wrd@tekigm2.UUCP (Bill Dippert) (12/12/85)
Given that you use two dice in a game and that you are trying to figure
the odds of getting a particular combination: do you use the combinations
or the permutations to determine the odds?
Permutations Combinations
Number Ways Number Ways
2 1 2 1
3 2 3 1
4 3 4 2
5 4 5 2
6 5 6 3
7 6 7 3
8 5 8 3
9 4 9 2
10 3 10 2
11 2 11 1
12 1 12 1
--- ---
36 21
What we are trying to determine is in the game of Railbaron, what the odds
are of reaching each city. To do this requires a throw of three dice.
(One red and two white.) The first throw determines region by the formula:
Red even or odd
White total of spots
The second throw then determines the city by the formula:
Red even or odd
White total of spots
It is obvious that the red die has 50% odds of reaching either chart, the even
chart or the odd chart. From there it becomes less obvious, so I am appealing
to those other Mathematicians out there whose schooling is much more recent
then mine. I say that you look at the ways you can get the combinations and
the other person is arguing for using the permutations. The problem is further
complicated by the fact that some cities have more then one possible way of
getting picked. I.e. odd 7, even 3, even 10.
The point being: does it make any difference to throw an 11 whether one die
is a 5 and the other is a 6 or the reverse? Are the odds of throwing an
11 the same as a 12 or l/2? Are the odds of throwing an 10 l/2 of a 12 or
l/3?
I am cross posting this to the interested group as well as to the "expert"
group.
Thanks,
--Bill--ark@alice.UucP (Andrew Koenig) (12/13/85)
Let's disregard the red die and color the other two green and blue. You will see that the green die can give any number from 1 to 6 with equal probability and so can the blue die. Thie means that there are 36 equiprobable outcomes. What are the chances of rolling, say, 9? You can have: green blue 6 3 5 4 4 5 3 6 In other words, there are four equiprobable ways of getting 9, so the chance of rolling a 9 is 4/36 or 1/9. And so on.
jeff@hpcnoe.UUCP (12/15/85)
> Given that you use two dice in a game and that you are trying to figure > the odds of getting a particular combination: do you use the combinations > or the permutations to determine the odds? > > Permutations Combinations > Number Ways Number Ways > 2 1 2 1 > 3 2 3 1 > 4 3 4 2 > 5 4 5 2 > 6 5 6 3 > 7 6 7 3 > 8 5 8 3 > 9 4 9 2 > 10 3 10 2 > 11 2 11 1 > 12 1 12 1 > --- --- > 36 21 > Actually, in the strict mathmatical sense, this is neither a "permutations" problem nor a "combinations" problem--factorials are not involved. The correct answer is the left column e.g. there is two ways of throwing a three (1 2) and (2 1), the two are equally likely because the dice have no knowledge of what the other has thrown. -- Jeff Wu
darin@ut-dillo.UUCP (Darin Adler) (12/15/85)
<> > Given that you use two dice in a game and that you are trying to figure > the odds of getting a particular combination: do you use the combinations > or the permutations to determine the odds? You use the permutations. There are a number of ways to convince yourself that this is correct, but one of the best ways is to roll a bunch of dice (or use a random number generator, if you "trust" computers). -- Darin Adler {gatech,harvard,ihnp4,seismo}!ut-sally!ut-dillo!darin "Such a mass of motion -- do not know where it goes" P. Gabriel
carl@aoa.UUCP (Carl Witthoft) (12/16/85)
In article <4699@alice.UUCP> ark@alice.UucP (Andrew Koenig) writes: >Let's disregard the red die and color the other two green and blue. >You will see that the green die can give any number from 1 to 6 >with equal probability and so can the blue die. Thie means that there are 36 equiprobable outcomes. What are the chances of rolling, ...and so on. There is an interesting fact he glossed over here. What happens if one uses two identical dice, i.e. same color and so on? In one line of reasoning, you'd say that rolling 5,6 is indistinguishable from rolling 6,5. This screws up the probality table. In fact, IT IS ONLY BY EXPERIMENT that it has been shown that dice always act as though they are distinguishable. By way of contrast, certain photon-photon scattering experiments show that two photons with identical energy and quantum numbers are NOT distunguishable. This is neat stuff. Darwin's Dad ( Carl Witthoft @ Adaptive Optics Associates) {decvax,linus,ihnp4,ima,wjh12,wanginst}!bbncca!aoa!carl 54 CambridgePark Drive, Cambridge,MA 02140 617-864-0201x356 "Selmer MarkVI, Otto Link 5*, and VanDoren Java Cut."