Is there a stop to Internet Backgammon cheaters?

Anonymous
2010-04-09T18:01:10+00:00

So I have enjoyed playing Backgammon in the past, but nowadays is just IMPOSSIBLE.  I am sure many others have noticed this.  There are players out there that can control the dice at least for them, and some can even move more chips than the allowed by standard rules.

For example, not 10 minutes ago, I rolled 6 and 5, then the guy doubled me on his first roll, and got double three, then I rolled and I got 4 and 1, then he rolled and got 3 and 1, then I rolled and got 3 and 2, then he rolled and got double 5 eating my lonely chip in position 1 and covering position 3 from position 8.  Do you call that a lucky roll?  I call that an engineered roll, if you know Backgammon.

Yesterday, a guy moved 6 chips using a double 5.  That same guy rolled 4 consecutive pairs.

Yesterday, I had 5 out of 6 positions occupied in my side and I ate one of the other player's chips.  I did this repeatedly.  The guy just lost 1 roll out of 4 chips I ate.  He also had 5 of 6 positions occupied.  He ate a chip of mine twice.  I had to roll some 10 times before I could get out each time.

If you know Backgammon, you know this is NOT right.  Losers out there have found a way to cheat, and that ruins the fun for us, the ones with a brain.

Microsoft, please put an end to this.  Your Windows license is expensive, and I paid for it, and I expect to be able to enjoy all the features in the OS, and that includes the included Internet gaming.  Roll a patch that can end this blatant cheating!


MCP

Windows for home | Other | Gaming

Locked Question. This question was migrated from the Microsoft Support Community. You can vote on whether it's helpful, but you can't add comments or replies or follow the question.

0 comments No comments
Answer accepted by question author
Anonymous
2010-04-10T13:55:17+00:00

Hi,

If you have any feedback’s, then I would suggest you to post here:

https://connect.microsoft.com/?wa=wsignin1.0

 Please respond back with the status of the issue and let us know if you have any issues.

 Azam – Microsoft

Was this answer helpful?

30+ people found this answer helpful.
0 comments No comments

175 additional answers

Sort by: Newest
  1. Anonymous
    2011-11-03T03:40:55+00:00

    Get out your probability math book and do some studying.

    "Get the same X number in each dice with two dices= 1/36" -- WRONG!

    There are 36 different combinations of 2 dice and 6 ways for equals - (1,1) (2,2) (3,3), (4,4), (5,5) and (6,6).

    6/36 = 1/6

    To get a 6 from one die (come out roll) = 1/6

    The probability for either one of the above to happen = 1/6 + 1/6 = 1/3.

    Actually it is 1/6 + 1/6 - 1/36. = 1/3 - 1/36 = .305%

    (You have to subtract the probability of both happening)

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

    Was this answer helpful?

    0 comments No comments
  2. Anonymous
    2011-11-03T03:32:18+00:00

    This is NOT a political forum.  Please refrain from spouting your one sided politics and keep to the issues here:  BACKGAMMON.

    Was this answer helpful?

    1 person found this answer helpful.
    0 comments No comments
  3. Anonymous
    2011-11-03T03:23:19+00:00

    You are right for you interpretation of what I said.  But you misunderstood.  I said that the probability of a come out roll of a six OR equal come out rolls is 33%

    1. Rolling a single six with one die (the come out roll)  = 1/6
    2. Both your come out roll and you opponent's come out roll are equal = 1/6 (you agreed in your second paragraph)

    Therefore 1 OR 2 is 1/6 + 1/6 = 1/3  (you add probabilities if the are OR'ed.  (You multiply them if they are ANDed)

    I counted both situations (1 and 2 above) and got 242 out of 344 games starts.

    Yeah, I also noticed many players moving a pip back and forth to waste time.

     

    I think you are totally wrong about it, if you 1/6 + 1/6 you are saying that is easier to get two 6 with two dices than one 6 with one dice, the correct is 1/6 X 1/6 = 1/36, so:

    Get X number with one dice = 1/6

    Get the same X number in each dice with two dices= 1/36

    Was this answer helpful?

    0 comments No comments
  4. Anonymous
    2011-11-03T03:11:09+00:00

    You are right for you interpretation of what I said.  But you misunderstood.  I said that the probability of a come out roll of a six OR equal come out rolls is 33%

    1. Rolling a single six with one die (the come out roll)  = 1/6
    2. Both your come out roll and you opponent's come out roll are equal = 1/6 (you agreed in your second paragraph)

    Therefore 1 OR 2 is 1/6 + 1/6 = 1/3  (you add probabilities if the are OR'ed.  (You multiply them if they are ANDed)

    I counted both situations (1 and 2 above) and got 242 out of 344 games starts.

    Yeah, I also noticed many players moving a pip back and forth to waste time.

    Was this answer helpful?

    0 comments No comments