[bushindo]

10 friends are going to a party. They check in their hats at the door. The hatchecker lost the receipts for those 10, so upon their leaving, he gave them back the 10 hats, but in a randomly shuffled order.

What is the chance that exactly 2 people have the correct hats?

Edited May 22, 2009 by bushindo

[Guest]

.287 or 28.7%

urs is a little harder then mine nice job tho

[Guest]

[spoiler=not sure but I like to try ]To start everyone has a 10% chance to get the right hat. So to have exactly 2 people to have their hat, and the rest to have different hats, the chance is 25%. ??

I have'nt taken any math class in over 8 years, so I am very rusty, but I used to enjoy math problems. If anyone knows any good websites so I can brush up on some formulas or whatever that would be great.

[Guest]

urs is a little harder then mine nice job tho

.287 or 28.7%

How did you set that up?

o start, 1 in 10, but if first person gets correct hat, then 1 in 9 for the next. But if first person gets incorrect hat, then the person that hat belongs to has no chance of getting the correct hat, so cant be figured as 1 in 9 for the next nor as 1 in 10. Can someone explain how to do this?

Edited May 22, 2009 by chicory

[Guest]

I think the coatchecker has a 1 in 45 chance to give exactly 2 people the right hat and the rest be wrong. So my new answer is .0222...%

[Guest]

my new answer that i have some confidence in

33%

just finished havent checked which is the perfect time to post your answer so you look like more of an idiot... its fun

if i think its right ill try to explain it (its really not that hard once i worked on it for ....)

[Guest]

Hi ... Ma first post ... I'm a B.Tech 3rd year guy from India , and my stream is Mathematics and Computing ... So, pretty much know the answer mathematically .. dont think there's a logical way to solve this

Firstly, the total no. of possibilities of hats getting exchnaged is 10!.

- Now, we first select the two guys who r gonna get the correct hat ... So ¹⁰C₂.

- So, among the remaining 8 guys, noone gets his own hat. The mathematical formula for this problem is

8!*[ 1 - (1/2) + (1/3) .... + (1/7)]

or something like that ... not sure the last one is precise.

So, the answer is ( 8!*[ 1 - (1/2) + (1/3) .... + (1/7)] ) * ( ¹⁰C₂ ) / (10!).

Too lazy to calculate that ....

[Guest]

nCr(10,2)*8!/(e*10!) = 18.39% (Using the approximation for derangements of 8 objects (8!/e).)

[bushindo]

nCr(10,2)*8!/(e*10!) = 18.39% (Using the approximation for derangements of 8 objects (8!/e).)

yongsu got it. Well done.

[Guest]

So, the answer is ( 8!*[ 1 - (1/2) + (1/3) .... + (1/7)] ) * ( ¹⁰C₂ ) / (10!).

Too lazy to calculate that ....

If you put in the factorials, that is the formula for derangements.

Dn = n!(1/0! - 1/1! + 1/2! - 1/3! + 1/4! ... 1/n!) where (1/0! - 1/1! + 1/2! - 1/3! + 1/4! ... 1/n!) can be approximated by e^-1, the inverse of the Euler number, 2.71828...

Because (1/0! - 1/1! + 1/2! - 1/3! + 1/4! ... 1/n!) is the power series for e^-1. So the number of derangements of n objects is approx. n!/e. This can be proven using

the principle of inclusion/exclusion.

[Guest]

yongsu got it. Well done.

Well, yeah its basically the same thing that I posted .... the term after 8! was nothing but the expansion of e ..

But I'm afraid yongsu made a mistake approximating it ... I mean That term equals e only when the number of objects are very large ... cant be done for 8 objects to get the exact probability

[bushindo]

Well, yeah its basically the same thing that I posted .... the term after 8! was nothing but the expansion of e ..

But I'm afraid yongsu made a mistake approximating it ... I mean That term equals e only when the number of objects are very large ... cant be done for 8 objects to get the exact probability

This approximation approaches the true value very fast. The true number of derangement for n=8 is 14833. yongsu's approximation is 8!/e = 14832.9. Very small difference.

Edited May 22, 2009 by bushindo

[Guest]

This approximation approaches the true value very fast. The true number of derangement for n=8 is 14833. yongsu's approximation is 8!/e = 14832.9. Very small difference.

well .. yeah ... guess so !!!

[Guest]

so heres how i did it wrong and wondering why

(10C2 *8! -10C3*7!+10C4*6!-10C5*5!+10C6*4!-10C7*3!+10C8*2!-10C9*1!)/10!

did i just do the math wrong, im going to do the math over again but i thought i did it right the logic was had 10 choose 2 to definitely be right arrange other 8. then subtract the ones that have been counted double by being at least three right and so on

[Guest]

Just thinking out loud, hoping somebody corrects any errors in my logic

successive incorrect hat is drawn, another person's chance of getting the correct hat drops to zero. So if person1 gets incorrect hat x, personx to whom the hat rightfully belongs has zero chance of getting his hat. He has a 100 percent chance of getting wrong hat. The other 8 people still have a 1 in 9 chance. But each time an incorrect hat i is handed out, the number of people with zero chance increases by one. person n odds = 1/(10-n). Will the odds change if 2 correct hats are given at the start, versus 2 incorrect hats? The pool of possibly correct hats is 10-n

The pool of possibly correct people remains 10-n also, versus 10-(i+1) after an incorrect hat is handed out. But it is the pool of hats that determine each persons odds, so the odds dont change whether the 2 correct picks occur early, later or spread apart.

[Guest]

oh my answer is off by a factor of 2 for some reason who knows y? anyone

anyway i think it does kind of matter because then what if the second person gets the first guys incorrect hat no ones chances have gone down theve actually gone decently up but maybe im wrong i chose to do it from the other direction

[Guest]

I was under the assumption the coatchecker had 1 chance at giving exactly 2 people the correct hat and the rest the wrong one. There are 45 possibilities of this happening, so 1/45=.0222%. Can someone breifly explain what I did?

[Guest]

i dont know exactly whatt you did but there are 45 pairs of two which you normally would use except for the dependency issue of seperate trials stops you from solving just with that.

the 1 in 45 would be the chance that two specific people got the right hats knowing that only two people got their hats. thats the only thing i can think of.

dont know if this helped but once im sure i understand the anwser myself ill try and help if you still need help

[Guest]

oh my answer is off by a factor of 2 for some reason who knows y? anyone

anyway i think it does kind of matter because then what if the second person gets the first guys incorrect hat no ones chances have gone down theve actually gone decently up but maybe im wrong i chose to do it from the other direction

Ow, Ill have to think about that.

I recognize ! and e but dont really know what they mean, just their names.

[bushindo]

I'll explain the steps for the answer that was given earlier by krashe and yongsu.

The total number of permutations that 10 hats could be given to the 10 gentlemen is 10*9*8*....*1 = 10!

Out of those, if we can compute the number of permutations where exactly 2 people get the correct hat (the other 8 each get the incorrect hat), and divide that by 10!, then that's the answer.

The number of permutations where exactly 2 people get the correct hat = (number of ways to choose a group of 2 out of 10)*(number of ways none of the remaining 8 gets the correct hat)

(number of ways to choose a group of 2 out of 10) = combination(10,2) = 10!/( 2!*8!)

(number of ways none of the remaining 8 gets the correct hat) = this is a famous problem in statistics called derangement. Essentially, this is equal to n!/e.

Putting all that together, we get

answer = combination(10,2)*8!/(e * 10!) = .18

[bushindo]

I was under the assumption the coatchecker had 1 chance at giving exactly 2 people the correct hat and the rest the wrong one. There are 45 possibilities of this happening, so 1/45=.0222%. Can someone breifly explain what I did?

You have the correct reasoning here. The number 45 corresponds to the number of ways 2 people can get the correct hat. However, you need to multiply that by the chance that the remaining 8 people would all have the wrong hat. So the answer should look something like this

(number of ways 2 people can get the correct hat = 45) * (chance that the other 8 all have the wrong hat)

[Guest]

You have the correct reasoning here. The number 45 corresponds to the number of ways 2 people can get the correct hat. However, you need to multiply that by the chance that the remaining 8 people would all have the wrong hat. So the answer should look something like this

(number of ways 2 people can get the correct hat = 45) * (chance that the other 8 all have the wrong hat)

I see it now. Thanks for the explanations.

[Guest]

so i forgot to add 10C10*0! to the end but thats one anyway my thing gives exactly twice and i was taught that is how your formula was derived so i have to be doing something so close to almost in the decent vacinity of corretedness

i get .3678794643 and ive got not a clue y the answer is off by a factor of 2 if .183939 is the answer

anyway i didnt see where ! C and e were explained so here

! is factorial it is a recursion method or more simply itself times all the numbers lower than it to zero (in positive integer cases) so 5!=5*4*3*2*1

e is well i dont remember it by anything other than e but its 2.71ish and its a number used for stuff its also related to hyperbolic sins but in my opinion its pretty useless unless u use it. like trig or physics

c or choose is more complicated (A)choose(B) is A!/B!(A-B)! so 10C2=10!/2!8!=10*9/2=45 so you have ten people choose 2 of them you have 45 ways to do this

anyway please help me with my problem this is gonna bug me forever

[Guest]

approximately 0.5

Edited May 22, 2009 by fruitless.tree

[Guest]

so i doubt anyone cares anymore but i found the error in my formula and i found the answer without approximation

i first i thought the formula was as below

(10C2*8!-10C3*7!+10C4*6!-10C5*5!+10C6*4!-10C7*3!+10C8*2!-10C9*1!+10C10*0!)/10!

but it doesnt count for the multiple recounts of each occurrence (just 1) so i made an adjustment

and it became

(10C2*8!-10C3*7!*3+10C4*6!*6-10C5*5!*10+10C6*4!*15-10C7*3!*21+10C8*2!*28-10C9*1!*36+10C10*0!*45)/10!

which becomes

10!(1/2-1/6*3+1/24*6-1/120*10+1/720*15-1/5040*21+1/40320*28-1/362880*36+1/3628800*45)/10!

which the 10! cancel and the answer is .1839409722

now i think this is the answer you guys got but i just like to know where formulas and approximations come from so here it is

10C2 finds two to give the correct hats. 8! arrange the others how you like. but this just calculates >=2 correct

so now you need to subtract the >=3 well 3's are counted 3C2 times so subtract 310C3*7!

but now the 4's have been counted 4C2-3*4C3=-6 so add 6 4's or 6*10C4*6!

but now the 5's have been counted 5C2-3*5C3+6*5C4=10 times so subtract 10*10C5*5!

and so on youll notice a trend of its the sum of all numbers below it anyway there it is gotta love math

[Guest]

so i doubt anyone cares anymore but i found the error in my formula and i found the answer without approximation

i first i thought the formula was as below

(10C2*8!-10C3*7!+10C4*6!-10C5*5!+10C6*4!-10C7*3!+10C8*2!-10C9*1!+10C10*0!)/10!

but it doesnt count for the multiple recounts of each occurrence (just 1) so i made an adjustment

and it became

(10C2*8!-10C3*7!*3+10C4*6!*6-10C5*5!*10+10C6*4!*15-10C7*3!*21+10C8*2!*28-10C9*1!*36+10C10*0!*45)/10!

which becomes

10!(1/2-1/6*3+1/24*6-1/120*10+1/720*15-1/5040*21+1/40320*28-1/362880*36+1/3628800*45)/10!

which the 10! cancel and the answer is .1839409722

now i think this is the answer you guys got but i just like to know where formulas and approximations come from so here it is

10C2 finds two to give the correct hats. 8! arrange the others how you like. but this just calculates >=2 correct

so now you need to subtract the >=3 well 3's are counted 3C2 times so subtract 310C3*7!

but now the 4's have been counted 4C2-3*4C3=-6 so add 6 4's or 6*10C4*6!

but now the 5's have been counted 5C2-3*5C3+6*5C4=10 times so subtract 10*10C5*5!

and so on youll notice a trend of its the sum of all numbers below it anyway there it is gotta love math

Thankyou Final, and the others who posted how they get their answers. I'm still trying to learn this, so I appreciate explanations and feedback.