Six balls are at the front of the classroom, and six students are each assigned a different colored ball.

Then they are asked to go up one at a time and take the ball they were assigned.

However, the first student doesn't like the color he was assigned, so he picks randomly from the remaining five.

After that, each successive student takes the color they were assigned if it's available, otherwise they choose randomly from the remaining balls.

What is the probability that the last student gets the ball they were assigned?

(In reply to

re: related problem by Charlie)

1/n + ((n-1)/n) * x = 1/2 , n=6 ----> x = 0.4 What a cool way to solve this problem.

Thanks! I will write it out as I understand it:

p(a) = p(a and b) + p(a and not b)

p(a) = p(a|b) p(b) + p(a|not b) p(not b)

Now, I had shown that when the 1st guy

draws randomly and all the rest either

find theirs or else choose randomly,

the probability of the last guy getting his own is

0.5. Call the last guy getting his "a". P(a)=0.5

On the first random draw, the first guy either

picks his own (call this "b") with probability 1/n or

doesn't ("not b"), with probability (n-1)/n.

If the first draws his own, then the nth

will get his own with probability 1. p(a|b)=1

If he does not draw his own, the likelihood

the nth will get his own is x. p(a|not b)=x

So: p(a|b) p(b) = 1 * 1/n and

p(a|not b) p(not b) = x * (n-1)/n.

p(a) = p(a|b) p(b) + p(a|not b) p(not b)

1/2 = (1) * (1/n) + (x) * ((n-1)/n)

x = (n/2-1)/(n-1)

for n= 2, 3, 4, 5, 6, 7...

this is 0, 1/4, 2/6, 3/8, 4/10, 5/12,...

0, 0.25, 0.333..., 0.375, 0.4, 0.4166...

*Edited on ***April 26, 2019, 9:53 am**