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  1. #1
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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
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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
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    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
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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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    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
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    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; 03-01-2021 at 06:50 PM.

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

  9. #9
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    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?

  11. #11
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    Quote Originally Posted by bundabergdevil View Post
    This better not be heading toward a Ymm, Probabilities thread...
    What are the odds of that?

  12. #12
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    Quote Originally Posted by YmoBeThere View Post
    Would Ycch, Probabilities be more popular?
    Would a Ymm, Probabilities and Ycch, Probabilities thread be mutually exclusive?

    Quote Originally Posted by camion View Post
    What are the odds of that?
    You use puns with frequency...

  13. #13
    Quote Originally Posted by sagegrouse View Post
    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).
    This was my first approach when I found this thread.

    Would take much longer to actually find 2 dies in my house.

    However, I have an App that will do that. But I expect that the algorithm that creates random numbers is not completely random, so there would be some round-off error..

    Are they completely cubic dice, or the rounded edges like playing Stratego? Need to know to calculate fringe effects.

    Larry
    DevilHorse

  14. #14
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    Quote Originally Posted by sagegrouse View Post
    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).
    In a genetics course I co-teach, we instruct the students to calculate conditional probabilities in an additive way: chance of getting 1 or 2 on first die = 1/3. A 2/3 chance that they don't get it on the first die, and same probability that they get 1 or 2 on the second die (1/3), so probability that they get it only with the second die is 2/3 x 1/3 = 2/9. Summing the mutually exclusive events gets you the overall probability: 1/3 + 2/9 = 5/9.


    Quote Originally Posted by DevilHorse View Post

    Are they completely cubic dice, or the rounded edges like playing Stratego? Need to know to calculate fringe effects.

    Larry
    DevilHorse
    Trick question, there are no dice in Stratego! The gig is up, "Larry", if that's even your real name...

  15. #15
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    Quote Originally Posted by DevilHorse View Post
    This was my first approach when I found this thread.

    Would take much longer to actually find 2 dies in my house.

    However, I have an App that will do that. But I expect that the algorithm that creates random numbers is not completely random, so there would be some round-off error..

    Are they completely cubic dice, or the rounded edges like playing Stratego? Need to know to calculate fringe effects.

    Larry
    DevilHorse
    random.org apparently creates truly random numbers, so there's that....which is nice.
    "We are not provided with wisdom, we must discover it for ourselves, after a journey through the wilderness which no one else can take for us, an effort which no one can spare us, for our wisdom is the point of view from which we come at last to regard the world." --M. Proust

  16. #16
    Quote Originally Posted by rsvman View Post
    random.org apparently creates truly random numbers, so there's that...which is nice.
    What? Truly random numbers?
    I'd like to see how that is demonstrated.
    To be useful you have to at least have boundaries/range/constraints[Integers] to make the output useful to you.

    I recall making a rudimentary random number generator, and it worked fairly well. Simple to implement, but someone had to create the algorithm. It had limits. Had to have limits. I've seen "tests" or questions on the internet, "which of these pictures (of data points in a 2-D field) looks random?" It is a tough thing to quantify.

    In the case of physical systems, similar types of discussion can apply to Chaos, but that has some interesting, and different bounds/properties.. But I digress.

    Larry
    DevilHorse

  17. #17
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    Quote Originally Posted by DevilHorse View Post
    What? Truly random numbers?
    I'd like to see how that is demonstrated.
    To be useful you have to at least have boundaries/range/constraints[Integers] to make the output useful to you.

    I recall making a rudimentary random number generator, and it worked fairly well. Simple to implement, but someone had to create the algorithm. It had limits. Had to have limits. I've seen "tests" or questions on the internet, "which of these pictures (of data points in a 2-D field) looks random?" It is a tough thing to quantify.

    In the case of physical systems, similar types of discussion can apply to Chaos, but that has some interesting, and different bounds/properties.. But I digress.

    Larry
    DevilHorse
    Which look random? The answer is almost always the 2-D plot that has patterns -- which, in case after case, form naturally and randomly. Which is why there are conspiracy theories.
    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

  18. #18
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    Quote Originally Posted by DevilHorse View Post
    What? Truly random numbers?
    I'd like to see how that is demonstrated.
    To be useful you have to at least have boundaries/range/constraints[Integers] to make the output useful to you.

    I recall making a rudimentary random number generator, and it worked fairly well. Simple to implement, but someone had to create the algorithm. It had limits. Had to have limits. I've seen "tests" or questions on the internet, "which of these pictures (of data points in a 2-D field) looks random?" It is a tough thing to quantify.
    That site claims they are, anyways. They are supposedly generated from atmospheric radio noise. There are others that use radioactive decay events and the like - I read about one somewhere (can't find it now) where someone made a machine that rolled actual dice (in a hopper or something) for hours a day and stored the results for retrieval as needed over the interwebs.


    Quote Originally Posted by sagegrouse View Post
    Which look random? The answer is almost always the 2-D plot that has patterns -- which, in case after case, form naturally and randomly. Which is why there are conspiracy theories.
    That would be my thought too, but the random.org site says a plot of their results looks like:
    randbitmap-rdo.jpg

  19. #19
    Quote Originally Posted by crimsondevil View Post
    That site claims they are, anyways. They are supposedly generated from atmospheric radio noise. There are others that use radioactive decay events and the like - I read about one somewhere (can't find it now) where someone made a machine that rolled actual dice (in a hopper or something) for hours a day and stored the results for retrieval as needed over the interwebs.




    That would be my thought too, but the random.org site says a plot of their results looks like:
    randbitmap-rdo.jpg
    Well, that explains everything right there!
    Who could argue with that picture?
    Although if you flip it, I believe it says, "I buried Paul."

    All seriousness aside..
    I do believe that if you took some random thing (like the time of day that you query for the input) and incorporate it into the seed of a random algorithm, you could get randomized results.
    Unless of course, they give you an API to extract random data and you sample it at a rate that is synchronized with their random input (like the time of day that you query for the input).
    But of course you would have other randomizing factors baked into the algorithm.

    At the root of it you have an algorithm, being the selection of previously thrown dice, or input to a seed in a function, there is some deterministic aspect which if known suggests a degree of non-randomness.

    But it is OK for Government Randomness or Conspiracy Theories (as suggested above).

    In the words of Edward Teller, "If something is Foolproof, the Fool is bigger than the Proof."

    Larry
    DevilHorse

  20. #20
    Quote Originally Posted by crimsondevil View Post
    That site claims they are, anyways. They are supposedly generated from atmospheric radio noise. There are others that use radioactive decay events and the like - I read about one somewhere (can't find it now) where someone made a machine that rolled actual dice (in a hopper or something) for hours a day and stored the results for retrieval as needed over the interwebs.




    That would be my thought too, but the random.org site says a plot of their results looks like:
    randbitmap-rdo.jpg
    I can't be the only one who sees the rabbit above the "...---.-.---..." ? Are you telling me it's just a coincidence that spells S O R O S in Morse Code and looks like a bat flying upside down? So who's the rabbit here? How clear do they have to make it?? Wake Up Sheeple!!!

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