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  1. #1
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    Feb 2007
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    Probability Formulas - Anybody?

    There was a time in my life when I knew basic probability formulas. That time is gone. So I was playing backgammon with my teenage son the other night and he had a man knocked off and needed to get him back in the game. I had two open spaces for him to attempt to land on. Let's say they were spaces 1 and 2. Whether you know how to play backgammon or not, the question is: when rolling two 6-sided dice, what are the odds/probabilities that he rolls a 1 or a 2 with at least one of those dice?

    Because he's a teenager and everything is simple and obvious, he thought it was simple as 2 out of 6, or 1 of 3. I didn't think so.

    I'm pretty sure the odds of rolling at least one 1 (or any number of course) when rolling two dice are 11 of 36. But I don't know what formula would yield that result.

    I think the odds of rolling a 1 or a 2 (or both) when rolling two dice are 20 of 36, but again, I have no idea what formula would yield that result.

    I'll be shocked if there isn't somebody who posts on these boards who knows these answers instantly.

  2. #2
    Join Date
    Feb 2007
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    Quote Originally Posted by tommy View Post
    There was a time in my life when I knew basic probability formulas. That time is gone. So I was playing backgammon with my teenage son the other night and he had a man knocked off and needed to get him back in the game. I had two open spaces for him to attempt to land on. Let's say they were spaces 1 and 2. Whether you know how to play backgammon or not, the question is: when rolling two 6-sided dice, what are the odds/probabilities that he rolls a 1 or a 2 with at least one of those dice?

    Because he's a teenager and everything is simple and obvious, he thought it was simple as 2 out of 6, or 1 of 3. I didn't think so.

    I'm pretty sure the odds of rolling at least one 1 (or any number of course) when rolling two dice are 11 of 36. But I don't know what formula would yield that result.

    I think the odds of rolling a 1 or a 2 (or both) when rolling two dice are 20 of 36, but again, I have no idea what formula would yield that result.

    I'll be shocked if there isn't somebody who posts on these boards who knows these answers instantly.
    It is 1- 2/3 x 2/3, or 5/9.

  3. #3
    Join Date
    Oct 2007
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    Cabbagetown, Atlanta, GA
    Quote Originally Posted by CDu View Post
    No, it is 12 of 36 (or 22 of 66). You can roll a 1:1, 1:2, 1:3, 1:4, 1:5, 1:6, or a 2:1, 2:2, 2:3, 2:4, 2:5, or 2:6. There are 24 other combinations you can roll, none of which help you.

    Note: if you want to allow for inverting the “order”, then you have 66 combinations and 22 possible successful outcomes (not 36).
    12 out of 36, or 1 in 3, so the teenager's answer was correct.
    Hard at work making beautiful things.

  4. #4
    Join Date
    Feb 2007
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    Quote Originally Posted by CDu View Post
    It is 1- 2/3 x 2/3, or 5/9.
    Ha! I KNEW it would be you CDu who responded quickly.

    So the possible combinations when rolling two dice are:

    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

    That is 36 combinations. The six going across the top first row all have a one (the first of the two digits) and then the other five in the first column (with the one being the second digit) seems to me to add up to 11 of 36, as you don't want to count "double ones" twice. What I am missing so far?

  5. #5
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    NC
    Quote Originally Posted by Edouble View Post
    12 out of 36, or 1 in 3, so the teenager's answer was correct.
    No, I misread the question. See my edit. It is 5/9. The probability of NOT rolling a 1 or 2 on two separate rolls.

  6. #6
    Join Date
    Feb 2007
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    NC
    Quote Originally Posted by tommy View Post
    Ha! I KNEW it would be you CDu who responded quickly.

    So the possible combinations when rolling two dice are:

    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

    That is 36 combinations. The six going across the top first row all have a one (the first of the two digits) and then the other five in the first column (with the one being the second digit) seems to me to add up to 11 of 36, as you don't want to count "double ones" twice. What I am missing so far?
    Yes, 2 columns of 6 and 2 columns of 4.

    The probability of rolling at least one 1 or 2 with two dice is the complement of the probability of failing to roll a 1 or 2 at all. The probability of rolling 1 or 2 with one die is 1/3, so probability of not rolling 1 or 2 is 2/3. The probability of not rolling a 1 or 2 on two rolls (or one roll of 2 dice) is 2/3*2/3, or 4/9. So the probability of a any 1s or 2s is 1 - 4/9 = 5/9.

    Sorry for my shorthand explanation earlier!

  7. #7
    Quote Originally Posted by tommy View Post
    There was a time in my life when I knew basic probability formulas. That time is gone. So I was playing backgammon with my teenage son the other night and he had a man knocked off and needed to get him back in the game. I had two open spaces for him to attempt to land on. Let's say they were spaces 1 and 2. Whether you know how to play backgammon or not, the question is: when rolling two 6-sided dice, what are the odds/probabilities that he rolls a 1 or a 2 with at least one of those dice?

    Because he's a teenager and everything is simple and obvious, he thought it was simple as 2 out of 6, or 1 of 3. I didn't think so.

    I'm pretty sure the odds of rolling at least one 1 (or any number of course) when rolling two dice are 11 of 36. But I don't know what formula would yield that result.

    I think the odds of rolling a 1 or a 2 (or both) when rolling two dice are 20 of 36, but again, I have no idea what formula would yield that result.

    I'll be shocked if there isn't somebody who posts on these boards who knows these answers instantly.
    A few ways to get the answer (answering in detail so you can refresh the principles involved):

    Event A: Get a 1 or 2 on die 1.
    Event B: Get a 1 or 2 on die 2.

    Event of interest Y: Get a 1 or 2 on either die.

    First approach, using OR probabilities:
    Getting a 1 or 2 on either die is the same as P(A or B).
    In general P(A or B) = P(A)+P(B)-P(A&B).
    For independent events (here the outcome of each die is independent) P(A&B) = P(A)*P(B) [See note at bottom]

    So P(Y) = P(A)+P(B)-P(A&B) = 1/3 +1/3 - (1/3*1/3) = 1/3+1/3-1/9=5/9 (or 20/36 as you surmised).

    The second approach is calculating the odds it doesn't happen and subtract from 1.
    Getting a 1 or 2 on either die is the same as NOT getting 3->6 on both dice. So if
    Event C: Getting a 3->6 on die 1
    Event D: Getting a 3->6 on die 2
    then P(getting a 1 or 2 on either die) = 1 - P(C & D) = 1 - (4/6 * 4/6) = 1 - 16/36 = 1 - 4/9 = 5/9

    Third approach is counting combos:
    There are 6 combos of {1,x}, 6 combos of {2,x}, 6 combos of {x,1} and 6 combos of {x,2}. That would be 24 combos but four of the combos ( {1,1},{1,2},{2,1},{2,2} ) are counted twice, so total # of ways to get {1 or 2, 1 or 2} = 24-4 = 20/36 or 5/9.

    Note for P(A & B):
    In general P(A&B)=P(A)*P(B|A) where last term is Prob of B given A has occurred. But when events are independent P(B|A) is just P(B) so formula for both occurring simplifies to P(A)*P(B).
    Last edited by Skydog; Yesterday at 06:50 PM.

  8. #8
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    This better not be heading toward a Ymm, Probabilities thread...

  9. #9
    Join Date
    Feb 2007
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    Steamboat Springs, CO
    I didn't read all the answers -- but here's a common-sense approach. You calculate the chance of NOT rolling a 1 or 2 on either die. That chance is 4/6 for one die 0.667 or (4/6)^2 for two, or 16/36 or 0.444. Then it's the complementary event or 1.0 - 0.444 = 0.5555, or 5/9, as at least one person said above.

    As in much of real life, if you pose the question correctly, the answer is obvious.

    Gawd, I loved it when life was as simple as p or (1-p).
    Sage Grouse

    ---------------------------------------
    'When I got on the bus for my first road game at Duke, I saw that every player was carrying textbooks or laptops. I coached in the SEC for 25 years, and I had never seen that before, not even once.' - David Cutcliffe to Duke alumni in Washington, DC, June 2013

  10. #10
    Quote Originally Posted by bundabergdevil View Post
    This better not be heading toward a Ymm, Probabilities thread...
    Would Ycch, Probabilities be more popular?

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