wcaclimbing wrote:We both did it differently, but i dont understand how you did it.
The probabilities i did were done using the idea of "what is the probability that YOU get all 6", not the probability that one of them gets all 6.
The formula should be X*(1/X)^Y where X is the number of players and Y is the number of countries.
Each player has a 1/X chance of holding one of the continents. you do that 1/X times itself for each country.
The (1/x)^y works only if there is actually a 1/x chance for each territory. However, if you think of the territories as being assigned to each person in turn (with 3 people), the first territory has a 24/72 chance of being assigned to you. Assuming that it is, the next territory has a 23/71 chance of being assigned to you. If it is, the next one has a 22/70 chance. And so on. So for all 6, you have 24x23x22x21x20x19/(72x71x70x69x68x67), which is 24P6 / (72! / 66!).
Another way to look at it is to count the total number of possible layouts, which is 72! (treating duplicate results in different orders as a different layout) Then how many qualifying layouts are there? Each of the qualifying layouts has to let you have the 6 territories in the continent, with the other 66 being in any order (so 66!). The number of ways in which you can be assigned 6 territories in the continent is again 24P6, or 24x23x22x21x20x19. (Try it with an 8-country map and 2 players, with a 2-country continent. One of the players has territories ABCD, the other has 1234. The continent is the first two. You can have ABxxxxxx, ACxxxxxx, ADxxxxxx, BAxxxxxx, BCxxxxxx, BDxxxxxx, CAxxxxxx, CBxxxxxx, CDxxxxxx, DAxxxxxx, DBxxxxxx, and DCxxxxxx, which is 4P2, 4x3 or 12.)
Putting that together, I get the same result of 24P6 x 66! / 72!.