need help from some mathmatical geniuses...

mibi · Post by **mibi** » Fri May 25, 2007 11:24 pm

Ok so there is some debate going on in the D-Day map thread, and we are too dumb to figure out the odds.

So here is the situation,

There are 72 territories on the map, and there are two 'continents' that when held, give a bonus of +10. They are 6 territories each.

So the question is, what is the likelihood that someone will start with all six territories in a 3 person game. And maybe you could throw in 4,5 and 6 person game stats too.

Thanks!

darkspy · Post by **darkspy** » Fri May 25, 2007 11:35 pm

I've always thought there were safeguards against starting out with continents (with the exception of 3 player 2.1). Maybe i've just never got that lucky i guess.

mibi · Post by **mibi** » Fri May 25, 2007 11:38 pm

darkspy wrote:I've always thought there were safeguards against starting out with continents (with the exception of 3 player 2.1). Maybe i've just never got that lucky i guess.

no safeguards. I've seen people start with australia or south america a few times.

wcaclimbing · Post by **wcaclimbing** » Fri May 25, 2007 11:49 pm

probability that all 6 are owned by a single person in the start of the game:

3 players= 0.137%
4 players= 0.0244%
5 players= 0.0064%
6 players= 0.00214%

as you can see, those are rediculously small percents.
You wont have to worry about someone getting all 6 in the start of the game.

Any other questions?

mibi · Post by **mibi** » Sat May 26, 2007 12:09 am

wcaclimbing wrote:probability that all 6 are owned by a single person in the start of the game:

3 players= 0.137%
4 players= 0.0244%
5 players= 0.0064%
6 players= 0.00214%

as you can see, those are rediculously small percents.
You wont have to worry about someone getting all 6 in the start of the game.

Any other questions?

how do birds have sex?

hwhrhett · Post by **hwhrhett** » Sat May 26, 2007 12:11 am

with tiny bird condoms and tiny bird tequila.

chessplaya · Post by **chessplaya** » Sat May 26, 2007 4:48 am

hwhrhett wrote:with tiny bird condoms and tiny bird tequila.

the funniest shit i've heard over the last few weeks

DiM · Post by **DiM** » Sat May 26, 2007 9:06 am

same 72 terit map.

what are the odds of a player starting with 4 or 5 terits from those 6?

mibi · Post by **mibi** » Sat May 26, 2007 9:17 am

also,

what are the odds of someone staring with 4 or 5 territories and all the other players just let the lucky player take over the whole map.

DiM · Post by **DiM** » Sat May 26, 2007 10:42 am

mibi wrote:also,

what are the odds of someone staring with 4 or 5 territories and all the other players just let the lucky player take over the whole map.

mibi don't bring that in here. if you really want to disrespect the foundry opinion go to landgrab and shut up.

here the community is important and all map makers have to kindly apply all the suggestions or refuse them with valid reason.

oh and the "do this my way or i'll take my map and leave" attitude is not gonna work. i really expected more from you.

wcaclimbing · Post by **wcaclimbing** » Sat May 26, 2007 10:45 am

mibi wrote:also,

what are the odds of someone staring with 4 or 5 territories and all the other players just let the lucky player take over the whole map.

the way to solve ANY of these "how likely is it that a player owns X territories in a Y player game" is:

(1/X)^Y

where X is the number of players and Y is the number of countries they have to hold at once.

and the answer to that question i quoted is:
that depends on the skill of the people in the game. If everyone else deadbeats, the probability is 100%. If everyone in the game is really good, then the probability is really low.

Coleman · Post by **Coleman** » Sat May 26, 2007 11:47 am

DiM wrote:
mibi wrote:also,

what are the odds of someone staring with 4 or 5 territories and all the other players just let the lucky player take over the whole map.

mibi don't bring that in here. if you really want to disrespect the foundry opinion go to landgrab and shut up.

here the community is important and all map makers have to kindly apply all the suggestions or refuse them with valid reason.

oh and the "do this my way or i'll take my map and leave" attitude is not gonna work. i really expected more from you.

It is gonna work. We're done, people don't like the map, then it's over. Not much too it. I'm not planning on running around and whining about it.

DiM · Post by **DiM** » Sat May 26, 2007 12:01 pm

Coleman wrote: It is gonna work. We're done, people don't like the map, then it's over. Not much too it. I'm not planning on running around and whining about it.

ok so just because it can't be done your way you abbandon the map. smart move

well i guess qwert must be really happy now cause he has to followers

Bean_ · Post by **Bean_** » Sat May 26, 2007 12:19 pm

That formula doesn't seem right. Someone check me since I haven't done math in a long time, but I think the chance with 3 players (for one particular group of 6) is:

3 x 24P6 x 66! /72! ~ 0.2584%

The chance with 4 players is:

4 x 18P6 x 66! /72! ~ 0.0475%

The chance with 5 players is:

5 x 14P6 x 66! /72! ~ 0.0096%

The chance with 6 players is:

6 x 12P6 x 66!/72! ~ 0.0036%

If there are two groups of 6, the chances are roughly double (have to throw out the possibilities that 1 person has both groups of 6 or 2 people have groups of 6, but these are very remote).

Hence, with 3 player, roughly once in every 200 3-player games, and with 4 player, roughly once in every 1000 4-player games.

For purposes of the above, xPy = x!/(x-y)!

wcaclimbing · Post by **wcaclimbing** » Sat May 26, 2007 1:02 pm

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.

Bean_ · Post by **Bean_** » Sat May 26, 2007 1:28 pm

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!.

maniacmath17 · Post by **maniacmath17** » Sat May 26, 2007 1:43 pm

yeah, bean is right on this one.