probability

[merlinmynx]

There are 100 people who had left their shirts in a pile, while having a shower. They all came back and picked up a shirt at random,
What is more likely :
a) everyone picked a wrong shirt; or
b) at least 1 person picked up his own shirt.

Please give a brief explanation of the answer.

[magicdice]

imagine there were ten people, A-J(principles are same & it's a lot easier to explain)

there are two ways of going about this - firstly that they all return from the shower one after the other & pick a shirt, and secondly that they all pick a shirt at the same time but somehow none pick the same one. the first alternative is a better model of the situation

 

for the first situation, A has a 9 in 10 chance of picking the wrong shirt. then, B has an 8 in 9 chance of picking the wrong shirt. C is 7/8, D 6/7, E 5/6 etc etc so we end up with

9x8x7x6x5x4x3x2x1 divided by 10x9x8x7x6x5x4x3x2 which leaves us with 1/10

for the second situation, if at least one person picks up their own shirt then that means that any number of people can pick up the right shirt except for ZERO people. the above gives us the odds of nobody getting the right shirt, so 1 - 0.1 = 0.9

0.9 > 0.1 so it's nine times as likely that at least one person has picked up the right shirt

 

secondly, each has a 9/10 chance of picking the wrong shirt, so we have 9/10*9/10*9/10*9/10 etc to give us 0.9^10, which is 0.3487

as before, subtract this from one to get the likelihood of at least one of them getting the right shirt this time - it works out as about 2/3. so you're twice as likely to have at least one person get the right shirt even in this model of the situation!

 

PS next time try the science pages - that's where the maths students like myself loiter so you're more likely to get a good, quick response from there

[Marsh]

I didn't agree with part of magicdice's logic in his first approach (see separate posting below) and ran into a snag with the second, when I tackled the problem differently.

Assume a is the shirt belonging to A, b the shirt belonging to B, etc.
For two people you have only two possible outcomes:
Aa:Bb   (2)
Ab:Ba   (0)
In the first line A has picked up his own shirt; in the second he has picked up B's shirt.  (The number in brackets is the number of people with the right shirt.)
The chance of neither getting the right shirt is 1/2.

For three people you have six possibilities:
Aa:Bb:Cc (3)
Aa:Bc:Cb (1)
Ab:Ba:Cc (1)
Ab:Bc:Ca (0)
Ac:Bb:Ca (1)
Ac:Ba:Cb (0)
The chance of none getting the right shirt is 2/6 or 1/3.

For four people, twenty four possibilities, which I'll post seperately so that my counting can be checked ifd anybody wants.  The chance of none getting the right shirt comes out as 9/24  or 3/8.

I haven't checked  5 people but so far none of the answers fit 0.9^10 for the chance of nobody getting the right shirt and only the first two fit 1/10.

It still seems that for 100 people it is more likely than not that at least one person will get the right shirt.

Could you do us a favour?  If you are told that the no-shirt figure is (99/100)^100, could you ask why that approach doesn't work for 2, 3 and 4 people?  If you are given an approach that copes with 2, 3 and 4, would you post it so that we can know too?

Let us know how it goes.

[Marsh]

For the keen ones, here are the twenty-four possibilities.

Aa:Bb:Cc:Dd (4)
Aa:Bb:Cd:Dc (2)
Aa:Bc:Cb:Dd (2)
Aa:Bc:Cd:Db (1)
Aa:Bd:Cb:Dc (1)
Aa:Bd:Cc:Db (2)

Ab:Ba:Cc:Dd (2)
Ab:Ba:Cd:Dc (0)
Ab:Bc:Ca:Dd (1)
Ab:Bc:Cd:Da (0)
Ab:Bd:Ca:Dc (0)
Ab:Bd:Cc:Da (1)

Ac:Ba:Cb:Dd (1)
Ac:Ba:Cd:Db (0)
Ac:Bb:Ca:Dd (2)
Ac:Bb:Cd:Da (1)
Ac:Bd:Ca:Db (0)
Ac:Bd:Cb:Da (0)

Ad:Ba:Cb:Dc (0)
Ad:Ba:Cc:Db (1)
Ad:Bb:Ca:Dc (1)
Ad:Bb:Cc:Da (2)
Ad:Bc:Ca:Db (0)
Ad:Bc:Cb:Da (0)

[Marsh]

Hi magicdice,
My problem with your first approach is this
"For the first situation, A has a 9 in 10 chance of picking the wrong shirt then B has an 8 in 9 chance of picking the wrong shirt".
If A picked up B's shirt in the first place, then B has a 9/9 chance.  Adding that possibility in, I make it that if A picks the wrong shirt, the probability of B getting it wrong is 1/10 x 9/9 + 9/10 x 8/9 or 9/10. 

It looks as though it's heading for 0.9^10, like your second approach but I haven't pursued that.  I thought that your second approach was right - until I started playing with the alphabet.

 

[(2^2)n3!]

There is a 99/100 chance of 1 person choosing the wrong shirt and (99/100)^100 chance of 100 people choosing the wrong shirt (0.336). The chance of 1 or more people choosing the correct shirt is 1-0.366.