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# Tough probability riddle

anurag-1234

1
Thu 12 Jan, 2012 07:08 am
@KSiebert,
the probability of first person not taking your coat is (n-1)/(n).

For the second person the same will be different cause there are two scenarios for him:
first: where somebody before him has taken his coat (ie he doesnt find his coat): in this the probability of the last person to have taken his coat is(cause he cannot take your coat also that is why its not 1/n): (1)/(n-1)
So in this case the probability of this person also not taking your coat is : (1)(n-2)/(n-1)(n-1)
Second case:
this persons coat is still there : in this case the probability of the first person to not have take his coat is (n-2)/(n-1) and as he will take his own coat now the probability of him not taking your coat is 1. so joint probability (n-2)/(n-1)

SO probability of your coat still being there after two people is :
{ (n-1)/(n) } { [(n-2)/(n-1)] + [(1)(n-2)/(n-1)(n-1)] }

which comes out to be { (n-1)/(n) } { (n-2)/(n-1) } { (n)/(n-1) }

which is (n-2)/(n-1)

this way if you keep on going you will keep getting values that cancel each other out until you reach 1/2.

that means the probability is :

1/2 if n>1
1 if n=1
NS for n<1
0 Replies

dna2blackhole

0
Sun 5 Feb, 2012 02:26 am
@KSiebert,
I think the probability of nth man taking his own coat after the party is
1/n!
0 Replies

sandy111

1
Fri 27 Apr, 2012 11:21 pm
@vishva,
u may b right dear...solve my riddle.
0 Replies

spendius

1
Sat 28 Apr, 2012 10:47 am
0 Replies

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