# Simple probability question. Involves drawing name for gift.

• Oct 9th 2009, 07:10 AM
moris7
Simple probability question. Involves drawing name for gift.
We had a dinner with my family where we draw names to give gift this christmas. The only rule was to not pick our name.

We did it a couple of time because we kept picking our name.

Finally we succeded and i started to think about what was the probability of someone picking his own name thus making this fail.

I built an excel simulation program which simulate drawing and got a consistent 50 % failure chance for 2 persons, about 66 % for 3 persons...and so on...but it's like there is a somekind of a maximum of about 68 % as i add people to my simulation.

For 6 persons i get 68 % failure and for 100 persons, it's about the same. I don't understand why that is.

So i started thinking...and the figures that i think it should be is about :

success rate should be 1/number of person, so 1/6 for 6 persons so a failure rate of 5/6....

Can anyone explain me what i should be getting ?
• Oct 9th 2009, 08:39 AM
Plato
Quote:

Originally Posted by moris7
We had a dinner with my family where we draw names to give gift this christmas. The only rule was to not pick our name
We did it a couple of time because we kept picking our name.
Finally we succeded and i started to think about what was the probability of someone picking his own name thus making this fail.

Just to be clear on the question: “Six people draw names from the hat.
What is the probability that no one draws his/her own name?
If that is the question, there is a well-known answer. These are derangements.
The number of derangements on six objects is $\displaystyle D(6)=265$.
So the probability no one has a match is $\displaystyle \frac{265}{6!}=0.3681$.
Subtracting that number from 1, gives the probability of at least one match.

BTW: If $\displaystyle n\ge 3$ then a very close approximation of $\displaystyle D(n) \approx \frac{{n!}}{e}$.
$\displaystyle \frac{6!}{e}=264.873197643438$
• Oct 9th 2009, 09:03 AM
moris7
Thank you very much for your answer...this seems like the right answer...i will try to see why my simulation does not point exactly toward this number.
• Oct 9th 2009, 09:09 AM
moris7
It's because of excel random function which is clearly not random.
• Dec 28th 2009, 06:52 AM
yossarian
a descriptive title
I solved it once without derangements... think of this name-drawing as a permutation where each spot represents a person... and then those spots get filled in with their names (ie 1 through 6) in some order.... 123456 143265 654312 etc... being in the correct spot represents drawing your own name.

• Number of permutations where all 6 draw their name: 1

• Number of permutations where exactly five draw their own name: 0 (it's impossible)

• Number of permutations where exactly four draw their own name: 6C4*1 = 15

• Number of permutations where exactly three draw their own name: 6C3*2=40 (6C3 ways to choose which three draw their own names and 2 ways for the remaining people to not draw their names)

• Number of permutations where exactly two draw their own name: 6C2*Q where Q=the number of ways for four people to not draw their own name.

To get Q you have to recursively use the previous steps... think permutations like 1243 1234 etc.

permutations (length 4) where all 4 draw their own name: 1
permutations where exactly 3 draw their own name: 0
permutations where exactly 2 draw their own name: 4C2*1=6
permutations where exactly 1 draws their own name: 4C1*2=8
permutations where none draw their own name: (4x3x2x1)-(1+6+8)=9
therefore Q=9
6C2*Q=6C2*9= 135 and therefore there are 135 permutations (length 6) where exactly 2 people draw their own name.
• number of permutations where exactly 1 person draws their own name: 6C1*L where L is the number of ways for 5 people to not draw their own names

permutations (length 5) where all 5 draw their own name: 1
permutations where exactly 4 draw their own name: 0
permutations where exactly 3 draw their own name: 5C3*1= 10
permutations where exactly 2 draw their own name: 5C2*2= 20
permutations where exactly 1 draws their own name: 5C1*Q=5C1*9= 45
permutations where none draw their own name: (5x4x3x2x1)-(1+10+20+45)=44
therefore L=44
6C1*L=6C1*44= 264 and therefore there are 264 permutations (length 6) where exactly 1 person draws their own name.
Therefore, the probability of no one drawing their own name in a group of six people is:
1-P(1 or 2 or 3 or 4 or 5 or 6 people draw their own name)
=1- (1+15+40+135+264)/(6x5x4x3x2x1)
=0.36805555555