Can Brain Teasers Enhance Problem-Solving Skills?

[Promethium147]

-snip-

This is exactly what I got. The problem has missing information. I'll notify the writer. The solution calls for an assumption which is a bit of a stretch.

The four security methods are designated 1 2 3 and 4

The four hackers are designated A1, A2, A3, and A4.

The four lockers are designated B1, B2, B3, B4

The security methods don't have to match up with the lockers. For example, locker B1 could have security 3, and B3 can have security 2, etc.All four hackers designate one console to check on their first try:

A1 checks B1, A2 checks B2, A3 checks B3, A4 checks B4.

If the hacker sees his own console, good. If not, he will go to the console corresponding to the security method on that console. For example, A1 goes in, and checks console B1. Upon opening it, he finds B1 to be encrypted by security 3. The next console he will check, then, is B3.

This should prevent any two who were incorrect with their first guess ending up at the same locker their second guess.

Now, the probability calculations:

This will end up in all four being successful 10 out of 24 times.

- 1, when all four are right on their first guess.
- 6 when 2 are right on their first guess.
- 3 when none are right on their first guess, but the securities have been pairwise swapped.

10/24 ~ 41%.5 more solutions remain. 2 of those are straightforward solutions, like the one I've shown. The other 3 are a bit more tricky.

This is supposedly one of the solutions, but I don't see how this is any different than them just coordinating their attacks before hand. It seems like something is wrong here, perhaps the probability calculation.

 

[davee123]

This is exactly what I got. The problem has missing information. I'll notify the writer. The solution calls for an assumption which is a bit of a stretch.

I'm not sure I understand *why* it works probability-wise, but it works! But there is indeed missing information insofar as the hackers are able to determine which security system is on which console, if it's not the one they're an expert at.

However, the probability works out as stated. There are 24 possible alignments of the 4 security systems on the 4 consoles. And of those 24, following the stated logic, 10 scenarios will allow the hackers to succeed, yielding a 41.666667% chance.

Again, I'm not entirely sure why, but I think it has something to do with the fact that you're guaranteed not to waste time on a particular console. IE, with the stated system, at most two hackers will attempt to hack a given console, which is guaranteed NOT to be another hacker's successful first choice.

If console number 3 has security protocol number 3, then you're guaranteed that no other hacker will even attempt to *try* using console number 3, apart from the correct hacker. Hence, if console number 1 is *not* secured with security method 1, then hacker number 1 will *not* try console number 3, where he otherwise might in a "random" situation.

I don't know how you wind up with the 41.6667%, apart from just working it out, but it does indeed seem correct.

But yes, you should inform the author of the problem that there is information missing from the problem-- the hackers are able to determine the security system assigned to a console, even if they can't break it.

DaveE

 

[NateTG]

I don't know how you wind up with the 41.6667%, apart from just working it out, but it does indeed seem correct.

It goes something like this:
Hacker 1 goes in and tries a console A, and, instead of leaving a note, he communicates with the next guy by choosing which console to attack.

So there are three possibilities:
The first console works (25%)
The second console works (25%)
The hacker dies (50%).

So, now hacker two comes into the pit. If console B is hacked, then he knows that console A has security protcol 2, so he hacks that, and they're home free. If consoles C or D are hacked, then he knows A is not protocol 2, so they're guaranteed to succeed.

If, on the other hand, hacker 1 succeds, hacker 2 has a 2/3 chance of hitting the right terminal.

So, the chance of failure is going to be 25%+25%*2/3=41 2/3%

 

[davee123]

It goes something like this:
Hacker 1 goes in and tries a console A, and, instead of leaving a note, he communicates with the next guy by choosing which console to attack.

Actually, the logic works even without the communication! The logic would work the same way even if they had to work synchronously without the remotest possibility of relaying their status to one another.

But yes, I think your probability works out correctly:

- If the 1st hacker hacks his console correctly (25%), the 2nd hacker has a 2/3 chance of hacking his console correctly (combined 16.6667%)

- Using the logic provided, the only way for the hacker to succeed WITHOUT being successful on his first attempt is if his designated console is transposed with another hacker's. In that case, the other hacker, even without knowing that the first hacker was successful, tries (unsuccessfully) to hack his own console, but then necessarily successfully hacks his 2nd attempt. And the only option for the two remaining hackers is that their consoles are assigned correctly, OR they are similarly transposed. Hence, if the 1st hacker succeeds on his 2nd attempt (25% chance), all other hackers are guaranteed to succeed with a 100% chance.

So, as you stated, it is indeed 25% + 16.6667% = 41.6667%.

DaveE

 

[physicsmasta]

use a magnet to find a needle in a haystack