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There are 4 boxes. One of them contains a lottery ticket

  1. Aug 24, 2013 #1
    There are 4 boxes. One of them contains a lottery ticket....

    There are 4 boxes. One of them contains a lottery ticket. Others have nothing. My probability that if i select and i win is 1/4 = 0.25. But before me someone selects a box, and i do not know if he has won or lost. What is the probability that if now i go to select i will win.
     
  2. jcsd
  3. Aug 24, 2013 #2
    P(you win) = P(you win| first win)*P(first win)+P(you win|first lose)*P(first lose)=0*1/4+1/3*3/4=1/4=0.25

    You use the law of total probability, because, the events "first win" and "first loose" are mutually exclusive, and exhaustive.
     
  4. Aug 24, 2013 #3
    does that mean that after first person, second person picks a box, then i select still probability will remain 0.25 ? Because for third time it is 3/4*2/3*1/2 = 0.25 ?
     
  5. Aug 24, 2013 #4

    D H

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    The probability remains at 1/4 if you don't know whether those previous one, two (or even three) people picked the winning box. That someone picked a box before you adds no new information. On the other hand, there is new information if you do know the outcome of those previous picks. If one of those people has picked the winning box (and you know it), your a posteriori probability drops to zero. If you know that none of them has picked the winning box, your odds of winning have just increased.
     
  6. Aug 25, 2013 #5
    Does the first player leave his chosen box open, or will he close it so that you cannot tell what his choice was?
     
  7. Aug 25, 2013 #6
    Yeah, but that was if he kept the box open, so you knew which box he had chosen(but not if there was a prize or not in this). As someone else mentioned it will mean something if he closes the box or not. If he closes the box, and you can choose the same box again, then the probability is 0,1875.
     
  8. Aug 25, 2013 #7

    HallsofIvy

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    There is an ambiguity here. Do you mean that the other person selects a box and that box is no longer available to select? Or that the other person selects a box, takes the ticket, if any, out of the box without you knowing and leaves it for you to select?

    In either case, the probability the first person finds the ticket is 1/4. In the first scenario, if he does, you have 3 boxes to choose from and none have the ticket. The probability the first person does not find the ticket is 3/4. There are three boxes left to choose from and you have probability 1/3 of finding it. The a-priori probability of you finding the ticket is (1/4)(0)+ (3/4)(1/3)= 1/4.

    In the second scenario, the if he finds the ticket, all 4 boxes are left but none have the ticket. If the first person does not find the ticket, all 4 boxes are left and you have probability 1/4 of finding the ticket. The a-priori probability of you finding the ticket is (1/4)(0)+ (3/4)(1/4)= 3/16.
     
  9. Aug 26, 2013 #8
    I agree.

    My reasoning was also strengthened by doing a mind exercise where 1,2,..,n boxes were being offered and all but the first box were blocked and only the 1st was left for me to choose. Well, each box entity had from the get go 1/n, why change now?
     
  10. Sep 8, 2013 #9
    The first player takes away that box . Only 3 are available after he chooses. :)
     
  11. Sep 8, 2013 #10
    The first player takes away that box . Only 3 are available after he chooses. I think this question has a lot of practical applications. :)
     
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