So imo, combining these you have

Travel to 1
Q^3/(Q^3+P^7)
multiplied by
P^10/(Q+P^10)

Add to this
Travel to 11
P^7/(Q^3+P^7)
multiplied by
Q^10/(P+Q^10)

So

(Q^3)*(P^10)/((Q^3+P^7)*(Q+P^10)) + (P^7)*(Q^10)/((Q^3+P^7)*(P+Q^10))

Some math genius might want to see if that can be simplified

I agree with that solution. And I believe that the reason you can ignore the other scenarios (not going direct to an endpoint) is because they end up canceling out with themselves when finding the other endpoint.

A funny mistake btw

Also heads/coins

lol

I had coins in my brain.

3) You have 13 coins, one is fake and has a slightly different weight. You have a traditional balance and must pick out the fake coin with three weighings.

I know a similar one, but it's with 12 coins and you must also determine whether the coin is heavier or lighter.

You sure that it's 13 in this puzzle?

I know this one with 12 coins. I'll see if 13 is much different when I get a chance.

I know the one with 12 and found out you can do it with 13 as well. It will not be much extra work.

The 13 coins might require an extra step, but not needing to determine heavier or lighter removes one, so it's a wash.

There's another thread in here somewhere with a bunch of similar puzzles

Optimal Solution for the gold coin problem:

MORE PUZZLES

4). You're in a group of 12 prisoners and you're given a challenge by the warden which will either set you all free or get you all the electric chair. You are brought to a room randomly and are faced with 2 switches and you must toggle one of them. You can be brought to the room multiple times in a row or not for a very long time. Who is taken to the room is completely random each time and includes repeats. When someone declares that they know that everybody has been brought to the room at least once then the challenge is over. If the person who declares this is correct then everybody is set free. If they are wrong then everybody is executed. You can plan a strategy before hand but once the challenge starts there is no more communication by any of the prisoners. What strategy should the prisoners plan to guarantee success?

5). You are part of a group of 3 prisoners. A hat is put on each prisoner's head and it is either red or blue. You cannot see your own hat and you can see the hats of the other prisoners. You cannot communicate what you can see to the other prisoners. You all have the opportunity to guess what colour your own hat is. You do not have to guess. If you are wrong then you are all executed. If nobody guesses then you are all executed. Anybody who wants to guess must not announce this fact before hand and must guess simultaneously when a whistle is blown. If everybody who guesses is correct then you are all set free. You can plan a strategy before hand. What is the best strategy to give yourself the best chance of being set free?

Also the podcast that had the solution to the clock puzzle had a broken link and so I don't have the solution to that one.

Sorry bout that...

For number 5, do we know anything about the distribution of possible hats? Or it is just random? Or do we have no clue whatsoever?

5.

I am assuming that each prisoner is required to toggle exactly one switch, and that toggling a switch means leaving it in the state opposite to which it is found. I also assume that the prisoners do not know the initial state of the switches.

Correct.

I assume for 4 that this goes on endlessly until someone declares?

For 4, I think the only way has to involve appointing a leader and for him to be able to tell whether everyone has visited the room or not.

4.



edit: This assumes SW1 starts in down position, or more generally, that 1 of the switches' initial states is known.

If we don't know this, I think we'd have to change your "The first time" to "The first two times". There would be a ****load of redundancy, but I think that would 100% guarantee safety.

There may be a better (more efficient) way, but I think this works.

You're right, they can be unknown initially.

I think 5 is a variation of the red dot/ green dot trick. In that scenerio you can actually be certain of the colour of dot you have on your forehead simply based on the fact that the other two people can't figure it out.

But, in that trick some communication is required. for example , raising your hand if you can see at least one red dot on someone's forehead. Not sure how the hat trick can be solved without communication of some sort