dgreening

If person 1 cuts a small slice, no one will take it and person 1 gets the small slice

This is a variation on the problem of 2 people trying to split something. [where A cuts and B chooses which half]. One solution that solves this problem, but can be used for any practical value of N people

I may be missing something. some of these schemes seem to be based on knowing the weight of a regular coin. I didn't read the problem that way. I believe that we only know that 11 coins weigh some unknown amount and that the 12th coin weighs slightly more or less than thatt unknown amount depending on how it is positioned.. Our task is to find the magnetized coin without the additional information of what a regular coin weighs.

A few observations the 5 PM reference is just a red herring! the odds are the same all day

I'm not sure at the moment how exactly to tackle the generalization, although I have the feeling it is likely the same answer as for the first part.

Good job!

Some thoughts - unless I am missing something in round 1, there are 64 teams in 32 games. If you choose randomly, you should win about half the games [16 in round 1] for each win you will get your bet back + that same amount [2*$bet].

This is not getting a lot of attention, so I am going to jump in. I think there must be something I am missing. If the paths are no coplanar, then I don't think there are positions that will support 2 alignments. Let's simplify the problem, by making the paths of the 3 planets coplanar. If the 3 paths are parallel, then I think there can be only 1 point in time/ space that they will line up -- after that, they will never be in alignment again. If we imagine that the paths of the 3 planets intersect and that the planets are travelling on paths as follow [i can't figure out how to post a picture that I created, so I am going to revert to compass points for ease of illustration]: "Green" planet is travelling due east [90 degrees] towards the point of intersection "Red" planet is travelling due ESE [120 degrees] towards the point of intersection "Black" planet is travelling ENE [60 degrees] towards the point of intersection If they line up before the intersection, there is a possibility that they will line up again after the intersection. After that, they will continue to diverge. This second alignment requires just the right mix of relative speeds [probably Green is the middle speed while one of the others is faster and one is slower.. Not really a proof, but I think that is the answer. [sorry about having to revert to compass points].

Very interesting strategy. I was working on something like bet after so many [3 or 4] black in a row. The addition of a red component seems very counter intuitive - although in my small number of simulations, the red strategy paid off much better than the black strategy. Overall in 60 iterations [Excel sheet that is very labor intensive] I had the following results: The red component successfully bet 22 times The black component successfully bet 2 times Note: no instance went to the last card Total number of successful bets - 24 out of 60 The actual random return [no bet] for the cases I ran were 28 I reanalyzed the data using just the black component of the strategy The black component successfully bet 26 times 4 instances wne to the end of the sequence 3 of these ended in Red 1 ended in Black Thus using just the Black component of this strategy, resulted in 29 wins out of 60 [vs 28 based on no bets at all. Since your sim is probably much more automated, it might be interesting to see what happens if you use only the black component of your strategy.

it wasn't me. and I did find a much more practical interpretation of the solution Is the the interpretation you were looking for??

K-Man is right, you cannot organize in a way to have each team play 9 teams in 5 weeks But with different constraints, there is an answer. 8 team in 5 weeks You can set up 4 tournaments/ week and have each team play 8 of the other teams at 2 schools, 3 teams will compete School A hosts B&C D hosts E&F at 2 schools 2 teams compete G hosts H; and I hosts J Over 5 weeks each school will play 3 times in a 3 team event and twice in a 2 team event At the end of 5 weeks each school will have played 8 teams [2 teams x 3 weeks + 1 team x 2 weeks]. 9 teams in 6 weeks If you want to have everyone play each other team, you must add a 6th week. in which you will have five 2-team events

Nicely done Rainman! The break through was coming up with (x + y) (x - y)! I dabbled with variations on xn but found that too many of the combinations yielded a zero. So I came to the same conclusion that you had to add another prime number to the mix and that 3 yielded the lowest results. Interesting puzzle, nice elegant solution.

Thoughts on the 3x case

I think you are on the right track. I like your analysis. Nicely done!

Harey is absolutely correct. I started playing with some small values of n to get some sense for the problem. It turns out that the chances of hitting an particular square along that diagonal looks like "binary normal distribution" [i am sure there is a better description for this] Of all the possible paths available. The only way to get to one of the corners of the diagonal row [for corners that are not A or B] and that is to choose all of one type of turn [up, up, up ..., up]. So if that route was chosen, the odds of colliding are very low. Moving inward from the corners. the only ways to get to the next 2 spots is to use the same sequence as above, but replace an "up" with a "right". and so on Consider the 3x3 grid. There are 6 possible paths. 2 of them [1100 and 0011] will take you to the 2 diagonal corners 4 of the [1001, 1010, 0110, 0101] will take you to the center square so distribution [of ways to get to that spot] along the diagonal will be 1, 4, 1 in this case there is a pretty good chance of colliding on the center spot Now consider a 4x4. There are 20 combinations. The diagonal is now 4 squares long 2 of the [111000] and [000111] end up at the corners 9 sequences will end up at the "middle" square on the top left 9 sequences will end up at the "middle" square on the lower right Thus the distribution will be 1, 9, 9, 1 In this case the 2 players could pass each other on the adjacent squares and the likelihood of collision would be much lower. I would guess that cases where N is an even number will have a lower probability than adjacent odd numbers.

Yes, the total number of candies makes a difference [if very large, the answer approaches 0]

I agree!

I do not have an answer yet, but I did some calculations that might help get this started.

Another real world question ...

I like that! Any idea which side that is?

I just looked, nothing there.

I will happily share the spread sheet. I am sure how I can post it or send it to you?? There does not seem to be a way to attach a file to a post or a msg. pls send me a note.