Can Brain Teasers Enhance Problem-Solving Skills?

[arunbg]

Firstly you are not the one "distributing the typos".
If you open a book filled with typos at random, would you argue that it is more probable to find 1 typo in one page than say all typos in one page?
I don't think so.Unless more information is given regarding the actual distribution of typos, you just don't know.

This question came in the JEE exam this Sunday.
What is the probability that a nucleus decays during two half lives?

Whoa! I marked 3/4 too. I used the same line of reasoning as you did.
Hope FIITJEE is wrong for once.
Doesn't make a difference though, I am not going to get through anyway.

 

[Jimmy Snyder]

Firstly you are not the one "distributing the typos".

The issue is not who distributes them, but rather whether they are randomly distributed. I claim that by the procedure I laid out, they would be. Good luck on the test. I would have answered as you did even though the question doesn't say whether the half-lives are for the nucleus or for something else altogether.

Nate, I see no ambiguity in this puzzle. You could just as easily argue to me that the words book and page are ambiguous because both can also be verbs.

A book of 500 pages has 500 typos randomly distributed. What is the probability that page 29 has no typos ?

You could say that my hidden assumption is that the puzzler is using the English language in the same way that the rest of us do.

 

[Promethium147]

The four hackers made their way to the cooling control center, which has four consoles, each controlling a separate plasma vent. I've set a different security protocol on each plasma vent, each wildly different from the other. Unfortunately, the humans knew this, and so each elite hacker is an expert in bypassing each type of security, but only the hacker who's an expert at that particular security will be able to bypass it, and no one else.

Knowing this vulnerability, and as a last ditch effort to save my station, I flooded the control room with intense radiation. Since the control room is only big enough to allow one person to work on it, the four hackers are forced to go only one at a time. Each hacker must enter the control room via a series of tight and lengthy corridors (which are designed to stop radiation from leaking into the main station), and pick one of the four consoles to check. If he is capable of disarming that particular security, he will do it, and disable that particular vent, at which he will die knowing his task is done. (He will not have enough time to go back and inform his comerades about the result) If he is not, then he has time to check a second console before the radiation turns him into green goo.

All four vents must be shut down in order for the station to be destroyed.

The four hackers, of course, knew this all beforehand, and so started to devised a plan which will maximize their probability of success. They thought as follows: if all of them pick consoles 1 and 2, the probability of success is 0, because there will always be two failures. If all of them pick 2 random consoles, then the probability of success will be 1/2 * 1/2 * 1/2 * 1/2 = 1/16, or 6.25%. At last, a few minutes later, they came up with a plan that will allow them to have a probability of success of over 40%. To make things simpler, they designated the four consoles to be A B C and D, and the different security protocols to be 1 2 3 and 4.

What plan is this?

This is supposedly possible but I haven't figured anything out yet. See if you can get it.

EDIT: This comes from a story, but the rest is not necessary to solve.

 

[davee123]

This is supposedly possible but I haven't figured anything out yet. See if you can get it.

EDIT: This comes from a story, but the rest is not necessary to solve.

It seems that it's impossible. It nearly expressly states that no hacker can tell the others of his success. But even if he *could*, the maximum success rate that I can muster is 33%.

Effectively, the first hacker's decision is random. He has a 50% chance of success at shutting down 1 vent. If he fails, they all fail. Assuming the first hacker's success, if the next hacker can *tell* which console was successfully hacked, he has a 66% chance of successfully hacking, since he may try 2 of the remaining 3 consoles. Again, if he fails, so does the mission. The chance that both the first and second hackers succeed is now 33%. If the first two succeed, and the last two can again tell which consoles were successfully hacked, they each have a 100% chance at success, so they stay at 33% for overall success.

So, I'm pretty sure that you can't get more than 33%. And maybe there's a sneaky way of getting 33% without actually knowing the successes of your preceeding hackers, but I don't think you can get more than 40% success.

DaveE

 

[Jimmy Snyder]

I came to the same conclusion as davee123. I looked into the possibility of using some variation of the 'Monty Hall' problem to improve chances, but with no success. The conditions necessary for that problem do not prevail in this one.

If you consider that the first person has a 50% chance of success, then in order to achieve a better than 40% overall chance of success, the second person would need a better than 80% chance of succeeding.

I did assume that the second and subsequent hackers could identify the hacked consoles by some method, perhaps by the puddle of green goo next to it (if there is goo by the console then either the console is hacked or the jig is up). If there is no way to identify the hacked consoles, then the best chance for success is one in sixteen. To achieve this, the hackers simply arrange ahead of time to evenly distribute the attacks among the consoles.

 

[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