[Prof. Templeton]

Dave walked into Sue’s office. “So, I’ve collected all the secret Santa gifts for the office party, how do you plan on distributing them?”, he asked.

Sue replied, “We’ll just have every one of the nine people in the office reach in randomly and take a gift. Simple.”

“Um, won’t there be a chance that someone will get their original gift? I mean they all look about the same size and everyone used the same wrapping paper that was in the break-room. You can’t tell one gift from another.”

“Come on, Dave. I think there is a high probability that no one will receive their original gift”

What is the probability that no one receives their own gift?

A little while later Dave came back into Sue’s office. “The two guys from the mail-room want to come to the party also. They heard about the secret Santa thing, so they wrapped their gifts and I took them and put them with all the others. Now there are eleven gifts that all look the same. Does that change the probability of someone not getting their own gift?

What is the probability that no one receives their own gift now that two additional people are added?

[Guest]

The formula for finding the probability of any one person out of n selecting their own gift is p = (1-1/n)^(n-1). The reason the exponent is n-1 rather than n is that when the next to last person selects, the selection is fixed (i.e. each person has a gift assigned to them). The last person has no choice because only one gift is left. For 9, this formula gives 0.38974. For eleven, the probability is .38554. And it does indeed converge to 1/e for n approaching infinity.

[Guest]

The formula for finding the probability of any one person out of n selecting their own gift is p = (1-1/n)^(n-1). The reason the exponent is n-1 rather than n is that when the next to last person selects, the selection is fixed (i.e. each person has a gift assigned to them). The last person has no choice because only one gift is left. For 9, this formula gives 0.38974. For eleven, the probability is .38554. And it does indeed converge to 1/e for n approaching infinity.

Not quite. Your solution does not account for the possibility that your gift has already been selected by the time your turn comes. Agree that it converges to 1-1/e as n->/infty, but I think you'll find the actual solution is slightly more... er, deranged.

[Guest]

The way I see it is that we have the question: What is the probability that someone will chose his or her own gift, given that no others ahead have chosen that gift. If that is correct we can notate it

P(A and B)= P(A)*P(B|A) or The probability of one choosing a gift and it's their own = the probability of one choosing a gift given that they choose their own.

9/9 *9/9 = 64/64=1

The probability that one will chose his or her own gift is 1

11/11*11/11=111/111=1

The probability that one will chose his or her own gift when 2 more people and gifts are added is still 1

oohh I hope I'm right, I just finished a stats class

[Guest]

The way I see it is that we have the question: What is the probability that someone will chose his or her own gift, given that no others ahead have chosen that gift. If that is correct we can notate it

P(A and B)= P(A)*P(B|A) or The probability of one choosing a gift and it's their own = the probability of one choosing a gift given that they choose their own.

9/9 *9/9 = 64/64=1

The probability that one will chose his or her own gift is 1

11/11*11/11=111/111=1

The probability that one will chose his or her own gift when 2 more people and gifts are added is still 1

oohh I hope I'm right, I just finished a stats class

Sara- if you just finished stats class then my first advice is always think through the logic of your answer to see if makes any sense... if you've got 9 gifts in a sack and pull one out, there is NO WAY the probability of it being your own is 1 (100%). Just imagine yourself making the choice and having to bet your life (or paycheck) that you're going to pull your gift! Given that, your calculation approach can't be correct so you should revisit your approach. Also note that the assumption "given no others ahead have chosen that gift" is a key part of the problem in calculating the probability since odds are good that someone pulls yours before you get to go if you are at the end of the list.