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Thread: Pretty tricky probability "dice" question

  1. #1
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    Pretty tricky probability "dice" question

    Suppose a modified version of the dice game craps is played with two regular (i.e., perfectly symmetrical) dodecahedra. Each die has its sides numbered from 1 to 12 so that after each throw of the dice the sum of the numbers on the top two surfaces of the dice would range from 2 to 24. If a player gets the sum 13 or 23 on his first throw (a natural), he wins. If he gets 2, 3, or 24 on his first throw (craps), he loses. If he gets any other sum (his point), he must throw the dice again. 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?

  2. #2
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    Not that tricky. Draw yourself a 12x12 grid and start counting!
    "Unique Answers Don't Care How You Find Them." - Many may have said it, but I hear it most from me.

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    Does anyone have any ideas on how to approach this problem?

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    Quote Originally Posted by bennyJ View Post
    Does anyone have any ideas on how to approach this problem?
    Did you apply the "idea" provided above?
    “... mathematics is only the art of saying the same thing in different words” - B. Russell

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    Quote Originally Posted by bennyJ View Post
    Does anyone have any ideas on how to approach this problem?
    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.

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    Silly question:
    why not use 2 stacks of 12 cards, each labelled 1 to 12,
    and pick a card at random from each?
    I'm just an imagination of your figment !

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    Quote Originally Posted by Denis View Post
    Silly question:
    why not use 2 stacks of 12 cards, each labelled 1 to 12,
    and pick a card at random from each?
    Yeah, that would be a way to simulate the game, but how would it help solve the problem, short of playing the game thousands of times and seeing what the win ratio converges to?

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    Quote Originally Posted by Denis View Post
    Silly question:
    why not use 2 stacks of 12 cards, each labelled 1 to 12,
    and pick a card at random from each?
    Does that take into account the probability of drawing again (potentially infinitely) if you don't win on the first try?
    Last edited by bennyJ; 01-22-2018 at 12:18 AM.

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    Quote Originally Posted by bennyJ View Post
    Does that take into account the probability of drawing again (potentially infinitely) if you don't win on the first try?
    Sorry if that confused you Benny.
    All I meant was throwing a 12sided die is equivalent
    to picking at random a card from a 12card deck.
    Disregard my post. Thanks.

    I simply meant that my question was probably silly.
    Clearer would have been "May I ask a silly question".
    Sorry about the confusion I created.
    Last edited by Denis; 01-22-2018 at 09:22 PM. Reason: added apology.
    I'm just an imagination of your figment !

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    Is this question beyond everyone's powers?

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