Question

Three people are issued movie tickets with seat assignments, however, it was dark and the three people decide to just sit on the three seats without ascertaining if it is the correct numbered seat. What is the probability they are all sitting on the correct seats?

Answer 1

I would assume we would need to know how many seats are there O_o .

Answer 2

I would also assume that you would need to know then umberof seats

Answer 3

A lot needs to be thought off. You need to take into account the probability of the first guy of choosing the right seat in order for the second. I would say 1/6 is the answer.

Answer 4

1/3 chance of the first times 1/2 chance for the second?

Answer 5

@ChmE . How? You know nothing except that 3 people bough tickets. The probability changes as the number of seats changes.

Answer 6

Isn't it:
1/3!

Answer 7

I assumed the number of seats is 3

Answer 8

decide to just sit on the three seats. When I read this I interpret it as they see the 3 seets they bought but don't know their individual seat assignment

Answer 9

@sauravshakya Why? You don't know how many seats are there.

Answer 10

I guess. I am not sure if you can make that assumption though...

Answer 11

So the first guy walks up and chooses. He has a 1/3 chance. The second walks up and he now has a 1/2 chance. The last is grandfathered in given the first two were crrect

Answer 12

Because they have three tickets of three sits and the question said that they sit on the three sits.

Answer 13

What if there was 50 seats? it would be a 1/50 chance.

Answer 14

3/50 sorry.

Answer 15

1/50 time 1/49 times 1/48

Answer 16

If there was 50 sits then
1/50P3

Answer 17

Same as @ChmE did

Answer 18

I guess I just have to be intuitive in this question i guess :/ .

Answer 19

My interpretation of the problem is they bought 3 tickets. They can't see the individual seat assignment illustrated by "without ascertaining if it is the correct numbered seat" But they do see THE 3 seats ilustrated by "three people decide to just sit on THE three seats"

Answer 20

Ohh. If it just said "on" instead of "the" the question would change completely.

Answer 21

I get it.

Answer 22

@lgbasallote you've been silent. What are your thoughts?

Answer 23

sorry i was away

Answer 24

anyway..the number of seats doesn't matter

Answer 25

Wouldnt the probably just be then 1/nP3 ?

Answer 26

Do you know the answer or are you looking for help with this solution

Answer 27

Where n is the number of seats?

Answer 28

yes i do @ChmE

Answer 29

My answer is 1/6. Correct or no?

Answer 30

why so?

Answer 31

I've already explained it above

Answer 32

you were inferencing above that the number of seats affect the outcome

Answer 33

It has to.

Answer 34

@ChmE's answer should be correct... 1 in 6. The question doesn't provide the number of seats so it's assumed to be 3.\[\frac{1}{3}*\frac{1}{2}=\frac{1}{6}\]

Answer 35

Chance of first guy chooses correct 1/3. Given he is correct, there are 2 seats left and the chance for the second guy to be correct is 1/2. Given he is also correct, the last guy has a 1/1 chance of being correct

Answer 36

If the number of seats were higher, it would obviously affect the odds of each selection above.

Answer 37

oh so that's why

Answer 38

I got a question for you now @lgbasallote

Answer 39

shoot

Answer 40

@lgbasallote: What if the number of available seats was 5?\[\frac{1}{5}*\frac{1}{4}*\frac{1}{3}=\frac{1}{60}\]

Answer 41

Just to illustrate...

Answer 42

why not include 1/2?

Answer 43

You are on a game show. There are 3 doors. Behind one of the doors is a car. You choose door number 1. The host walks up to door 2 and reveals nothing behind it. He asks you if you want to change your guess. Is it in your best interest to change doors?

Answer 44

that's monty hall's problem

Answer 45

Ohh I remember that one. Behind two doors is a goat. Behind 1 is a car.

Answer 46

that urban myth states that you switch

Answer 47

there is no myth. Just an extra 33% chance of getting it correct if you switch.

Answer 48

It's shown in the movie "21"

Answer 49

all math for me is urban myth

Answer 50

haha

Answer 51

At that point in time you have a 50/50 chance of choosing the right door...switching doors won't change your odds *at that point in time*.

Answer 52

Wrong. In the beginning you have 1/3 chance is in door 1 and 2/3 chance of being in door 2 and 3. After he reveals door 2 is wrong, all 2/3 chance gets put on door 3. It is accounting for variable change

Answer 53

u.r.b.a.n m.y.t.h

Answer 54

I guess if you say at that exact point in time looking at only those 2 doors and forgetting there were 3. You would be correct

Answer 55

You won't be able to convince that switching doors will increase your chances at that point in time. It depends on how you look at it...but ultimately it boils down to a 50/50 chance that you were right at that point.

Answer 56

It's a tough concept to grasp, but in my stats class a couple years ago, we spent a couple days talking about these types of problems.

Answer 57

Yea, I've dealt with it before as well.

It's like trying to roll, say 23, on a theoretical die with 100 sides. With each successive roll your overall odds of rolling a 23 goes up in theory. However, each roll is a completely separate event with the same exact odds of hitting a 23.

Answer 58

The (false) assumption here is that previous rolls affect future rolls.

Answer 59

The die example you gave is independent events. My example the events are not independent

Answer 60

How are they not independent in your example?

Answer 61

I see it as going from 1 in 3 to 1 in 2...so I guess you could say they are not independent.

Answer 62

You chance of door 1 is 1/3. He tells you door 2 is wrong and you see, oh hey, if door 2 is wrong and my chance of 2 or 3 was 2/3. Then because of the event where he said door 2 is wrong, door 3 now has all the 2/3 chance. i should switch.