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?

alanback: That's not what I'm asking. I'm asking for the probability of actual rolling of endless sequence of 55s. What is the probability that you will roll 55s forever and nothing else.

Pedro Martínez: The probability that an infinite sequence of random rolls will include a sequence of N consecutive double fives approaches 100% . . . where N is an arbitrary integer.

Pedro Martínez: I'm more considering the infinite set of dice rolls as a fait accompli that we can just dip into and grab something out of. As the set already exists and it contains every dice roll sequence, then the 5-5... must be in there. 100%

As you head towards infinity, generating the 5-5s as you go, the probability shrinks by 1/36 each time and certainly tends towards zero. But when you eventually reach infinity (I know, lol, I know) then Hey Presto! the probabilities for all sequences suddenly jump to 100%! Strange things them infinities.

But I'm looking at it from a logical point of view, not that of a mathematician. Strange things them mathematicians. ;-)

The probability that I will keep rolling 55 for an infinite number of times is almost 100%?

grenv: "One piece each wouldn't work, it [the 2 pieces each case] relied on players hitting each other's block."

I can imagine a 2 pieces each case with the two blocks at a given distance apart ( say 3 points) and 3-3 is endlessly rolled. This will give a single infinite game, so maybe I haven't found the one you were thinking of.

The 1 piece case that I'm envisaging is similar to the 5-5 case mentioned below except that the 14 pieces on each ace point have been born off first. Then it's a question of combinations of rolls that ensure that a piece keeps getting hit.

Peeky: The amount of change in your rating also has something to do with the number of games that you have completed. Less games finished, more variation. Perhaps you've just played a few games and your opponent has played a lot of games?

Pedro Martínez: more correct to say it approaches 100% -- since the probability of completing an infinite series in finite time is zero ;-)

And you are saying that the probability of rolling a certain sequence from an infinite number of sequences is 100%?

Pedro Martínez: Yes, yes. Next?

grenv: Can we agree that 1 is 100% and 0 is 0%?

Pedro Martínez: Actually 1

playBunny: Almost 0.

Pedro Martínez: Lol. So if that infinite set contains everything then what's the probablity of a given thing being contained in it?

Peeky: The weaker player loses more than the stronger player in the case of a loss and gains less with a win. Or two players can play two games and after each having won one game both their ratings will have increased.

Can I explain? No. But I can join you in a complain. It's a terrible formula that we have.

playBunny: Are you sure about that?
Yes, I am...:)
Does not the infinite set contain everything?
Yes, it does...:)

Pedro Martínez: "The probability of any specific roll or sequence of rolls is lower than 1, no matter if you consider the number of moves finite or infinite."

Are you sure about that? Does not the infinite set contain everything?

Pbarb2: Yes, it is relatively new. For the future, just make sure you don't accept invites or sign up to tournaments that have a red dot next to their "time control"...:)

Pedro Martínez: Is this something new? I knew that was there but didn't know what it meant. I sure know now.LOL My first time out or use of the auto vacation. In this case non use. I will make sure I don't get one of those games again. In my case you never know what is happening health wise with me. Thanks for the quick reply.
BARB