Is there a stop to Internet Backgammon cheaters?

Below is one way that the stall cheat is performed, though there are variants. Everyone has a right to this information and not just the Microsoft engineers who want to cheat everyone else when they play backgammon.

Move and undo moves repeatedly starting before the appearance of the time warning message  in the left chat pane. When the opponent starts the clock continue to move and undo but make sure the moves are completed just under the five second mark. The opponent will then be frozen out and can be timed out by starting his clock when the opponent stall message appears. This method can probably be simplified by more precise timing with fewer undos since opponents have timed me out with nothing more than a couple of undos.

One thing is now very clear. Microsoft is fully aware of this cheat and knows exactly what part of their code allows it and knows exactly how to fix it.  Microsoft willfully wants this cheat to remain in existence. This famous cheat is so simple that a fix is obviously quite easy. The fact that Microsoft has allowed a simple to fix cheat that they are fully aware of to exist for years is evidence of a sociopathic attitude by Microsoft.

P.S.

Just got through playing a 2500 player. He doubled me immediately and when I finally ended up with a winning position and redoubled he quickly timed me out. He made a single move a single undo and then the single move again. When he delayed and I started his clock, I not he was the one that got timed out.

I tend to agree with most of your post (Shackle).

 

When I argued that there was cheating, I was only referring to the stall bug used, ie: opponent moves then rescinds move, does this a few times then stops; the next thing that happens is you are told that you are delaying your move and are eventually timed out.

 

I doubt anyone playing on msn has not come across this, and we will all agree that it is cheating, and I just wanted to know if there is a way around this.

 

WIth regards to the dice: there is no doubt that they are not random, and in some way react to what is needed, but I doubt the opponent can fix them. I have frequently lost 12 on the trot, and sometimes won 12 on the trot.  When on your side you frequently throw the perfect throw needed, and vica versa when on losing streak.

 

My only disagreement with Mr Shackle is saying odds on any paritcular throw are 18-1. Obviously I agree they are not 36-1, that is obvious, but the correct odds are 21-1, as there are 21 different thriw combinations. Look at the table below. I may be incorrect, and please tell me if I have missed something, but I don't think so. I do remember when I first started playing Backgammon (30 years ago) reading a book, and they commented on the 36-1 myth, and used the table below to explain the 21-1 odds

 

1-1     2-2      3-3     4-4     5-5     6-6

2-1     2-3      3-4     4-5     5-6

3-1     2-4      3-5     4-6

4-1     2-5      3-6  

5-1     2-6 

5-1    

 

6  +     5    +    4    +    3    +   2    +   1      =    21 

This is gobble-de-gook.  You must include the TOTAL event space, not a partial, when calculating probability.  The book you read was wrong if it told you to use 21 as the denominator of probability calculations.  This partial list is valid only to describe the possible rolls in backgammon, but not to be used in calculating probability.

Here are some an examples:

All events:

(1,1) (1,2) (1,3) (1,4) (1,5) (1,6) (2,1) (2,2) (2,3) (2,4) (2,5) (2,6)

(3,1) (3,2) (3,3) (3,4) (3,5) (3,6) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6)

(5,1) (5,2) (5,3) (5,4) (5,5) (5,6) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

Rolling a six:

(1,6) (2,6) (3,6) (4,6) (5,6) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

That is 11/36 or 30.55%

Rolling 2 fours

(4,4)

That is 1/36

Rolling a four and a two

(4,2) (2,4)

That is 2/36 or 1/18

Rolling 2 fours or (a four and a two)

(4,4) (4,2) (2,4)

That is 3/36 or 1/12

Be careful not to espouse erroneous information and  visit..

http://gwydir.demon.co.uk/jo/probability/calcdice.htm

Quite clearly as we are talking about backgammon, I was talking about the odds on a throw, which is 21-1.

 

Do you need advice on the Emglish language?

You are wrong.  I'm a retired math professor.  One of my courses was in probability.  Read a book if you can.  The odds are NOT 21-1!!!!!  Can you not follow the above or do you have a problem with Emglish? 

According to your convoluted logic you would argue that the odds of tossing two coins and getting 2 heads is 1/3  where the total event space of HH, HT, TT.   But, of course, it is 1/4 with an event space of HH, HT, TH, TT.

The probability of tossing 2 coins and getting HT or TH (either way) is

P(getting HT) + P(getting TH) = 1/4+1/4 = 1/2.  That's how you deal with a combination where the order does not count.  (P = "probability of")

The same with dice.  If, as in backgammon and craps, the order does not count about rolling a 3 and 4 (either way) then the probability is...

 P(getting a 3 and 4) + P(getting a 4 and 3) = 1/36 + 1/36 = 1/18

When you see an OR of two independent events, then the probabilities ADD.

The general formula for OR of any two events A and B, (independent or not) is...

P(A or B) = P(A) + P(B) - P(A and B)  where P(A and B) = P(A) X P(B)

Stop espousing erroneous information.  Your erroneous information also says the odds of 6-6 is 21-1.  Then why does a casino pay out 30-1???

For reference visit...

http://gwydir.demon.co.uk/jo/probability/calcdice.htm

and go down to the section called "Other conditions with two dice".  The denominator is always 36.

Now if you do not understand the above and what is on the web site and continue with the 1/21,  then I am not interested in discussing this with you any longer.  In my math classes, there was an occasional moron that could not learn; so I flunked him (he was a loser anyway) and concentrated on those who could learn.

I tend to agree with most of your post (Shackle).

 

When I argued that there was cheating, I was only referring to the stall bug used, ie: opponent moves then rescinds move, does this a few times then stops; the next thing that happens is you are told that you are delaying your move and are eventually timed out.

 

I doubt anyone playing on msn has not come across this, and we will all agree that it is cheating, and I just wanted to know if there is a way around this.

 

WIth regards to the dice: there is no doubt that they are not random, and in some way react to what is needed, but I doubt the opponent can fix them. I have frequently lost 12 on the trot, and sometimes won 12 on the trot.  When on your side you frequently throw the perfect throw needed, and vica versa when on losing streak.

 

My only disagreement with Mr Shackle is saying odds on any paritcular throw are 18-1. Obviously I agree they are not 36-1, that is obvious, but the correct odds are 21-1, as there are 21 different thriw combinations. Look at the table below. I may be incorrect, and please tell me if I have missed something, but I don't think so. I do remember when I first started playing Backgammon (30 years ago) reading a book, and they commented on the 36-1 myth, and used the table below to explain the 21-1 odds

 

1-1     2-2      3-3     4-4     5-5     6-6

2-1     2-3      3-4     4-5     5-6

3-1     2-4      3-5     4-6

4-1     2-5      3-6  

5-1     2-6 

5-1    

 

6  +     5    +    4    +    3    +   2    +   1      =    21 

This is gobble-de-gook.  You must include the TOTAL event space, not a partial, when calculating probability.  The book you read was wrong if it told you to use 21 as the denominator of probability calculations.  This partial list is valid only to describe the possible rolls in backgammon, but not to be used in calculating probability.

Here are some an examples:

All events:

(1,1) (1,2) (1,3) (1,4) (1,5) (1,6) (2,1) (2,2) (2,3) (2,4) (2,5) (2,6)

(3,1) (3,2) (3,3) (3,4) (3,5) (3,6) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6)

(5,1) (5,2) (5,3) (5,4) (5,5) (5,6) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

Rolling a six:

(1,6) (2,6) (3,6) (4,6) (5,6) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

That is 11/36 or 30.55%

Rolling 2 fours

(4,4)

That is 1/36

Rolling a four and a two

(4,2) (2,4)

That is 2/36 or 1/18

Rolling 2 fours or (a four and a two)

(4,4) (4,2) (2,4)

That is 3/36 or 1/12

Be careful not to espouse erroneous information and  visit..

http://gwydir.demon.co.uk/jo/probability/calcdice.htm

Quite clearly as we are talking about backgammon, I was talking about the odds on a throw, which is 21-1.

 

Do you need advice on the Emglish language?

You are wrong.  I'm a retired math professor.  One of my courses was in probability.  Read a book if you can.  The odds are NOT 21-1!!!!!  Can you not follow the above or do you have a problem with Emglish? 

According to your convoluted logic you would argue that the odds of tossing two coins and getting 2 heads is 1/3  where the total event space of HH, HT, TT.   But, of course, it is 1/4 with an event space of HH, HT, TH, TT.

The probability of tossing 2 coins and getting HT or TH (either way) is

P(getting HT) + P(getting TH) = 1/4+1/4 = 1/2.  That's how you deal with a combination where the order does not count.  (P = "probability of")

The same with dice.  If, as in backgammon and craps, the order does not count about rolling a 3 and 4 (either way) then the probability is...

 P(getting a 3 and 4) + P(getting a 4 and 3) = 1/36 + 1/36 = 1/18

When you see an OR of two independent events, then the probabilities ADD.

The general formula for OR of any two events A and B, (independent or not) is...

P(A or B) = P(A) + P(B) - P(A and B)  where P(A and B) = P(A) X P(B)

Stop espousing erroneous information.  Your erroneous information also says the odds of 6-6 is 21-1.  Then why does a casino pay out 30-1???

For reference visit...

http://gwydir.demon.co.uk/jo/probability/calcdice.htm

and go down to the section called "Other conditions with two dice".  The denominator is always 36.

Now if you do not understand the above and what is on the web site and continue with the 1/21,  then I am not interested in discussing this with you any longer.  In my math classes, there was an occasional moron that could not learn; so I flunked him (he was a loser anyway) and concentrated on those who could learn.  

   - Spoken like a true teacher. Now we know why their programs are the first cut.  If you want to learn how to cheat on MS BG or how to stop cheaters go to pastbin and search "IV/Binomial Rand." -but you have to go with a proxy server. MS cant stop people from cracking their simple codes. 75% of the people playing  MS BG are cheating. And that's the nice number which varies only depending on time of day. Many sites have cheaters - MS is particularly full of it, but hey. welcome to the world wide web.

What stumps me is when you're beginning a game, you roll, and the opponent wins the roll in astounding numbers, like 15 out of 16 tries.  In my experience, players who win rolls like that usually (if not always) win the game. 

And what's up with an opponent rolling, and then several unrelated stones move -- then, later on, the game quits, citing "error?"