Oh no, don't get me wrong, I know it's incredibly easy, I'm just having trouble believing that you haven't seen it before.

As for the spider problem though, I take very much offense to your calling it easy. Do you remember it from the old thread, have you seen it somewhere else, or do you have a new answer that you just came up with?

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A blind folded man sits at a table. On the table there are a large number of coins. 37 of these coins are heads, the others are tails. The solution to this problem must be valid for any number of coins lying on the table.
The man has three "operations" that he can perform on the coins, but he can't feel whether the coins are heads or tails, they have no texture.
1. He can count the coins
2. He can group the coins as he pleases
3. He can turn them around.

The man has to arrange the coins into two groups, both containing an equal amount of heads.

Good luck

uclabb, could the strategy be that the first guy says hell pick, say, all the boxes but the one furthest to the left. If they are not all killed, then the second guy knows to include the box furthest to the left in his 99 and thus has a slightly better chance of surviving, so he picks that box and all the others except the one second furthest to the left. if he survives, the next includes those 2 and 97 others besides for the one on the 3rd to left. Same logic for the rest, until the last one, who just knows to pick all 99 except the one furthest to the right, which for sure has his name. although i wouldnt be too hyped about this groups chances.

@yaks, that strategy has a 1/100! success rate, which is better than the naive approach, but not even close to 1%.

Just checking: All of the coins have to be contained in the two groups, right?

Yup.

And the answer gives a 100% chance of success, or are we trying to maximize probability?

Both.

I assume "he can turn them around" means he can flip them so..flip all the tails to heads then count the total coins, split into two even piles. if you have an odd number, flip on heads to a tail and pull it out of both groups.

Ok, I think I know the answer. You flip 37 coins, and then those 37 coins are one of your groups, and the rest are the other group. Let me think a bit to make sure that always works.

Yeah, that works. Yay!

That's correct uclabb. shikari, you forgot the blind-fold! :-)

uclabb:

Assuming no communication is allowed once the first prisoner enters...

The easiest way to solve a problem like this is to reduce it to a far easier case. Say only 3 boxes and 3 prisoners.

If all 3 prisoners choose randomly the odds of survival are:

2/3 (P#1) * 2/3 (P#2) * 2/3 (P#3) = (2/3)^3 = 30%, not so good

If, however, Prisoner #1 chooses the farthest left box (the others know he will, prearranged)

P#1 = 2/3

Assuming survival, Prisoner #2 knows box 1 is now 50% likely to contain his name (it didn't have Prisoner #1's name, so only 2 other options). Of the two remaining boxes, each is equally likely (25% each), so he'll choose one (the next left, prearranged)

P#2 = 3/4

Prisoner #3 knows that the only way box 3 contains his name is if prisoner #1 and prisoner #2 had their names swapped in boxes 1 and 2. The odds of that occuring are 1/3 * 1/2 = 1/6 (probability that prisoner #3's name lands in box #3 and prisoner #1's box lands in box #2, forcing Prisoner #2's name in box 1, so:

P#3 = 5/6

The cumulative chance of survival then =

Ps = 2/3 * 3/4 * 5/6 = 42%



You can quickly see how this will map to the 100 man 100 boxes challenge.

Probability of survival choosing randomly:

Ps = (99/100)^100 = 36%

Probability of survival choosing sequentially:

Better :P I hate permutations and always fuck 'em up so I won't try.

Bah... I just reread the problem and realized I answered the wrong question but i believe the answer doesn't change.

Assuming these prisoners are lucky enough to make it, to round 2, we know that if prisoner #1 opened all but the leftmost, then sequential prisoners KNOW that one is safe.

If they make it to round 3, prisoner #3 knows the left two are safe...

all the way to round 100, where there is a 100% chance of survival... though they'll really never make it that far.

LoL this is a lot easier than the question I answered.

odds of surviving, btw, for those curious, are

1/100 * 1/99 * 1*98 ...... 1/3 * 1/2 (i.e. piss poor)

Here's a simple YJ logic puzzle...

You have 100 tin cans. 1 of them has $1,000,000 dollars in it, the rest have syphilis for you.

You must pick one can to hold on to.

the cruel bastard hosting the game then removes one of the remaining 99 cans, assuring you that it definitely had syphilis (he's telling the truth).

You can then open your can, or choose a new one to open. What do you do?

Well it's highly unlikely you will fit $1,000,000 in a tin can so I would back out of the deal.

A check. You're bad at these "outside the box" thinking questions, aren't you shikari? :P

you change your choice YJ

More like purposely evading providing an answer while making people reply.
I would retry because the original guess had a 1/100 chance of being 1mill while the second guess is 1/98 so your chances have improved.

shikari, no one asked you to put anything in a tin can. Grouping things doesn't equal putting them into containers. You lose. :P

Redhouse: Doesn't seem to be a guaranteed solution unless there are no tails at all at the start. Obviously probability of winning decreases as number of coins increases.

This one comes down to guaranteeing that you have an even number of heads after you flip, the fewest possible number of heads after you flip, and then splitting the coins into two piles as equal as you can get. The way I see it, depending on count, you either flip 1, flip everything, or flip everything but one. Doing anything else results in a 50% chance of you having an odd number of heads before you split them up, so you'd be fucked no matter how the split lands.

If there are an odd number of total coins, and the total coin count (N) is less than 37*2 = 74, flip everything, this guarantees that there are now an even number of heads on the table, and fewer heads than you started with. Then take (N-1)/2 coins and put them in a different pile. If N = 39 you've ensured a 50% chance of winning. If N = 41 you've ensured a 3/8 chance, and decreasing so on, but splitting (N-1)/2 always has the best expectation value.

If there are an odd number of total coins, and the total coin count (N) is more than 37*2 = 74, just flip one coin (flipping all of them now will guarantee you have MORE than 37 heads, and the more heads on the final board the worse the odds your distribution of heads falls into the desired even split). Now you know you have either 36 or 38 heads, you hope it's 36 but it doesn't really matter. Then split them as before, (N-1)/2 in a new pile.

If there are an even number of total coins, shit gets a bit easier. If N < 74, flip everything but one. Even number of heads guaranteed - fewest possible heads in final board - maximum odds of even distribution (this time just split them into 2 equal piles).

If there are an even number of total coins, and N > 74, flip only one. Even number of heads guaranteed (either 36 or 38 again), you've minimized the number of possible heads on the final board, and you'll again make two equal piles.

If N = 74, you could flip all but one or only one, probability is the same

Also, fuck you redhouse for making me think that one through :)

Fuck you double if I'm wrong :)

@YJ, I should be doing quantum chemical calculations right now and not probability theory, so I'm going to leave you to work out the example of 4 coins on the table and 2 heads, and work your way to larger numbers and then generalize to m coins and n heads. GL.

@YJ- Or you could look up a couple posts and see that I already posted the answer to redhouse's question. ;)

And your answer to my question is ok, I guess.... You can do about 99! (more than the number of molecules in the universe) times better, though.

surely the flip 37 and that's your pile,the rest is the other pileis right it we are starting with only 37 heads?

This is a pretty good simpler one:

You and a friend each has a fair coin. You can decide on a strategy and then play the following game, without any further communication with each other. You flip your coin and then write down a guess as to what your friend's coin will say. Meanwhile, your friend flips her coin and writes down a guess as to what your coin says. There's a third person involved: The third person collects your guesses and inspects your coins. If both you and your friend correctly guessed each other's coins, then your team (you and your friend) receive 2 Euros from the third person. But if either you or your friend (or both) gets the guess wrong, then your team has to pay 1 Euro to the third person. This procedure is repeated all day. Assuming your object is to win money, are you happy to be on your team or would you rather trade places with the third person?

team

pretty sure that's the right answer, even though i think most people will say the third person

Isn't that pretty much exactly what I did? I don't write in math :)

Am I wrong?

I don't see how there can be a guaranteed win. (I'm waaay off if there is)

with your minimal example, you can improve your odds of winning by flipping 2 coins, but that doesn't hold as m -> infinity. as soon as m > 2*n you want to minimize your number of coins flipped.