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Thread: really need some help solving this! probability

  1. #1
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    Question really need some help solving this! probability

    There are 12people. Each person buys 1 present each andwraps it and puts it under the tree (each person will know what present theywrapped).

    All names of the 12people go into a hat. The first persondrawn has the option to pick a present from under the tree (whilst there are 12presents – you know your one, so you have 11 to pick from)

    This goes on with therest of the people.

    What are the chancesof the last person being left with their present that they wrapped.
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  2. #2
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    Re: really need some help solving this! probability

    So, the first 11 people pick the first 11 people's presents, but not their own. This was out of 11 possible gifts. That's

    \dfrac{D_{11}}{11!} =\dfrac{!11}{11!}\approx 36.79%

    Where $D_{11} $ is the 11th derangement number.
    Last edited by SlipEternal; Nov 15th 2017 at 08:57 PM.
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  3. #3
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    Re: really need some help solving this! probability

    That's not quite right.
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    Re: really need some help solving this! probability

    It will be

    \dfrac{!11}{!12-1}=\dfrac{1}{12}
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    Re: really need some help solving this! probability

    Quote Originally Posted by c2017 View Post
    There are 12people. Each person buys 1 present each and wraps it and puts it under the tree (each person will know what present they wrapped). All names of the 12people go into a hat. The first person drawn has the option to pick a present from under the tree (whilst there are 12 presents – you know your one, so you have 11 to pick from)
    This goes on with the rest of the people.
    What are the chances of the last person being left with their present that they wrapped.
    This is a poorly written question. The way it is written everyone could possibly get the gift that s/he warped.
    There are $12!$ ways to distribute the gifts. There are $11!$ ways in which the last person gets her/his on gift.
    Can you finish?
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    Re: really need some help solving this! probability

    Quote Originally Posted by Plato View Post
    This is a poorly written question. The way it is written everyone could possibly get the gift that s/he warped.
    No, the first post specifically said that each person knows his own gift so will not pick it. As long as there is more than one gift, a person will NOT choose his own gift.

    There are $12!$ ways to distribute the gifts. There are $11!$ ways in which the last person gets her/his on gift.
    Can you finish?
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    Re: really need some help solving this! probability

    The total number of outcomes is given by the number of derangements where only the last person can get their original gift. The specific outcomes are where the last person does get their original gift. We will use Derangement Numbers to calculate. I came close in my previous post, but I was still wrong. Here, I actually figure it out.

    For the count of the numerator, we know the last person gets his own gift, so we assign that. That leaves 11 gifts to permute, but no person can get their own gift. That is the number of derangements of 11 people: D_{11} = !11.

    For the denominator, we can have that the last person gets his own gift, but none of the other people do, or no one gets their own gift. That is D_{11}+D_{12} = !11+!12.

    So, the probability that the last person receives his own gift is:

    \dfrac{!11}{!11+!12} = \dfrac{14684570}{190899411}

    Wolfram|Alpha: Computational Knowledge Engine

    This assumes that outcomes are equally likely which I did not calculate to see if it is true or not. This is the probability that of all of the possible outcomes being equally likely that the outcome you are considering occurs.

    Edit: I just ran it through a simulation 1000 times, and it gave me 8.7% chance of the last person getting his own gift. Next 1000 tries gave 7.5% chance. I think my calculation is correct this time.
    Last edited by SlipEternal; Nov 16th 2017 at 09:41 AM.
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    Re: really need some help solving this! probability

    Quote Originally Posted by HallsofIvy View Post
    No, the first post specifically said that each person knows his own gift so will not pick it. As long as there is more than one gift, a person will NOT choose his own gift.
    I am fully aware that the poster said that each person knows which gift is the one s/he supplied. But nowhere did it say that a person cannot or must not choose the gift s/he supplied, just should not.
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    Re: really need some help solving this! probability

    Quote Originally Posted by Plato View Post
    I am fully aware that the poster said that each person knows which gift is the one s/he supplied. But nowhere did it say that a person cannot or must not choose the gift s/he supplied, just should not.
    Actually, the poster does not say that the person should not take their own. Instead, the poster says that s/he has fewer choices (out of the 12 gifts, s/he has 11 choices). This seems to be an explicit statement that the person cannot choose his/her own gift.
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    Re: really need some help solving this! probability

    Quote Originally Posted by SlipEternal View Post
    Actually, the poster does not say that the person should not take their own. Instead, the poster says that s/he has fewer choices (out of the 12 gifts, s/he has 11 choices). This seems to be an explicit statement that the person cannot choose his/her own gift.
    I would argue that is not explicit. There are several versions of this question. I think I saw something similar in Marx's text. It may have been something like this. Twelve men place a sum of money into a seals envelope. Each writes his name where the return address should be. The envelops are placed in a container. One member the mixes the content. He then drawn out an envelope (using alphabetical order) gives it to the first member if that member's name is on the envelope. If then the envelope is returned to the container and the process continues.
    If the last envelope in the container belongs to the man doing the drawing, he then must exchange with any other member.
    What is the probability that the member giving out the envelopes must do an exchange?
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  11. #11
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    Re: really need some help solving this! probability

    Quote Originally Posted by Plato View Post
    I would argue that is not explicit. There are several versions of this question. I think I saw something similar in Marx's text. It may have been something like this. Twelve men place a sum of money into a seals envelope. Each writes his name where the return address should be. The envelops are placed in a container. One member the mixes the content. He then drawn out an envelope (using alphabetical order) gives it to the first member if that member's name is on the envelope. If then the envelope is returned to the container and the process continues.
    If the last envelope in the container belongs to the man doing the drawing, he then must exchange with any other member.
    What is the probability that the member giving out the envelopes must do an exchange?
    Without further input from the OP, we may never know the intent. My solution in post #7 is correct if the OP is looking for permutations where none of the first 11 people were allowed to take their own gift, but the last one does. Your solution in post #5 is correct if the OP is looking for any permutation for the distribution of presents resulting in the last taker to take his/her own gift.
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    Re: really need some help solving this! probability

    Boy oh boy...much ado about nuttin' Billy Shakespeare would say!
    Thanks from romsek
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    Re: really need some help solving this! probability

    Quote Originally Posted by DenisB View Post
    Boy oh boy...much ado about nuttin' Billy Shakespeare would say!
    Have you ever taught this material?
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