# Probability Question

## Homework Statement

Suppose that the integer values 1 2 and 3 are written on each of three different cards. Suppose you do not know which number is the lowest (you do not know beforehand what the values on the cards are). Suppose that you are to be offered these cards in a random order. When you are offered a card you must immediately either accept it or reject it. If you accept a card, the process
ends. If you reject a card, then the next card (if a card remains) is offered. If you
reject the first two cards offered, then you must accept the final card.

(a) If you plan to accept the first card offered, what is the probability that you will
accept the lowest valued card?

(b) If you plan to reject the first card offered, and to then accept the second card if
and only if its value is lower than the value of the first card, what is the probability
that you will accept the lowest valued card?

## The Attempt at a Solution

A) The answer is obviously 1/3 for this question.

B) this is a conditional probability question. Given that P(E|F) = P(EF) / P(F) then i must first figure out what P(E) and P(F) stand for.

P(E) is the probability that i will accept the lowest card.
P(F) is the probability that the second card i choose is lower than the first rejected card.

If i reject the card with 1 on it, then i have no chance of selecting the lowest card next.
If i reject the card with 2 on it, then there is a 1/2 chance i will select the lowest card next.
If i reject the card with 3 on it, then i have a 1/2 chance i will select the lowest card next.

So there is a 2/3 * 1/2 = 1/3 chance that the second card is lower than the first rejected card, right?

I am not sure how to proceed after this.

CompuChip
Homework Helper
If you reject the card with 2 on it, then you will either pick the card with 1 next (which is lower, so you will accept it and have the lowest card) or you will pick that card with 3 (which is higher, so you will reject it and be left with the last one).
So in fact, the probability of getting the lowest card there is 1.

Ah right, dunno how i missed that.

So how do i calculate the chance that the second card will be lower? I have 0, 1 and 1/2 as the probabilities, depending on which card is rejected first.

Hi Rooski.
Let L be the event that you accept the lowest card.
Let $$F_{n}$$ be the event that the value of the first (and rejected) card is n, n=1,2,3.

Events $$F_{n}$$ are disjoint.

Try to write down p(L) using the total probability rule.

Assuming L is the event i accept the lowest card,

P(L) = P(L|A)P(A) + P(L|B)P(B) + P(L|C)P(C)

Where A,B,C denote cards 1,2,3 respectively. Is that right or am i off? It seems wring since P(A), P(B) and P(C) would all be 1/3.

Ok.
According to your notation A is the event that the first (and rejected) card is the lowest one.
So, if you reject A, the probability of accepting the lowest card is zero.
So P(L|A) = 0.

What about P(L|B) and P(L|C) ?

P(L|B) = 1 since you will reject 3 if it appears, or accept 1 when it appears.
P(L|C) = 1/2 since you will accept 2 if it appears or accept 1 when it appears.

Have i calculated P(A) P(B) and P(C) wrong?

A is the event that the first card is A. So pA=1/3.
The same apply to B and C.
Just calculate P(L).

P(L) = 0 * 1/3 + 1 * 1/3 + ½ * 1/3 = 3/6 = 1/2

So there is a 50% chance that we will end up with the lowest card if we reject the first random card.