Tough probability riddle

http://www.able2know.org/forums/viewtopic.php?p=2987106#2987106

Ugly. here are the solutions per the rules for group sizes 1 -> 6 people

S(1) = 100% (problem collapses)
S(2) = 0%
S(3) = 1/3
S(4) = 2/6 = 1/3
S(5) = 7/14 = 1/2
S(6) = 14/28 = 1/2

Fun, fun, fun!

Einstein's stumped. I've been reading the other thread.

Even if N was the square of the speed of light in Roman numerology there surely must be a chance of you getting your own coat. I'll admit the chance may be infinitessimal but it's there.

If nobody chose your coat only your arithmetic could make it disappear.

A party of 4 yields a 3-1 chance first up, then a 2-1 chance next and finally an evens chance. A $ on a treble yields 23-1. Just under 4%.

It's probably been set in a party cloakroom because the teachers don't like mentioning gambling and card games to their charges but, of course, it is on those that it is based.

g_day, you're assuming that the first man grabs the wrong coat. The problem doesn't specify that everyone grabs the wrong coat, only that they grab a random coat.

Suppose the "wrong coat" they grabbed was the one that looked more expensive and better fitting than the one they came in with.

One can hardly imagine Oliver Hardy grabbing Stan's coat and going out into a north-eastern ice-storm in it.

In modern weather conditions one imagines that fur coats were grabbed first and so the question might resolve it self into the gender mix at the party. It could even have a class factor.

Was it a gay party, on the sword side I mean, or a gaggle-of-geese one.

That's it!

It's an asexual party. Of course. Teachers tend to be a bit hung up about sexual parties.

Let required probability be p(n) for n>1
There are two possibilities (i) first person A gets his own coat, prob.=1/n
(ii) first person A randomly takes someone else coat but not the coat of last person,for this n-2 possible choices out of n. Now the person B whose coat was taken by A comes at kth (k=2,3,4 . . .,n-1) position (its probability 1/(n-2)). For him there will be n-k+1 coats available including the coat of A and of the last person. In this case we can rename the coat of A as the coat of B and B is randomly selecting one of the coat out of n-k+1 coats
=> p(n) = (1/n) +[(n-2)/n]*sum from 2 t0 n-1 of [{1/(n-2)}*p(n-k+1)]
= (1/n) +[1/n]*sum from 2 t0 n-1 of p(k)
=> np(n) = 1+sum from 2 t0 n-1 of p(k) --------(1)
=>(n+1)p(n+1) = 1+sum from 2 t0 n of p(k) --------(2)
from (2)-(1), (n+1)p(n+1) - np(n) = p(n)
=>(n+1)p(n+1) = (n+1)p(n)
=>p(n+1) = p(n)
As p(2)=1/2, p(n) = 1/2 for n>1

I believe it is ((N-2)!(N-2) + 1)/((N-1)!(N-1) + 1)
This, however this is not needed if the first person chooses his/her jacket

hey i probably have an answer to your question . See,
there are a total of 'n' people and so there will obviously be 'n' coats.
Now the 1st person has 'n' choices to grab a coat , the second person has therefore 'n-1' choices to do so and so on .
thus the no. of events in the sample space will be 'n!'
now the event that you are talking about states that the last person must grab his/her coat . so let the event happen or to say that the last person has his own coat . now the first person will be having 'n-1' choices to grab a coat the 2 nd person will have 'n-2' choices and so on. so the no. of elements in the event space is equal to '(n-1)!'.
thus the final probability for the event to occur is (n-1)!/9n)!=(1/n)

you can also manually verify this by considering n=3,4,5.

thank you

The answer is 1/2 for all values of N>1. The logic is in the other thread linked.

Another variation of this problem is to ask "What is the probability everyone gets the wrong coat if everyone grabs a coat at random?"

Assuming there are N people, the answer is !N/N!
where !N= the number of derangements of the set of N people, which is the number of permutations where no element is in its right position; and N! is the factorial of N.

As N --> infinity, !N/N! --> 1/e

I think the ans is 1/N.
P(geting my coat) = 1-P(lossing it)
P(lossing it)=P(lossing it in any of the N-1 chances).
=[ 1/N + (N-1)(1)/(N)(N-1)+ LIKE THIS]
So u will get N-1/N. We need 1-wht we gt which results 1/N

With just three coats, you can see that 1/N doesn't work.

Let people A, B and C own coats 1, 2 and 3. The possible outcomes are:
A-1, B-2, C-3
A-2, B-1, C-3
A-2, B-3, C-1
A-3, B-2, C-1

If you are C, you have a 50% chance of going home with your own coat.

If N asymptotes infinity the chance asymptotes zero. Which leads to the idea that mathematical numbers start losing touch with reality at some indeterminate point.

And that the mathematics beyond that indeterminate point are only possible with a Gothic Christian conception of God and are outside the range of the atheist mind which likes to think it could have thought it up when it is there ready thought up but cannot show how it would have done had it not been.

@engineer you have left two outcomes
A - 1,B - 3,C-2
A-3,B-1,C-2

Now it becomes 1/3 and the ans 1/N is correct

Neither of those are valid outcomes since person B would pick out their own coat if it was available. Only person A is random.

In the original problem, there is a 33% chance that A will pick correctly and everyone gets their correct coat. There is a 33% chance that A will take C's coat and C cannot go home with the correct coat. There is a 33% chance that A will take B's coat. B will then take A's coat 50% of the time and C's coat 50% of the time. Of the six possible options, three lead to C taking home his own coat.

u people are dumb it form and infinite geometrical prodression sort of a question
though the question is great...

Answer : ( 1/2 ),
Solution : First we should calculate the total no. of possibilities,
The followings are the possibilities and no. of possible cases :
# "1st person grabs his own coat", No. of ways = 1;
.
# "1st person grabs r'th person's coat and r'th person grabs 1'st person's coat",
// where ( 2 <= r <= n);
No. of ways to select 'r' = (n-1).
.
# "1st grabs r1's, r1 grabs r2's and r2 grabs 1st's coat",
// where ( 2 <= r1 < r2 <= n ),
No. of ways to select r1 and r2 = C(n-1,2)
.........
.........
So total no. of ways = 1+C(n-1,1)+C(n-1,2)+....+C(n-1,n-1) = 2^(n-1);
.
If the last person gets his own coat :
Then r's (r1, r2....) are selected from 2,3,4...(n-1),
So no. of ways the last person gets his coat = 1+C(n-2,1)+C(n-2,2)+...+C(n-2,n-2) = 2^(n-2);
.
So the required probability = (2^(n-2)) / (2^(n-1)) = (1/2)...

i think the answer is (1/n)*(1+1/2+1/3+.....+1/n-1)