Steal-a-gift odds
For those who don't know, "steal-a-gift" is known by many names but works like this. You buy a gender-neutral gift for the "grab-bag." No one is supposed to know who brought what. You draw lots and person 1 picks any gift out. They open it and show it. Person 2 may "steal" that gift or open a blind gift. If the gift is "stolen" the person who had it picks another, wrapped gift. This repeats until the last person opens. When the last person opens person #1 may then force a switch with ANYONE. There is a limit on how many times a gift may be "stolen" and is usually 2-3 times depending on how many people are playing. However, I dount that much affectes your chances of being forced to get your own gift back.
So, math-oriented-folks, what is the chance and formula for "x" people? Assuming no one would be a bad sport and take their own gift my thought is you would have to pick last and only your gift left. What cannot be figured mathematically is if #1 gets your gift and wants to switch, so lets say they have a 50/50 chance of wanting to steal any gift then equal odds any particular gift gets stolen over the others (ie: 10 other gifts and they want to steal each person has a 1 in 10 chance no matter the desirability of any gift.)
What you describe was known a "White Elephant" in Baltimore, and on "The Office" it was also referred to as "Yankee Swap." So if you never chose your own gift, and I assume you could recognize the wrapping, there are two ways to get your own gift:
1. The last person trades with you, and you are forced to pick your own wrapped gift, because it is the only one left.
2. The first person has your gift, and forces a trade with you.
Do I at least have the problem right, before I start to try to do the math?
Quote: WizardI think I answered a simplified version of this in an "ask the wizard" question many years ago. If you simply get a gift at random, with no stealing, then the probability you get your own is 1/x, with x people in the office. The probability that nobody gets their own approaches 1/e as x approaches infinity.
What you describe was known a "White Elephant" in Baltimore, and on "The Office" it was also referred to as "Yankee Swap." So if you never chose your own gift, and I assume you could recognize the wrapping, there are two ways to get your own gift:
1. The last person trades with you, and you are forced to pick your own wrapped gift, because it is the only one left.
2. The first person has your gift, and forces a trade with you.
Do I at least have the problem right, before I start to try to do the math?
Do "locked" gifts figure in? Most early opened gifts will be off the board by the end of the game, reducing the number of active choices.
Also, we play a variation where if your gift is, "stolen", you may, "steal" another (no stealing back), up to three, "thefts" per round. The third person stolen from must open a new gift.
Quote: WizardI think I answered a simplified version of this in an "ask the wizard" question many years ago. If you simply get a gift at random, with no stealing, then the probability you get your own is 1/x, with x people in the office. The probability that nobody gets their own approaches 1/e as x approaches infinity.
What you describe was known a "White Elephant" in Baltimore, and on "The Office" it was also referred to as "Yankee Swap." So if you never chose your own gift, and I assume you could recognize the wrapping, there are two ways to get your own gift:
1. The last person trades with you, and you are forced to pick your own wrapped gift, because it is the only one left.
2. The first person has your gift, and forces a trade with you.
Do I at least have the problem right, before I start to try to do the math?
Yes, you have it right and indeed one place I worked called it "White Elephant." I couldn't for the life of me remember that name when I posted it.
Quote: AyecarumbaDo "locked" gifts figure in? Most early opened gifts will be off the board by the end of the game, reducing the number of active choices.
Also, we play a variation where if your gift is, "stolen", you may, "steal" another (no stealing back), up to three, "thefts" per round. The third person stolen from must open a new gift.
Last time I played any gift could be stolen a max of three times then it was as you say, "locked." I took one from my boss and locked it, btw. There was no max on "thefts per round" but in practice there were just a few gifts that were "hot" so we never went more than 2 thefts before someone took their chances with an unopened gift.
You can't do the math on this part, but the first time I played a person could easily see that about 20% if the gifts were wine or other bottles and another bunch were gift cards, Barnes and Noble being popular that night. I felt very bad for the guy next to me who got some real crappy gift that no one would ever steal.
Generally I hate grab-bags and secret-santa formats, but this is the one format I participate in at the office.
Quote: WizardThe exact answer looks pretty complicated, but it would be slightly less than (n+1)/n^2.
Thanks for the math work. Sadly this is the same group of people who refused to accept the smoker probability a few months back so I will enjoy knowing the answer myself. And the other readers of course.
