- 28 May '06 22:46depends on the relative probability of each rating, which you could either look at as i. a power law (many poor players, exponentially fewer good players) or as ii. a gaussian curve (few poor players, many average players, few good players). i'd imagine its closer to ii, and based on the start rating i'm guessing its centred somewhere around 1200, but we still need to know how the frequency of a rating varies with the size of the rating...
- 29 May '06 03:00

You are playing Darvlay in a clan challenge. That means the matchups are very biased towards close ratings. Probability? Not that small.*Originally posted by sonhouse***I just got paired with Darvlay, and it turns out our ratings are identical, at 1571 each. So what is the probablity of two players paired together here to have the exact same rating?** - 29 May '06 09:10 / 1 edit

Twu twu, that would make it an artificial problem. What about the more general case of non-clan pairs? Funny part is, before two moves were made, we were not equal any more.*Originally posted by XanthosNZ***You are playing Darvlay in a clan challenge. That means the matchups are very biased towards close ratings. Probability? Not that small.** - 30 May '06 17:07

The probability that there are two players on the site with the same rating who are playing against each other? I'd guess pretty high, as there are an awful lot of players.*Originally posted by sonhouse***I just got paired with Darvlay, and it turns out our ratings are identical, at 1571 each. So what is the probablity of two players paired together here to have the exact same rating?**

The probability that any given game will be against a player with the same rating. Probably still pretty good, as most people will tend to look for games against similarly rated players. - 30 May '06 20:15

Wow... your math is amazing...ly wrong. Even your assumptions are horrible. I believe it is impossible to have a score of 0 and at any rate, why not use the existing range of scores. Also, the distribution of scores will not be uniform at all. First you have to understand the distribution and then you can do the calculations.*Originally posted by uzless***come on wankers...easy question to answer**

Just find out how many players are at this site (I'll guess 10,000?) and what the upper and lower limit of the ratings can be. I assume the low rating is 0 and the upper rating is, well is there an upper rating? If there is, lets make it 3000 just to have a number.

Therefore you have 3000 possible rating and 1 ...[text shortened]... th other than figuring out how much a 2-4 costs so someone might want to rip my answer apart. - 30 May '06 20:43 / 1 edit

Well, no shiete sherlock but who's going to go to all that research for this one?????*Originally posted by Gastel***Wow... your math is amazing...ly wrong. Even your assumptions are horrible. I believe it is impossible to have a score of 0 and at any rate, why not use the existing range of scores. Also, the distribution of scores will not be uniform at all. First you have to understand the distribution and then you can do the calculations.**

At least my answer gives you, well maybe not you, a reasonable estimate to work with, or at least a reasonable template to work from should anyone figure out how many people are on the site and what the range of ratings are.

why don't you just try and answer the question instead of just saying what's wrong with someone else's solution....what are you, GreenPeace? - 30 May '06 21:53Assuming pairs are made totally random and most people are averagely smart and there are few people really smart and few people really unintellectual, gives us quite a "normal distribution" for this. Probability of choosing 2 people at random that are average smart and have the same IQ is much more likely than 2 people be put together that are really smart with the same IQ. Search for Probability of (standard) normal distributions on the internet. I don't know these formulas by heart but I think you'll figure out a solution that way? I think... Correct me if I'm wrong, I'm just a 20 year old student
- 30 May '06 22:21 / 2 edits

nah you're right, thats were i was going with post #2. but we are short of some information.*Originally posted by 1ted***Assuming pairs are made totally random and most people are averagely smart and there are few people really smart and few people really unintellectual, gives us quite a "normal distribution" for this. Probability of choosing 2 people at random that are average smart and have the same IQ is much more likely than 2 people be put together that are really smart with a solution that way? I think... Correct me if I'm wrong, I'm just a 20 year old student**

But, from the player tables we know there are 11 667 players at rhp, so the median player as around position 5834. i can't be bothered sorting through the tables, but i know a player ranked 5794th, which is close enough. he has a rank of 1274, so the median ranking is around 1250 ish. the highest is 2430, so making the assumptions (a) median roughly = mean; (b), the distance between the highest and lowest is 3 standard deviations (both fairly wild assumptions, but they will give us an estimate), gives us a standard dev of around 395. Using a java Z-table shows that roughly 0.073% of people have a ranking of 1570, so if you play one million randomly selected games on rhp, around 730 of them will be against people with the same ranking. qed. - 30 May '06 22:47 / 1 edit

Which works out to roughly one in a thousand chance of getting randomly paired with the same rating?, one chance in 1369 it looks like. Better odds than the lottery anyway!*Originally posted by xcomradex***nah you're right, thats were i was going with post #2. but we are short of some information.**

But, from the player tables we know there are 11 667 players at rhp, so the median player as around position 5834. i can't be bothered sorting through the tables, but i know a player ranked 5794th, which is close enough. he has a rank of 1274, so the median rank ...[text shortened]... y selected games on rhp, around 730 of them will be against people with the same ranking. qed.

How would the odds change if we polled say, the USCF, I think there are about a half million player there. But I think the average USCF rating is more like 1400 or so not 1270 that you are assuming here. - 31 May '06 08:00 / 1 edit

By this calculation there should be around 8.5 active players with a rating of 1570 (11667*0.073*0.01). Actually there are 5 as I'm writing this.*Originally posted by xcomradex***nah you're right, thats were i was going with post #2. but we are short of some information.**

But, from the player tables we know there are 11 667 players at rhp, so the median player as around position 5834. i can't be bothered sorting through the tables, but i know a player ranked 5794th, which is close enough. he has a rank of 1274, so the median rank ...[text shortened]... y selected games on rhp, around 730 of them will be against people with the same ranking. qed.

Also note this is a much more useful figure as it doesn't talk about playing games so it removes all discussion of bias due to game selection.

Question: How unlikely is this result (5 if the average is 8.5)?