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# Tough probability riddle

Tue 11 Dec, 2007 08:04 pm
I have been trying to figure out this riddle for hours. I'm completely stumped!

"You are attending a fancy party. Upon arriving, each of the N partygoers (yourself included) puts his or her coat on the coatrack. After the party is over, the first person to leave is in a rush and grabs a coat at random from the coatrack. As the rest of the partygoers leave, they try to grab their own coat, but if their coat has already been taken, they take someone else's coat at random from the rack. You are the last person to leave the party. What is the probability that you will go home with your own coat?"

If you can get THAT, you're impressive.
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Type: Discussion • Score: 5 • Views: 6,553 • Replies: 23
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TTH

1
Tue 11 Dec, 2007 10:11 pm
0 Replies

g day

1
Mon 17 Dec, 2007 07:15 pm
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!
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spendius

1
Tue 18 Dec, 2007 08:50 am

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.
0 Replies

Vengoropatubus

1
Tue 18 Dec, 2007 04:15 pm
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.
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spendius

1
Tue 18 Dec, 2007 06:16 pm
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.
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spendius

1
Tue 18 Dec, 2007 06:19 pm
That's it!

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

1
Fri 30 Jan, 2009 12:06 pm
@KSiebert,
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
0 Replies

holarjc

1
Fri 12 Jun, 2009 02:46 pm
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
0 Replies

chandangang

1
Tue 24 Nov, 2009 08:26 am
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
engineer

1
Tue 24 Nov, 2009 08:39 am
@chandangang,
The answer is 1/2 for all values of N>1. The logic is in the other thread linked.
0 Replies

dallasmath

1
Sat 10 Apr, 2010 07:27 pm
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
0 Replies

bincy

1
Mon 10 May, 2010 10:10 am
@KSiebert,
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
engineer

1
Mon 10 May, 2010 11:18 am
@bincy,
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.
spendius

1
Mon 10 May, 2010 11:53 am
@bincy,
Quote:
I think the ans is 1/N.

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.
0 Replies

cosester

1
Fri 14 Jan, 2011 01:52 pm
@engineer,
@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
engineer

1
Fri 14 Jan, 2011 02:20 pm
@cosester,
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.
0 Replies

vishva

1
Sun 13 Mar, 2011 06:29 pm
u people are dumb it form and infinite geometrical prodression sort of a question
though the question is great...
yogi1729

1
Thu 9 Jun, 2011 09:48 pm
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)...
0 Replies

nivrutti

1
Fri 10 Jun, 2011 12:03 pm
@KSiebert,
i think the answer is (1/n)*(1+1/2+1/3+.....+1/n-1)
0 Replies

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