**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--

**hes@ecsvax.UUCP (Henry Schaffer)** (12/13/85)

> 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 > The permutations is the simpler approach, because it provides easier, more direct computation of the probabilities. The reason is that the 36 different events are all equiprobable, and so the calculation of probabilities (or odds) reduces to just counting events. (The crux of the matter is that you see that there are 36 different events, even though we reduce them to the 12 different numberical sums. if you have trouble seeing that there are 36, consider that one die is white, and one is off-white, and each can come up with 6 different faces.) > ... > (One red and two white.) The first throw determines region 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 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? > No it doesn't make any difference which die is a 5 vs a 6 when you throw an 11 - however since it can happen two different (equiprobable) ways, vs. the one way a 12 can result, then the probability is twice as great. > Thanks, > --Bill-- --henry schaffer

**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.

**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."