The Cup Problem

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hallewiblen

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Dec 9, 2021
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Hello,

I am currently trying to figure out the different probabilities of winning in an activity I completed.

There are four cups with one cup having money underneath it. If four people chose a cup, one after another, what is the probability that each person would choose the cup with the money underneath it.

For example, if person 1 does not select the right cup, then person 2 will select another cup, and if person 2 selects the right cup, the activity is over.

I believe that each person has an equal chance to win (25%).

My logic of this is:
  • Person 1 has a 1/4 chance of winning.
  • Person 2 then has a 1/4 chance of winning as that is the product of person 1's chance of losing (3/4) and person 2's chance of choosing the right cup (1/3).
  • Person 3 then has a 1/4 chance of winning as that is the product of person 1's chance of losing (3/4) and person 2's chance of losing (2/3) and person 3's chance of choosing the right cup (1/2).
  • Person 4 then has a 1/4 chance of winning as that is the product of person 1's chance of losing (3/4) and person 2's chance of losing (2/3) and person 3's chance of losing (1/2) and person 4's chance of choosing the right cup (1).
However, I am unsure if this is the correct answer and I would like either confirmation of this or an explanation of the correct answer.

Thank you!
 
hallewiblen said:
Hello,

I am currently trying to figure out the different probabilities of winning in an activity I completed.

There are four cups with one cup having money underneath it. If four people chose a cup, one after another, what is the probability that each person would choose the cup with the money underneath it.

For example, if person 1 does not select the right cup, then person 2 will select another cup, and if person 2 selects the right cup, the activity is over.

I believe that each person has an equal chance to win (25%).

My logic of this is:
  • Person 1 has a 1/4 chance of winning.
  • Person 2 then has a 1/4 chance of winning as that is the product of person 1's chance of losing (3/4) and person 2's chance of choosing the right cup (1/3).
  • Person 3 then has a 1/4 chance of winning as that is the product of person 1's chance of losing (3/4) and person 2's chance of losing (2/3) and person 3's chance of choosing the right cup (1/2).
  • Person 4 then has a 1/4 chance of winning as that is the product of person 1's chance of losing (3/4) and person 2's chance of losing (2/3) and person 3's chance of losing (1/2) and person 4's chance of choosing the right cup (1).
However, I am unsure if this is the correct answer and I would like either confirmation of this or an explanation of the correct answer.

Thank you!
Click to expand...
Your answer and your reasoning are correct.

You should not have submitted the same question twice:

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Dr.Peterson said:
Your answer and your reasoning are correct.

You should not have submitted the same question twice:

READ BEFORE POSTING

Welcome to FreeMathHelp.com! Please take the time to read the following before you make your first post. It will help you to get your math questions answered promptly and in the most helpful manner. A summary is available here, but please read the complete guidelines below at your earliest...
www.freemathhelp.com

Have patience. There is no paid staff waiting on-hand to give instant replies. Many of the volunteer tutors have "real" jobs, and they all have to sleep from time to time. The people viewing your posts may be fellow students. Don't be offended if you see "views" but no replies. It may take hours, even days, for a tutor qualified in your topic's area to reply. Also, new members won't see their first five posts appear right away, due to SPAM control (explained here).​
Click to expand...
Thank you Dr Peterson!

Sorry about posting it twice as when I posted it the first time, it disappeared and I thought I had not posted it right so then I reattempted later on. After I realised my mistake, I was unsure how to delete it.
 
Another way of thinking about this problem.
Suppose before anyone looked under any cups the following happened.
The 1st person picked their cup number.
The 2nd person picked a cup number different from the 1st person.
The 3rd person picked a cup number from the two cup number remaining.
The 4th person get the last cup.
Then the process you described above starts. They all have a 1/4 chance of winning.

This is the same as the classic key problem. You have a key ring with n keys on it and exactly one opens the door. What is the probability that the kth key (k< n) will be the one that opens the door?
 
-Pr(1st win) = 1/4
-Pr(2nd win; given 1st did not win) = Pr(1st did not win)* Pr(2nd person win)= 3/4*(1/3) = 1/4
-Pr(the 3rd person wins; given the 1st & 2nd did not win) = Pr(1st person did not win)*Pr(2nd person did not win)*Pr(3rd person wins)=
(3/4)*(2/3)*(1/2)= 1/4
-Pr(the 4th person wins; given the 1st, 2nd and 3rd did not win)= (3/4)*(2/3)*(1/2)*(1)= 1/4
 
BigBeachBananas said:
-Pr(1st win) = 1/4
-Pr(2nd win; given 1st did not win) = Pr(1st did not win)* Pr(2nd person win)= 3/4*(1/3) = 1/4
-Pr(the 3rd person wins; given the 1st & 2nd did not win) = Pr(1st person did not win)*Pr(2nd person did not win)*Pr(3rd person wins)=
(3/4)*(2/3)*(1/2)= 1/4
-Pr(the 4th person wins; given the 1st, 2nd and 3rd did not win)= (3/4)*(2/3)*(1/2)*(1)= 1/4
Click to expand...
This is the OP's method, and it is fine. But Jomo is pointing out (as I chose not to immediately) that there is a quicker way to see it.

Suppose each person points to a cup before any are turned over. As long as they are all different (no matter who went first in picking a cup), each has the same chance of having pointed to any particular cup. And each cup has an equal chance on containing the prize. So your chance of winning, as the fourth person to pick, is the same as anyone else's: 1/4.
 
I was not showing my method because it was faster but rather because (1) I think that it is important to know the method I showed and (2) to always be able to check your result by solving the problem a different way (this was drilled into my brain by Dr Peterson-thanks).
 
I wonder whether one person could choose the same cup as the previous one did. If not, the later will have fewer chances.
 
brocule said:
I wonder whether one person could choose the same cup as the previous one did. If not, the later will have fewer chances.
Click to expand...
No, we're not allowing that. Why do you think it might happen?

And the whole point of the question is that the later people do have fewer choices, and so naively appear to have less chance of winning. But it turns out that everything cancels out and they are all the same. One might say that this is because, having fewer choices, their chance of winning given that the first didn't is actually greater! (That is, the probability that the second wins is 1/3 of 3/4, which is again 1/4.)
 
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