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

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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

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  1. Anonymous
    2011-11-03T18:03:53+00:00

    Just to set the record straight - you are correct based on you assumptions.  But that is not the problem I posed.

    I posed 2 outcomes on the first roll.

    1. Your opponent rolls a 6.  That is a probability of 1/6
    2. You and you opponent both roll the same number.  That is the probability of 1/6.  This in fact is the same probability of rolling doubles.

    Now the probability of either #1 or #2 happening is the sum of the probabilities because of the "OR" minus the probability of both happening = 1/36

    So, as I said,  the probability of your opponent rolling a 6 OR you both roll the same number is about 1/3 or precisely 1/3 - 1/36.

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

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  2. 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)

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  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

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  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.

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