0
$\begingroup$

Let's say there are 4 people and 5 boxes. One of the boxes has a reward inside while the others are empty.

The 4 people will take turns to remove a box with the first having 5 boxes to choose from and the last only 2 boxes to choose from. Who is likely to remove the box with the reward inside?

What is the formula to solve this? I have no idea how to put it on paper. By intuition I guess that the 1st person has the better odds as the more boxes that are taken the more likely that one of them had the reward.

So, what are Person A, B, C & D odds? How to put it on paper and what is the name of the formula to solve this?

  • 1
    Any box has an equal chance of being the winning box, thus each of the people stand a $\frac 15$ chance of getting the reward. – 2017-02-21
  • 0
    And you can derive lulu's correct comment as follows. The first person clearly has p=0.2 of choosing the winning box, and therefore 0.8 of not choosing it and leaving 4 boxes. So the second person has 0.25 chance of choosing the winning box **iff** the first person didn't already choose it, i.e., p=0.8*0.25=0.2 again. Similarly, for the third person, p=0.8*0.75*0.3333=0.2 yet again, etc. And I don't know "the name of the formula". But, hey, let's call it "lulu" (sounds more fun than "john"). – 2017-02-21

2 Answers 2

0

What is the formula to solve this? I have no idea how to put it on paper. By intuition I guess that the 1st person has the better odds as the more boxes that are taken the more likely that one of them had the reward.

Your intuition is wrong.

Person A has a choice of 5 boxes, 1 of which is the winning box. $$P(A)=\tfrac 15$$

Person B has a choice of 4 boxes, 1 is the winning box if person A did not choose it else none are. $$P(B) = P(B\mid A^\complement)P(A^\complement)+P(B\mid A)P(A) \\= \tfrac 14\cdot\tfrac 45+ 0\cdot\tfrac 15 \\ = \tfrac 15$$

This is called the Law of Total Probability, and we can continue in a like vein with C and D.

However, as lulu commented: there are five boxes and everyone of the four people has equal probability of selecting any one of them, so the probability for any particular person selecting the winning box is $1/5$.

1

Not a probability buff, so let me know if correct or where wrong.

A : Probability of success = 1/5,

B : Probability of success: 4/5 × 1/4 = 1/5,

C: Probability of success: 4/5 × 3/4 × 1/3 = 1/5,

D: Probability of success: 4/5 × 3/4 × 2/3 × 1/2 = 1/5.

  • 0
    Yeap. That is okay. That is the probability of the $k-1$ earlier picks being from the 4 wrong boxes and then the person picking the winner. $$\mathsf P(W=k) ~=~ \dfrac{{}^4\mathrm P_{k-1}}{{}^5\mathrm P_k}=\frac 15$$ – 2017-02-21
  • 0
    Thanks for reading and checking. – 2017-02-21