Secret Easter Bunny

[jfgw]

Some office workers come up with the idea of a Secret Easter Bunny. (This is similar to a Secret Santa but the shops are less busy.) There is only one shopping weekend to go and they need your help.

Each person is to buy one present for someone-else, wrap it and label it with the recipient’s name. Each person is to receive one present and must not know who bought it. They have one lunch break to decide who buys a present for whom and to inform everyone whom he or she is buying for (so a “keep trying until it works” solution is not acceptable, the method must work first time). Everyone in the office takes part and there is no-one else to call upon to help. Using a computer or an online "Secret Santa" service is not an option, it has to be done manually.

How can this be done?

Julian F. G. W.

[UncleEbenezer]

Um, I suspect I haven't grasped the problem. Put everyone's name into a hat (fold the paper so it has to be opened out to read the name), then draw out names? Appoint a proxy for anyone not present at the time, and put the name picked on their desk.

Your description sounds like a ghastly ritual. Wonder if you could be done for harassment if someone is uncomfortable participating and also uncomfortable saying no?

[jfgw]

The problem with putting everyone's name into a hat and drawing randomly is that it is possible for someone to get his or her own name.

Julian F. G. W.

[Gengulphus]

The problem with putting everyone's name into a hat and drawing randomly is that it is possible for someone to get his or her own name.

Whereupon that person knows who gave them their present, breaking the conditions of the problem. And by the way, it's not merely possible, but quite likely - without actually having proved it or looked it up, the chance of nobody getting their own name has all the hallmarks of being about 1/e = ~37% for large numbers of people. I.e. it's more likely than not that someone does draw their own name.

I take it that one should assume there are more than three (*) people in the office? ;-)

(*) Edited from "two" in the version I first posted!

Gengulphus

[psychodom]

how about:
1) split the office in half into equal sized groups A and B.
2) Group A put their name into hat A, group B put their name into hat B
2A) if number in office is odd, then let group A have 1 more person and let one person from A choose a name from hat A, replacing name if it is their own
2B) alternative to 2A, exclude the most scrooge-like person in the office (there's always one) from the secret santa to get an even number
3) Everyone else then from group A draws a name from hat B
4) Everyone from group B draws a name from hat A

-Dom

[chas49]

"Thinking aloud here.."

How about two hats and two lists. Divide the names randomly into two lists - written or printed on two different colours of paper (say, blue and yellow). Everyone inspects the lists to see which colour they are.

Label each hat with one or other of the two colours, and place the matching names in the right hat.

All the blue list people pick from the yellow hat, and vice versa.

This means you can't pick your own name and you pick randomly from the other list.

Obviously this won't work for less than four people, and at small numbers you have a pretty good idea who picked your name...

[EDIT: I see I was beaten to this solution while I was thinking....]

[UncleEbenezer]

The problem with putting everyone's name into a hat and drawing randomly is that it is possible for someone to get his or her own name.

Julian F. G. W.

In which case, that person speaks up: Re-draw all tickets. Repeat as necessary.

If that's not good enough, you might resort to lists. Each ticket bears a list of numbered names, ordered so that every permutation of people in the office is represented at some position on the list, and every name appears on exactly one ticket at any given number. Then everyone draws once, and looks at the list. If anyone sees their own name at #1, they say so, and everyone moves to #2. Repeat until noone has an own-name (since the papers bear every permutation, this must happen). Now you have your mapping.

Code: Select all

1. Alice   1. Bob   1. Charlie
2. Alice   2. Charlie   2. Bob
3. Bob   3. Alice   3. Charlie
4. Bob   4. Charlie   4. Alice
5. Charlie   5. Alice   5. Bob
6. Charlie   6. Bob   6. Alice


If one of the participants draws up the tickets, they should add in some random extra noise, so that if Alice gets the first ticket, her rejecting both rows 1 and 2 doesn't tell the participants for certain who is buying for whom. Or, better, anonymise who it is who's rejecting a row.

[UncleEbenezer]

h
2B) alternative to 2A, exclude the most scrooge-like person in the office (there's always one) from the secret santa to get an even number
-Dom

THank you. That's uncommonly considerate of you.

My take on office humbuggery:

Moderator Message:
Redsturgeon: Link to blog removed. UncleE, you have been asked several times to stop posting links to your blog site. If you have a pertinent comment to add that is from your blog then please just post it straight here, without the link. Thank you.

[UncleIan]

2A) if number in office is odd, then let group A have 1 more person and let one person from A choose a name from hat A, replacing name if it is their own

Or if it's odd, then have x hats, where x is divisible by the total number in the office. So then that works every time except for prime numbers.

[jfgw]

how about:
1) split the office in half into equal sized groups A and B.
2) Group A put their name into hat A, group B put their name into hat B
2A) if number in office is odd, then let group A have 1 more person and let one person from A choose a name from hat A, replacing name if it is their own
...

That would work and is a method I hadn't thought of. The "odd" person would have to draw a second name (if he drew his own name first) before putting his own name back for it not to be a "keep trying until it works" condition.

Julian F. G. W.

[Gengulphus]

In which case, that person speaks up: Re-draw all tickets. Repeat as necessary.

Not a solution to the problem posed - from the OP: "... (so a “keep trying until it works” solution is not acceptable, the method must work first time) ..."

If that's not good enough, you might resort to lists. Each ticket bears a list of numbered names, ordered so that every permutation of people in the office is represented at some position on the list, and every name appears on exactly one ticket at any given number. Then everyone draws once, and looks at the list. If anyone sees their own name at #1, they say so, and everyone moves to #2. Repeat until noone has an own-name (since the papers bear every permutation, this must happen). Now you have your mapping.

That still has a "keep trying until it works" aspect to it. It's also totally impractical if there are more than about 5 people in the office, due to the length of the lists - even at 5 people in the office, each ticket bears a 120-entry list.

That's because you said "every permutation", of course - and it's possible to get away with fewer permutations. For instance, you listed the six permutations of three people, but it's possible to get away with four:

Code: Select all

Ticket X   Ticket Y     Ticket Z
1. Alice   1. Bob       1. Charlie
2. Alice   2. Charlie   2. Bob
3. Bob     3. Alice     3. Charlie
4. Bob     4. Charlie   4. Alice


No matter who draws each ticket, your procedure will work by #4 at the latest - no great insight there, just list the six possibilities for who draws which ticket and check (the worst case is that Alice, Bob and Charlie draw tickets X, Y and Z respectively). It's not possible to prune it down to three permutations - but I haven't worked out the minimum numbers of permutations for larger numbers of people!

But even if the minimum numbers are quite small compared with the total number of permutations, going through the numbers until one works does not seem to me to be a solution.

Not that anything will produce an actual solution for a 3-person office, because it's an impossible problem in a 3-person office! That's because no matter what procedure is used, the moment Alice sees she's buying for Bob, she knows that Bob must be buying for Charlie and Charlie for her - or the moment Alice sees she's buying for Charlie, she knows that Charlie must be buying for Bob and Bob for her. So either way, she knows who is buying her a present.

Gengulphus

[jfgw]

Your description sounds like a ghastly ritual. Wonder if you could be done for harassment if someone is uncomfortable participating and also uncomfortable saying no?

No office workers were harmed in the production of this puzzle.

[Gengulphus]

how about:
1) split the office in half into equal sized groups A and B.
2) Group A put their name into hat A, group B put their name into hat B
2A) if number in office is odd, then let group A have 1 more person and let one person from A choose a name from hat A, replacing name if it is their own
2B) alternative to 2A, exclude the most scrooge-like person in the office (there's always one) from the secret santa to get an even number
3) Everyone else then from group A draws a name from hat B
4) Everyone from group B draws a name from hat A

Looks correct to me, and more-or-less what I had thought of - but had been beaten to it by the time I got around to posting. The only difference was that my method of dealing with an odd number of people was to have one person who was in neither group. So their own ticket is in neither hat to start with - but they start the process by picking a hat, picking a ticket from that hat and replacing it with their own ticket. It's equivalent to yours, just a slightly different way of doing it that avoids the "keep trying until it works" aspect of them possibly repeatedly drawing their own ticket.

Must say though that I don't regard the solution as very satisfactory, because while nobody knows who the person who gave them their present is, they do know some people it is not. Avoiding that wasn't a requirement of the original problem, so it's a valid solution, but it leads me to pose a more difficult problem: at the end of the process, nobody should be able to logically deduce "that person did not give me my present" about any of the other people in the office. Assume that there are at least four people in the office, since as indicated above it's an impossible task with just three (or fewer) people.

By the way, I haven't yet got a solution to that more difficult problem - not that I've thought about it for very long yet...

Gengulphus

[chas49]

Everyone in the office is given a slip with their name printed on one side. They also secretly pick a label (out of a hat) with a unique number on it. The label is stuck to the blank side of the slip, and the slip is pinned label-side out on a notice board (without others seeing who pinned which label).

Now everyone can pick a slip other than the one bearing their own secret code.

?

[Halicarnassus]

I think the name in a hat works. Simply, the rules are, if you pull out your own name then you show everyone and put it back and pull another. If you have the unlikely scenario that one of the last two pull their names, thus compromising identity, then just do a redraw. Simples.

[ReformedCharacter]

I think the name in a hat works. Simply, the rules are, if you pull out your own name then you show everyone and put it back and pull another. If you have the unlikely scenario that one of the last two pull their names, thus compromising identity, then just do a redraw. Simples.

It's a puzzle, with rules, not a real question about how to do a Secret Easter Bunny, not quite so simples!

RC

[Halicarnassus]

I think the name in a hat works. Simply, the rules are, if you pull out your own name then you show everyone and put it back and pull another. If you have the unlikely scenario that one of the last two pull their names, thus compromising identity, then just do a redraw. Simples.

It's a puzzle, with rules, not a real question about how to do a Secret Easter Bunny, not quite so simples!

RC

Thanks. I noticed the thread title but not the forum it was in!

[jfgw]

Everyone in the office is given a slip with their name printed on one side. They also secretly pick a label (out of a hat) with a unique number on it. The label is stuck to the blank side of the slip, and the slip is pinned label-side out on a notice board (without others seeing who pinned which label).

Now everyone can pick a slip other than the one bearing their own secret code.

?

A good answer except for one thing: The last person to pick may be left with his or her own ticket. It is similar in some respects to the answer I had in mind, however.

Julian F. G. W.

[jfgw]

Must say though that I don't regard the solution as very satisfactory, because while nobody knows who the person who gave them their present is, they do know some people it is not. Avoiding that wasn't a requirement of the original problem, so it's a valid solution, but it leads me to pose a more difficult problem: at the end of the process, nobody should be able to logically deduce "that person did not give me my present" about any of the other people in the office. Assume that there are at least four people in the office, since as indicated above it's an impossible task with just three (or fewer) people.

By the way, I haven't yet got a solution to that more difficult problem - not that I've thought about it for very long yet...

I don't have a solution to that problem either but I have something close. My solution does not allow two people to buy for each other but, with that exception, it provides no clues to the recipients as to who purchased their individual presents.

Julian F. G. W.

[psychodom]

Assuming in our motley office we can identify a scrupulous individual, how about:

1) everyone writes their name on a piece of paper and on an envelope, then folds the paper and puts it into an envelope
2) a trusted individual thoroughly shuffles all the envelopes with the name-side-down
3) once shuffled our trusted individual lines up the envelopes (still name-side-down) and takes each slip of paper and places it in the next envelope)
4) envelopes are then returned to each individual with their name on

-Dom