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Thread: Pretty tricky probability "dice" question

  1. #11
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    Quote Originally Posted by bennyJ View Post
    Is this question beyond everyone's powers?
    Hmm? Are you simply waiting for somebody to complete it for you?

    You were given a suggestion, in post #2.

    You were asked if you tried that suggestion, in post #4. You haven't responded to that question, yet.

    There's also a lengthy reply, in post #5. Did you see it?
    "English is the most ambiguous language in the world." ~ Yours Truly, 1969

  2. #12
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    Quote Originally Posted by bennyJ View Post
    Is this question beyond everyone's powers?
    Nope... we are waiting for you to show your work....
    “... mathematics is only the art of saying the same thing in different words” - B. Russell

  3. #13
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    Quote Originally Posted by j-astron View Post
    I would take an inventory of possible outcomes and see which ones lead to a win. What are the possible outcomes?

    - win on first throw
    - lose on first throw
    - have to continue after first throw

    These events are all mutually-exclusive, and cover all the possibilities, so their probabilities must add up to 1 (can you see why?)

    P(win on 1st) + P(lose on 1st) + P(continue) = 1

    What's the probability of winning? It's the sum of the probabilites of all the (mutually-exclusive) outcomes that lead to winning:

    P(win) = P(win on 1st) + P(win if continuing)

    Can you compute these probabilities? P(win on 1st) is straightforward, because it's just the probability of getting a couple of specific dice throws on your first try, and as Subhotosh said, that's just counting. Helpful things for that include

    - what's the total number of outcomes for any given dice throw? 12 possibilities for the value of the first die, and 12 possibilities for the value of the second die. So the total number of outcomes = ?
    - what Subhotosh said about constructing a 12x12 grid to count up the frequency of different sums more easily

    P(win if continuing) is more interesting to compute...

    I'm also wondering if it's easier to compute P(lose) and then take 1 - P(lose) to solve the problem. These are some thoughts on how to approach it.
    Well, the odds of getting a point on the first throw appear to be 7/72 since there's a 1/12 chance of getting a 13 (12/144) and a (2/144) chance of getting a 23. The chances of losing on the first throw by getting a 2, 3, or 24 appear to be 4/144, or 1/36.

  4. #14
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    So, is the probability of making a point on the first throw 1 - 7/72 - 2/72 = .875% ?? If so, how does one calculate the probability of ultimately winning at the start of the game?

  5. #15
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    Quote Originally Posted by bennyJ View Post
    Well, the odds of getting a point on the first throw appear to be 7/72 since there's a 1/12 chance of getting a 13 (12/144) and a (2/144) chance of getting a 23. The chances of losing on the first throw by getting a 2, 3, or 24 appear to be 4/144, or 1/36.
    Yeah, both of these probabilities you computed are correct. P(win on first throw) = 7/72

    You are also correct that the probability of "his point" is 126/144 = 0.875.

    What is P(win on second throw)?

    What is P(win on third throw)?

    What is P(win on fourth throw)?

    You have to add all of these up...forever. Can you think of what kind of sequence you're going to end up with that has infinitely-many terms that add up to a finite sum?

    Hint: P(win on second throw) = P(continue on 1st AND his point on second) = P(continue on 1st)*P(his point on second), where you take the product since these are independent events.

    The Probability of winning on subsequent throws will consist of a similiar product of probabilities (continue n-1 times and get his point on the nth time). Except one more factor will be added on to that product in each subsequent term...

    EDIT: This problem was a lot of fun! Slightly tricky until you see the pattern.

  6. #16
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    Quote Originally Posted by j-astron View Post
    Yeah, both of these probabilities you computed are correct. P(win on first throw) = 7/72

    You are also correct that the probability of "his point" is 126/144 = 0.875.

    What is P(win on second throw)?

    What is P(win on third throw)?

    What is P(win on fourth throw)?

    You have to add all of these up...forever. Can you think of what kind of sequence you're going to end up with that has infinitely-many terms that add up to a finite sum?

    Hint: P(win on second throw) = P(continue on 1st AND his point on second) = P(continue on 1st)*P(his point on second), where you take the product since these are independent events.

    The Probability of winning on subsequent throws will consist of a similiar product of probabilities (continue n-1 times and get his point on the nth time). Except one more factor will be added on to that product in each subsequent term...

    EDIT: This problem was a lot of fun! Slightly tricky until you see the pattern.
    Forever sounds daunting, but I would imagine that you're alluding to a converging infinite series.
    Last edited by bennyJ; 02-15-2018 at 04:49 PM.

  7. #17
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    Quote Originally Posted by bennyJ View Post
    Forever sounds daunting, but I would imagine that you're alluding to a converging infinite series.
    I was!

    Did you write out P(win on 2nd throw), P(win on third throw), and P(win on fourth throw)?

    It should be enough to establish a pattern

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    Quote Originally Posted by j-astron View Post
    I was!

    Did you write out P(win on 2nd throw), P(win on third throw), and P(win on fourth throw)?

    It should be enough to establish a pattern
    Since I'm not completely confident I wouldn't fudge one of the small calculations along the way (or even set them up properly), do you think there would be any way to calculate the win probability at the start of the game knowing the probability of making a point on the first throw (87.5%) and the probability of losing on any one of the subsequent throws (8.333333% per throw, 1/12, or the chances of getting 13 on each subsequent throw of the die)? I guess, by proxy, you know that the chances of winning on a subsequent throw or having to throw again would be 91.666666%).


    On this or any subsequent throw the player loses if he gets the sum 13 and wins if he gets his point but must throw both dice again if any other sum occurs. The player continues until he either wins or loses. To the nearest percent, what is the probability at the start of any game that a dice thrower will win?

  9. #19
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    Quote Originally Posted by bennyJ View Post
    Since I'm not completely confident I wouldn't fudge one of the small calculations along the way (or even set them up properly), do you think there would be any way to calculate the win probability at the start of the game knowing the probability of making a point on the first throw (87.5%) and the probability of losing on any one of the subsequent throws (8.333333% per throw, 1/12, or the chances of getting 13 on each subsequent throw of the die)? I guess, by proxy, you know that the chances of winning on a subsequent throw or having to throw again would be 91.666666%).
    As I mentioned earlier, you can try to calculate the total probability of losing instead, and take 1 - P(lose). Either way, you have to set up "small calculations", one for each round. Just do it! Do it the way I suggested first, and post your work here. We can point out any mistakes and provide hints along the way. It just requires being a bit methodical. In the end, if computing the win probability directly, and excluding the first round, there are literally only two numbers that go into each term.
    Last edited by j-astron; 02-19-2018 at 11:23 AM. Reason: typo

  10. #20
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    Quote Originally Posted by j-astron View Post
    As I mentioned earlier, you can try to calculate the total probability of losing instead, and take 1 - P(lose). Either way, you have to set up "small calculations", one for each round. Just do it! Do it the way I suggested first, and post your work here. We can point out any mistakes and provide hints along the way. It just requires being a bit methodical. In the end, if computing the win probability directly, and excluding the first round, there are literally only two numbers that go into each term.
    I must have made some mistakes in calculating the following outcomes because my numerator is more than 144 when you add them all up:

    2 - 2/144
    3 - 2/144
    4 - 4/144
    5 - 4/144
    6 - 8/144
    7 - 7/144
    8 - 8/144
    9 - 8/144
    10 - 10/144
    11 - 10/144
    12 - 10/144
    13 - 12/144
    14 - 12/144
    15 - 12/144
    16 - 12/144
    17 - 8/144
    18 - 8/144
    19 - 6/144
    20 - 6/144
    21 - 4/144
    22 - 4/144
    23 - 2/144
    24 - 2/144

    Whatever your point would be, I would guess you would take 1 - (probability of particular point) - 12/144 (lose on 13)
    Last edited by bennyJ; 02-19-2018 at 10:21 PM.

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