-probabilty " for all these values

Michael:

last m of Erik's OH solves and the last n of Rama's OH solves

What does that mean?

What about this "paradox"?
Mats has never beaten Erik so it is clear that based on this data Mats vs Erik will be 100% victory for Erik
Rama is "much" better than Erik so he will win "most" matches
So you would assume that if Mats always gets beaten by Erik, and if Erik isn't as good as Rama then Mats will always get beaten by Rama as well.
...but Mats actually has a (small) chance against Rama

 

[Swoncen]

Swoncen:

Int_-inf_+inf(D1(x)*D2(x)*dx). D1 and D2 are the distributions. Now it should be clear for all of us.

Yes, completely clear I also don't understand how you could do "point-probabilty " for all these values

Actually this equation is the general case and the calculation with the point-probability is a special case. So you don't understand why we have point probabilities? Because we have single values (The results) where there are some areas in the distribution which are 0 and other ones (most of them) are 1/n where n is the number of results from one person. What we can do is: feed the distribution with more data and calculate the probabilities for ranges (e.g.: how many results are between 15.6 and 15.8) - but then we loose some precision. Therefore we have to handle each result as one point probability and use the sum instead of the integral because we don't have a real curve but single samples.

I hope it's clear now?

Swoncen:

What about this "paradox"?
Mats has never beaten Erik so it is clear that based on this data Mats vs Erik will be 100% victory for Erik
Rama is "much" better than Erik so he will win "most" matches
So you would assume that if Mats always gets beaten by Erik, and if Erik isn't as good as Rama then Mats will always get beaten by Rama as well.
...but Mats actually has a (small) chance against Rama

With this statistical calculation you're right. Erik has a 100% chance of beating Mats but this is just because we have so less data about them. Erik has also results which are worse then Mats result in this list. We just need to look from result 101 to 10000 (for example) to see more of their results. In statistics you allways need alot of data to get more precise results.

 

[Swoncen]

So look at this case:

The red one is the best but his SD is very high. The green one is better then the blue one and the blue one has a better chance to win agains the red one then win agains the green one.

In our case it's a bit like this due to the short list of results. Let's look at this:

96: Rama
97: Erik
98: Mats
99: Rama
100: Rama
101: Erik
102: Erik
103: Rama

without the results from 101-103 Mats would have 0% chance in beating Erik but if we look at the data behind, we see that it's not true.

Easy example: Throw a dice twice. Let's say you had a 5 and a 1. Now you can say the prob. for 1 and 5 is 50% and for the others it's 0%. But you have to throw it much more often.

These results are "Elementarereignisse" http://de.wikipedia.org/wiki/Elementarereignis

Also take a look at this: http://de.wikipedia.org/wiki/Wahrsc..._Fall_diskreter_Wahrscheinlichkeitsr.C3.A4ume

(couldn't find the dutch or english version)

Once again I'm sorry for the bad drawings. lol

 

[trying-to-speedcube...]

It's more accurate than a 90% chance that noone wins ^^

What are you talking about? Who said 90% chance that no one wins?

Also, I wasn't talking about the average rank, not the average time.

:confused:

First one: you did, with your graph.
Second one: sorry, fixed

I will stop posting in this thread. I feel like a noob.

 

[Swoncen]

In my graph was no information that no one wins.. these are two distributions..

 

[qqwref]

Michael:

last m of Erik's OH solves and the last n of Rama's OH solves

What does that mean?

Suppose you don't think Erik has improved for the last 60 OH solves he did in competition... then you can effectively take those 60 solves to be 60 random OH solves from right now. The more solves you have from a given person, the more accurately you can determine how likely they are of beating something or getting sub-X second average, because you can get a better sense of how their times are distributed. So it's your benefit to look further back and take more times, but of course you don't want to go too far back, because for instance Erik got a 30 second OH average in 2007 and clearly he's not going to be getting a 30 second average now unless something terrible happens.

Anyway you'd take the last few of Erik's solves and the last few of Rama's solves, but the number you take of each person doesn't have to be the same. That gives you a large set of times from each person and then you can start comparing them.


EDIT: I ran the thing with Rama's solves since UK Open 2007 and Erik's solves since Dutch Open 2008. There are 101 of Rama's solves and 69 of Erik's and I get that Erik got a strictly better solve than Rama 1451/6969 of the time (~20.8%). If we just use Erik's solves since Belgian Open 2009 (his recent streak of sub25 averages) Erik gets a strictly better solve 550/2020 of the time (~27.2%). The probability that Erik can beat Rama in an avg5 contest is lower, though, because Rama averages lower than Erik does and we can expect avg5 times to be more consistent than single times.

 

[AvGalen]

Swoncen and Michael: You are changing the question. I specificaly said "Given these results....". You shouldn't change that by adding data from other solves or by adding date-information or "improvement" factors.

Swoncen: I understand WHY point-probability should be used, but I want to know HOW you would calculate that for this data.

And the "Mats can't beat Erik, but can beat Rama although Rama is better than Erik"-paradox can be simply explained like this:
[exaggeration]
Erik is a pretty good solver and he is consistent
Rama is an even better solver, but sometimes he sucks
Mats is a solver who sucks, but he doesn't suck as hard as Rama's suckiest solves
So normally the order would be Mats < Erik < Rama, but when everyone performs at his worst the order would be Rama < Mats < Erik.
As you can see Erik is always better than Mats, but Rama can be the best or the worst

 

[Swoncen]

Swoncen and Michael: You are changing the question. I specificaly said "Given these results....". You shouldn't change that by adding data from other solves or by adding date-information or "improvement" factors.

Swoncen: I understand WHY point-probability should be used, but I want to know HOW you would calculate that for this data.

And the "Mats can't beat Erik, but can beat Rama although Rama is better than Erik"-paradox can be simply explained like this:
[exaggeration]
Erik is a pretty good solver and he is consistent
Rama is an even better solver, but sometimes he sucks
Mats is a solver who sucks, but he doesn't suck as hard as Rama's suckiest solves
So normally the order would be Mats < Erik < Rama, but when everyone performs at his worst the order would be Rama < Mats < Erik.
As you can see Erik is always better than Mats, but Rama can be the best or the worst

You just take the number of results from a person (= n) and each result has a probability of 1/n. If there are 2 exact same results you would handle this result as 2/n but since we do that for each result we have twice 1/n which is also 2/n.

And for the data: I just said you cannot really tell who's better with a few results. Take the extreme: you have 3 results: 1. Rama, 2. Rama, 3. Erik.. Erik is much worse then Rama in this case and would have a 0% chance of beating Rama but we know there is alot of data behind that. But if you want to take only this data, just go ahead and use the point-probabilities.

But then you cannot expect results that can predict the chances of anyone beating another person in real life.

Now I really want to know what are my chances to beat you *g*. Would you be so kind to send me some of your recent 3x3x3 times? You have alot I know (maybe 100?).

 

[AvGalen]

Swoncen: I still don't understand how to calculate it. Maybe you can give an example with a very small set of data like this:

1. Rama: 17.00
2. Rama: 18.00
3. Erik: 19.00
4. Rama: 20.00
5. Erik: 21.00
6. Rama: 22.00

Now what would the chance be that Erik beats Rama based on just this data?
I would guess that it should be calculate like this:
Erik does 19.00: Chances of him beating Rama are 2 out of 4 (Rama's 20.00 and 22.00)
Erik does 21.00: Chances of him beating Rama are 1 out of 4 (Rama's 22.00)
So with this data I would say that Erik has a (2/4 + 1/4)/2 = 3/8 chance of beating Rama

Calculating this for Mats would be easy:
Mats vs Erik: 0/24 = 0% chance that Mats wins
Mats vs Rama: 1/75 = 1,3% chance that Mats wins

But I don't see any way this can be done with a formula. Just a loop that analyses each of Eriks solves and determines the chances of winning/draw/losing for that solve
So for this data that would be something like this:
22: (75 - 21) / 75 = n1
24: (74 - 22) / 74 = n2
36: (75 - 33) / 75 = n3
37: (75 - 33) / 75 = n4
38: (74 - 33) / 74 = n5
etc
98: (75-74)/75 = n24
Chances of Erik winning on a single solve: (n1 + n2 + n3 + n4 + n5 + etc + n24) / 24

For a corrected average of 5 this become much harder because then it is not just a matter of winning losing, but the actual times would matter. I guess that would mean that instead of doing 24 point analysis you would need to do point analysis for all 24^5 * 75^5 possible averages

Is there a program that can perform these calculations automatically (based on table-input) or is there a smarter way to do this?


I have dumped the Excel file with all 2295 results on
http://rapidshare.com/files/247177443/fifty_fifty.xlsx.html

 

[qqwref]

Swoncen and Michael: You are changing the question. I specificaly said "Given these results....". You shouldn't change that by adding data from other solves or by adding date-information or "improvement" factors.

You may have asked for us only to look at statistics given the top hundred single results, but I don't think that's a useful way of looking at things, which is why I didn't use it. The reason is that neither cuber is guaranteed to get in the top hundred on any given solve, so if you just look at those results you can only answer the question "if both cubers get 22.18 or better on a given solve, what's the probability Erik wins?" This is a valid question, but it can't be extended to looking at the probability Erik wins an avg5 contest unless you can guarantee that each of the ten solves will be 22.18 or better. (We don't have access to the probability distribution outside of that range, so you can't include it in your calculations if you hope to be accurate.) As I said you would get more accurate results (closer to the real-life probability of Erik beating Rama) if you took the past few competitions' worth of OH solves for both cubers and compared them.


As for the average of 5, suppose you chose n of Erik's times and m of Rama's. Instead of considering (n^5) * (m^5) possibilities (because that's a lot), what you would really have to do is consider the n^5 possibilities (not even that; you can divide by 120 because the order of the times is irrelevant) and for each one decide how many of the m^5 averages of 5 of Rama's would be worse than that. The sum of all of those, divided by (n^5 * m^5), is the probability Erik beats Rama. But remember that you have to choose a set of times from ALL the times each cuber gets, not just the ones under 22 seconds, because otherwise you will not have a distribution that looks like the distribution these cubers will get in real life, and then you will not have an accurate result.

 

[AvGalen]

Michael: I asked about these 100 results because I wanted to know how you would calculate this. Not because I really want to know what chances Erik has against Rama.

And for your second part I think you are wrong. Just calculating if Erik beats Rama on 1 or more solves isn't relevant for an average. You can beat someone on average even if you are only faster on 2 solves and are slower on 3 solves. So you really need to calculate all n^5*m^5 possibilities (we could debate about reducing that by 5!)

 

[JasonTriesToCube]

Mats will beat them 99/100 times.

Oh wait, was it the other way around?