The Box Problem That Baffled the Boffins...

[SYZ]

This puzzle has recently got attention among academic
mathematicians.

Make an intuitive guess at the answer before you try to
work it out—the answer is very surprising.

The 15 boxes

Andrew and Barbara are playing a game, in which fifteen
boxes are arranged in a grid as shown below.



Prizes are put in two randomly-chosen boxes. Andrew
will search the boxes row by row, so his search order
is ABCDEFGHIJKLMNO. Barbara will search column by
column, so her order is AFKBGLCHMDINEJO.

If Andrew and Barbara open their boxes together each
turn, that is, on the first turn, they both open A, on the
second, Andrew opens B and Barbara opens F, on the
third Andrew opens C, and Barbara opens K, and so on,
who is more likely to find a prize first?

a) Andrew.

b) Barbara.

c) Both equally likely.

    —Intuitively, I chose option c, both equally likely.    

[jerry mcmasters]

The "and so on" needs some clarification. When you get to round four Barbara is at box B but it has already been opened by Steve, who is still going through row one. Does Barbara just open the next available box in the column? I assume so, and then as Steve gets to row two he will find two boxes have been opened so he goes to box three on the row? If so it seems they will have an equal chance until they run out of boxes. At a couple boxes it seems they are forced to both open the same box as the box is the next available for both row and column.

[SYZ]

Intuitively, it feels like both players should get
to the box at the same time. If the prizes are in
randomly-chosen boxes, why should one search
method have an advantage over another?

Indeed, if there was only a single prize in a single
box, that would be the answer: they would be equally
likely to win. What changes everything is that that
there are two prizes, and the game is over when
the first of them is found.

Note that the puzzle instructions does say the game
is over when the first of the two boxes is found.

[Inkubus]

I don't know shit. But this looks to be an example the birthday problem thingy.

The birthday paradox is a veridical paradox: it seems wrong at first glance but is, in fact, true. While it may seem surprising that only 23 individuals are required to reach a 50% probability of a shared birthday, this result is made more intuitive by considering that the birthday comparisons will be made between every possible pair of individuals. With 23 individuals, there are ⁠23 × 22/2⁠ = 253 pairs to consider, far more than half the number of days in a year.

Or something like that.

[brewerb]

Once you get to box H, which one gets to pick it? It should be the next pick for both.

The game rule of both opening a different box at the same time fails. Game over.

[SYZ]

Once you get to box H, which one gets to pick it? It should be the next pick for both.

The game rule of both opening a different box at the same time fails. Game over.

Well spotted.   But there's a big "but"...  Andrew and Barbara
each have a set of their own boxes—two sets, as the puzzle
confirms. [...open their boxes together].

I guess if one of the prizes (the first one) is in box H, then
it's a draw?    Dunno.

[brewerb]

Once you get to box H, which one gets to pick it? It should be the next pick for both.

The game rule of both opening a different box at the same time fails. Game over.

Well spotted.   But there's a big "but"...  Andrew and Barbara
each have a set of their own boxes—two sets, as the puzzle
confirms. [...open their boxes together].

I guess if one of the prizes (the first one) is in box H, then
it's a draw?    Dunno.

Not the way I read it, two sets would be 30 boxes each set arranged in a grid of 15. 'Open their boxes together' could mean each take off the tops of B and F at the same time.

If there are two sets and they each have a separately random placement of the prizes then my guts says results would be 50/50 in the long run. If they only do it once then it's a crap shoot.

[Aractus]

The "and so on" needs some clarification.  When you get to round four Barbara is at box B but it has already been opened by Steve, who is still going through row one.  Does Barbara just open the next available box in the column?  I assume so, and then as Steve gets to row two he will find two boxes have been opened so he goes to box three on the row?  If so it seems they will have an equal chance until they run out of boxes.  At a couple boxes it seems they are forced to both open the same box as the box is the next available for both row and column.

That's right the rules are not properly explained, and SYZ needs to learn to link to his sources:

https://www.theguardian.com/science/2024...he-boffins

You would think that once a box is opened it's opened, and therefore skipped over and not opened again. That's how I read the puzzle.

This is how they justify their answer:



This shows the case for each opening boxes in sequence where there is only one prize. They are equally likely to win, but on average it takes Barbara longer to reach her winning box - but this is only true of they are opening boxes each of them already opened! That does not seem logical at all, the puzzle says they open box 1 together:

“If Andrew and Barbara open their boxes together each turn, that is, on the first turn, they both open A, on the second, Andrew opens B and Barbara opens F, on the third Andrew opens C, and Barbara opens K, and so on, who is more likely to find a prize first?”

That can also be taken to mean they are opening them in front of each other. Nowhere does it say an opened box is closed. If they don't reopen boxes already searched the number of rounds it takes to find a prize when it's in one box is as follows:



And therefore a second prize makes no difference.

[SYZ]

...That's right the rules are not properly explained, and SYZ needs to learn to link to his sources:

Thanks for spoiling this little brain-scrambler prematurely.
Of course I'd intended to post the Guardian's full solution
to the puzzle, but after letting it go for several days in order
to give people some time to think about it and post their
explanations.

But of course our resident Mr. Smartarse had to Google it,
thinking he was oh so clever—without even attempting to
figure it out himself.  Pathetic!

Even more telling is the fact that your comments about the
puzzle are so far off the mark, it was a waste of your time
even posting them LOL.

EG:  "You would think that once a box is opened it's opened,
and therefore skipped over and not opened again. That's how
I read the puzzle." —You've missed the fucking obvious point
that each of the contestants have their own set of 15 boxes.

      

[Aractus]

Thanks for spoiling this little brain-scrambler prematurely. Of course I'd intended to post the Guardian's full solution to the puzzle, but after letting it go for several days in order to give people some time to think about it and post their explanations.

But of course our resident Mr. Smartarse had to Google it, thinking he was oh so clever—without even attempting to figure it out himself.  Pathetic!

THIS is just the link to the “puzzle” not the solution. The so-called solution is here.

It's a bad logic puzzle because it doesn't make sense. If it's mathematically on a computer grid in front of each of them that's different, but the puzzle describes a physical presence and that's key to understanding most logic puzzles. Take for example the 100 prisoners and a light switch puzzle as an example, it's made explicit that they're allowed to confer with each other beforehand so they know what they're doing, and then that they don't see who flicks the switch etc.

It's not a valid riddle as it does not have the required information.

Even more telling is the fact that your comments about the puzzle are so far off the mark, it was a waste of your time even posting them LOL.

EG:  "You would think that once a box is opened it's opened, and therefore skipped over and not opened again. That's how I read the puzzle." —You've missed the fucking obvious point
that each of the contestants have their own set of 15 boxes.

      

Well the answer is entirely invalid if they have two sets of boxes as the prizes are randomised and not in the same locations in each set: “Prizes are put in two randomly-chosen boxes.” It doesn't matter how you select them if they're random and the two sets are not identical!

Sorry but the entire riddle is invalid in the way it is set up.

[SYZ]

Aractus... I'm not waste any further time debating
semantics with you.  You've not understood the
puzzle's foundation, its clear instructions, or its
mathematical resolution, and thus overthought
the whole thing.

   —Get over it.