playBunny: ty

FNA: One on the bar is the same as one in your opponent's home table - its a loss by backgammon.

(By "behind the board" I assume you mean in your opponent's home table)

question for the players out there. If you do not get a chip off the board its gammon, if you have a chip behind the board and none off its backgammon? what if you have one on the bar and none off?

Fencer:
OK Fencer just posted here because backgammon is the only game I play here, except for one experiment with dark battleboats. I do like your site!

ajtgirl: Such things happen but the backup system should be faulty anyway :-) However, since this board is for backgammon discussion only, it might be better to continue on General Chat.

Hi, I am new here, a refugee from the crashed site, It'sYourTurn.com. I have been playing there for over 2 years and they constantly have down time. This last one has caused the site to be unavailable for 8 consecutive days. They only have a staff of 3 people and obviously had a faulty backup system. They claim to have 2,000,000 plus members, paying an average of $15.00 annually, which comes to about $30,000,000...can anyone explain to me how a popular web site such as IYT can be run so shoddily? Also, has this site (BK) ever experience similar problems? My loyalty, perhaps misplaced, is still with IYT because they have a great game called Jamble, which I am addicted to, but your backgammon game here is damn good and you have numerous convenience features here that IYT does not have. Any thoughts?
Ajtgirl NY, USA

Ebru: Yeah, random dice can be quite cruel. ;)

whats the problem with these dices ???? They are making me MAD with the style they roll... lol

ArtfulDodger: Just for the record, the opponent did the sporting thing. I watched the match and Marfitalu .. 5-1 .. definitely got the last word. ;-)

Marfitalu: Just dont play any more moves until the times start again, and he will time out and lose, as he should. It will just take some more time.

Marfitalu: you're right but he's got to play the sport about it.

Marfitalu: You should know that it takes the system about five minutes to "recognize" a time-out. After the time runs out, the system will, with a 5-min delay though, terminate the game.

Marfitalu: it may be unsportsman like if you have an agreement on the time per move...I'd just be sure you win at least one more in the series. Then you'll have the last word. ;)

Marfitalu: he's winning the game and time is suspended...I don't get why he should resign.

Marfitalu: I don't see any reason why should he resign. I wouldn't resign either.

AbigailII: Oh, I guess I picked up the string in the middle, never mind then.

grenv: sorry .. my fault .. my message directly below your message which was a reply to me (phew! ;)) .. anyway .. that message of me was a reply to peeky

about the other rating issues .. i agree with you ;)

when i want to judge my opponents strenght i look at his bkr, the number of games he played, how many he won, and his graph to see how he did lately (he could have lost a lot by timeouts, and dropped momentarily :)) .. so i combine all those data (and rarely i look who were his last opponents :))

i would love more types of ratings, but i dont know much about the overhead it would create .. that depends on the database structure

grenv: *nod* i have seen that happen a lot as well .. but peeky's question was why high rating would change more than his opponents rating .. as peeky played more games and his rating should change less, and because of their rating, the higher rating should lose more rating points when he lost, than the one with the lower rating (in case he lost) ... this was the case though .. the one with the lower rating would lose more rating points upon losing .. the question was why

about the net gain in rating when you win and lose one game with the same opponent .. its weird indeed and talked about a lot before already .. i think thats a weird result as well .. it will eventually lead everyone (who isnt that bad :)) to a higher rating when they play enough games

alanback: But I specifically picked the infinite game to be a game with each site having 1 stone left, and both sides throwing 1-1s. Could you given a specific example of a position occuring in that infinite sequence where throwing 1-1 leads to the shortest way of finishing from that position?

AbigailII: Of course :-) For game n, the first n moves are moves 1 thru n of the infinite game. You then pick the shortest sequence to finish the game. Call the (n+1)th move X. X may or may not be the (n+1)th move in the infinite game. However, there will be an infinite number of values of n for which X is also the (n+1)th move in the infinite game. Therefore, the shortest sequence that finishes game n will also be the shortest sequence that finishes game (n+1).

WhiteTower: Maybe, just maybe, his "Pro Backgammon" initiative will include a backgammon ratings formula. :-) I wouldn't bet on it though. :-(

alanback: The method you describe will produce some duplicates.

Really? Care to give an example of such a duplicate?

We still haven't seen anything by Fencer on the issue - if he doesn't care, all we write here is pointless anyway :) If he does, keep going...

alanback: "Maybe if I were a better player I would care more "

You're ranked 4th of all the players with a decent number of games under their belt. You're not good enough yet? What hope the rest of us!! ;-))

WhiteTower: I like the wins/losses ratio as well but I only use it in conjunction with ratings and in situations where it has strong validity. In the VogClub tournaments, for instance, where all players get a wide variety of opponents who cannot be chosen, the ratio is indicative of strength amongst players within a given ratings range. Against the robots (strong, medium and poor) you need to know which robot the player prefers. But Grenv makes a good point when considering other players; it's hard to tell what the ratio means unless you keep track of who plays who. My ratio is 36:12 overall but against Walter it's 6:6.

I should point out that Vog maintains separate ratings and rankings for the three different ways of playing backgammon. It could be interesting for BrainKing to have players ratings and tournament ratings plus a combined rating and corresponding win/lose ratios.

WhiteTower: Win/loss is meaningless. Some players play the same group all the time. If 2 good players play a lot against each other that will obviously affect win/loss records.

However I agree this ratings system is a little flawed.

I actually think ratings is very important, especially when choosing an opponent.

WhiteTower: The question in my mind is how much trouble should be put into a ratings system for a recreational website. I might even agree that I would rather have no rating system than the one we have now; not because win/loss is a better measure of ability, but because I don't really care how precisely my ability is measured. Maybe if I were a better player I would care more :-)

alanback: Exactly - as long as we even consider more improvements, and can't even agree how to start off, it's dead in the water... I partly agree with what you say, but an inconclusive ratings system is worse than a win/loss record...

WhiteTower: I can't agree. Ratings are a much more accurate indicator than won-lost, because they take into account the strength of your opponents. This is not to say that the rating system here couldn't be improved!

grenv: Whatever the answer is, the morale stays the same: ratings aren't good enough however you calculate them. Win/loss/draw ratios are the real thing in the end...

Hrqls: I've seen this quite a lot, where I'm playing someone in 2 game, we split, and both our ratings end up higher than before. And no other games are completed between the 2 (which is the obvious thing to check first).

The explanations are not sufficient, if someone has played more then their rating would decrease and increase slower, but that doesn't explain the anomaly. Winning and losing streaks are not included in the calculation, only current rating.

So anyone have a mathematical reason why this might happen?

in this game i have rolled 2+5 .. of course i have to get off the bar first so i can only use the 2 as my first die ...

i still have the 'swap dice' link though .. when i click on it i get 2+5 again (as i have to move the 2 first) .. and the 'swap dice' link is gone now

this was what i was talking about before

just to let you know .. i will post it in the bug tracker :)

Peeky: i guess you played more games than he did ... so that would make you lose less (as your rating would be more established

you are lower in rating .. so you should lose less

then only one option remains ... you probably have move variation in your wins and losses ... and thereby your change will be higher .. he might have had a losing or winning streak lately .. and therefore his current game changes his rating less than it would have otherwise

AbigailII: Are you looking for an infinite number of finite games, or an infinite number of unique finite games? The method you describe will produce some duplicates.

AbigailII: True, well stated. It was the first time it was stated though.

grenv: Once you have an infinite game, it's easy to construct an infinite number of finite games from it. Do that as follows: number your games 1, 2, 3, .... For game n, the first n moves are the same n moves from the given infinite game. After the n moves, pick the shortest sequence that finishes the game.

playBunny:

Maybe we have to use one of those other infinities. How many are there? ;-))

Guess ;-)

Pedro Martínez: It was a direct answer to your post to me:
You should have said you were speaking of probability of possible "existence" of a certain sequence, not actual rolling it.
Why were you mentioning it in your reply to Chessmaster1000?
But never mind. It hardly matters.

Grenv: Your blocking example is the same the hitting one. Make as long a sequence of blocks(hits) as you like. Then tack on something different.

alanback: "at least if the infinity in question is the infinity that measures the number of integers (referred to I believe as aleph-sub-naught)."
Maybe we have to use one of those other infinities. How many are there? ;-))

Everyone: Perhaps the most important result of all this is that in googling something mathematical I chanced upon a link to some good jokes which you may enjoy. :-DD

grenv: Yup, that ought to clarify the whole shebang! Thank you very much. :)

I haven't time to read all the posts, but here goes:

The question I answered was "So if that infinite set contains everything then what's the probablity of a given thing being contained in it?"

1, since the infinite set contains everything, so any given thing has 100% chance of being in the set.

The other confusion was around infinte games. The example of 2 pieces on each side being next to each other works like this (assuming player is on 5 spot and opp on 6 spot of his own home):

Game 1: 6-6 GAME OVER
Game 2: 1-1 1-1 6-6 GAME OVER
Game 3: 1-1 1-1 1-1 1-1 6-6 GAME OVER

etc.. infinite number of games.

The example where players continually hit each other is an infinitely long game, not an infinite NUMBER of games.

Chessmaster1000: My math degree is 36 years old so I'm too rusty to be sure of this . . . but I think that, while an infinite sequence of random rolls would certainly contain any *finite* sub-sequence (indeed, an unlimited number of such sub-sequences), I don't think it's correct to conclude that it will contain any given *infinite* sub-sequence . . . at least if the infinity in question is the infinity that measures the number of integers (referred to I believe as aleph-sub-naught).

playBunny: I think we are tuned to a completely different frequence. I have no idea what your previous post has to do with our preceding discussion about the probability.

Pedro Martínez: I did, lol, 10 messages ago in Re: 100% vs 0%.

ChessM challenged my 5-5 example:
[playBunny: Both sides roll 5-5 ad infinitum]

1st)The probability that both sides will roll a 55 an infinite number of times is exactly zero!

2nd)Even if the game will continue with an infinite number of 55 (although this can never happen as i said), that game would be one single game and this doesn't help us in the question of how many Backgammon games exist? Finite or infinite?

This is correct when considering the production of the sequence but if you look at it from the viewpoint that you already have the infinite set of sequences then the sequence already exists, then you have a single infinitely long game.

Then, given that for each roll there are alternate sequences of rolls which will result in the same position (ie. one piece on the 5-point and one on the bar), there are an infinite number of games.

Wil and Abigail have already said much the same thing.

alanback: "As White Tower suggests, the laws of probability do not apply to infinite sequences. They are meaningful only in the context of a finite sequence."

Actually we can't blame the laws of probabilities for not being meaningful at an infinite number of rolls, but our brain's incapability to understand the infinite........


playBunny: "What is the probability that the sequence "Endless 5-5s" exists in the infinite set of all dice roll sequences?
A: 1 "

Since this infinite set contains ALL dice sequences, it's reasonable that it will contain and the "Endless 5-5s".......
So it's 1 or 100%....


Now, what AbigaiIII said about different possible Backgammon games was correct and his proof was correct, but i have found a link that states that the number is 10^140 and not infinite. Perhaps it defines with another way the "game". I will investigate this tomorrow..........

playBunny: You should have said you were speaking of probability of possible "existence" of a certain sequence, not actual rolling it.

Why were you mentioning it in your reply to Chessmaster1000?

Pedro Martínez:
Q: What is the probability that you will roll 55s forever and nothing else?
A: 0

Q: What is the probability that the sequence "Endless 5-5s" exists in the infinite set of all dice roll sequences?
A: 1

It's a viewpoint kind of thang.

alanback: Thank you. Now back to playBunny's post that led to this "debate":
Excuse my ignorance, I'm a logician more than a mathematician, but I would have thought that the probability of an endless sequence of 5-5s is exactly 1.

Pedro Martínez: The probability that I or anyone else will roll 55 forever is zero.

As White Tower suggests, the laws of probability do not apply to infinite sequences. They are meaningful only in the context of a finite sequence.

Please, guys, what you are debating is useless - rolling 5-5 all the time is a typical trivial case, especially as it leads to a hugely non-standard result (infinitely long game) - therefore let's concentrate on finite games...

anybody: What is the probability that you will roll 55s forever and nothing else?

Pedro Martínez: Since you will never be able to roll forever, why would you want to know that?