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

  • Context: Undergrad 
  • Thread starter Thread starter sahil_time
  • Start date Start date
  • Tags Tags
    lottery
Click For Summary

Discussion Overview

The discussion revolves around a probability problem involving four boxes, one of which contains a lottery ticket. Participants explore the implications of prior selections on the probability of winning when a new selection is made, considering various scenarios and the information available to the chooser.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants assert that the probability of winning remains at 1/4 if the outcome of previous selections is unknown.
  • Others argue that knowing the outcome of prior selections can change the probability, potentially dropping it to zero if a winning box was chosen.
  • One participant suggests that if the first player closes the box after their selection, the probability of winning changes, leading to a calculation of 0.1875.
  • Another participant raises a question about whether the first player's box remains available for selection, which affects the probability calculations.
  • Some participants discuss the law of total probability and how it applies to the scenario, emphasizing the mutual exclusivity of events.
  • A later reply indicates that the probability calculations differ based on whether the first player takes the box or leaves it available for the next chooser.

Areas of Agreement / Disagreement

Participants do not reach a consensus; multiple competing views remain regarding how prior selections affect the probability of winning, and the discussion includes various interpretations of the problem's conditions.

Contextual Notes

There is ambiguity regarding whether the first player's box is available for the next chooser and whether the outcome of the first selection is known. These factors influence the probability calculations and assumptions made by participants.

sahil_time
Messages
108
Reaction score
0
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.
 
Physics news on Phys.org
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.
 
bobby2k said:
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.

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 ?
 
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.
 
Does the first player leave his chosen box open, or will he close it so that you cannot tell what his choice was?
 
sahil_time said:
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 ?

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.
 
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.
 
D H said:
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.

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?
 
jostpuur said:
Does the first player leave his chosen box open, or will he close it so that you cannot tell what his choice was?

The first player takes away that box . Only 3 are available after he chooses. :)
 
  • #10
HallsofIvy said:
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.

The first player takes away that box . Only 3 are available after he chooses. I think this question has a lot of practical applications. :)
 

Similar threads

  • · Replies 45 ·
2
Replies
45
Views
5K
  • · Replies 25 ·
Replies
25
Views
2K
  • · Replies 47 ·
2
Replies
47
Views
5K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
8K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K