[net.games.board] Railbaron/Dice Odds Summaries

wrd@tekigm2.UUCP (Bill Dippert) (12/18/85)

The following are the comments summarized of all who have responded so
far on calculating dice odds for Railbaron:
------------------------------------------------------------------------
From: Paul Scherf <paulsc@orca.TEK>
Subject: Re: Dice "Odds"

In article <309@tekigm2.UUCP> you write:
>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

Use the permutations. Even if you can't tell the difference
between the dice, they are still not the same die. The
probabilities won't be different between rolling two white dice,
that rolling a green die and a blue die and pretending they are
both white.
-------------------------------------------------------------------------
From: eirik@tekchips
Subject: Re: Dice "Odds"

Use the permutations. As an analogy, consider the problem of
flipping a coin twice. There are four distinct events, of which two
are equivalent. Throwing dice is the same thing, with heads and
tails replaced by one through six.
-------------------------------------------------------------------------
From: stevev (Steve Vegdahl)
Subject: Re: Dice "Odds"

Permutations is the proper way of looking at it.  The result of each die
is independent of any other.  An easy way to look at this is to make
each die a different color, say blue and green.  There are two ways
to sum to 3, namely blue=1, green=2 or blue=2, green=1.  Why would the
odds be different if the dice happenned to be the same color?

Another argument against the "combinations" way of looking at things is
it implies (according to your chart) that there are 12 chances in 21 that
the sum will be even, and only 9 chances in 21 that the sum will be odd.
In fact, the chances of an even (or odd) sum is 50%.  (Given that the
blue die is odd, the chances of an even sum is 50%; given that the blue
die is even, the chances of an even sum is 50%.)

A simpler version of the same problem is to use coins rather than dice.
The "permutations" way of looking at it is that two heads will turn up
with probability 25%, two tails will turn up w/prob. 25%, and two tails
will turn up w/prob. 50%.  The "combinations" way would assign a probability
of ~33% to each, implying that two coins tossed independently have
probability ~67% of turning up the same--clearly incorrect!  (If you don't
believe it, try 1000 tosses of a pair of coins.)

>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.
>etc.......

In order to compute the probability of reaching a given city, take the total
number of ways that it can be reached, and divide it by the total number
of dice results.  Thus, if city X can be reached either by odd-5 followed
by even-3 or by even-6 followed by even-7:

First red odd: 3 ways -- 1, 3 or 5
First pair of whites sum to 5: 4 ways -- (1,4), (2,3), (3,2), or (4,1)
Second red even: 3 ways
Second pair of reds sum to 3: 2 ways

Thus there are 3x4x3x2 (=72) ways of getting odd-5 followed by even-3.
Similarly, there are 3x5x3x6 (=270) ways of getting even-6 followed by even-7.
Thus there are a total of 342 ways of rolling to get to city X, out of
6x36x6x36 (=46656) total ways of rolling the dice.  Thus the probability of
reaching city X is 342/46656, or about 0.0073 (i.e., about 3/4 of a percent)

>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?

As mentioned before, yes.
-------------------------------------------------------------------------
From: watmath!watdaisy!lmpopp (Len Popp)
Subject: Re: Dice "Odds"

You use the permutations rather than the combinations to determine the
odds.  For example, let's look at the odds of getting 11 on a roll of two
dice.  Obviously, this means getting a 6 on one die and a 5 on the other.
The two dice are independent of each other, so it does make a difference
which is 6 and which is 5.  The first die has a 1/6 chance of coming up 5
and a 1/6 chance of coming up 6.  If it is 5, there is a 1/6 chance of the
second die being 6 (for a total of 11).  If the first die is 6 there is
again a 1/6 chance that the second die is 5, for 11.  So the probability of
rolling 11 is (1/6 * 1/6) + (1/6 * 1/6) = 1/18.
 
Another way to arrive at this is to see that there are 36 possible rolls (6
possibilities for one die and, for each of these, 6 possibilities for the
other), and 2 of these rolls give 11 (6 and 5 or 5 and 6), so the
probability of getting 11 is 2/36 = 1/18.  difference which is 5 and which
is 6.

So the way to figure out the probability for dice rolls is to count the
total number of different rolls and the number of different rolls that give
a certain total, imagining that all the dice are different colours so that,
for example, 6 and 5 is different from 5 and 6.  By this reasoning, the
probability of rolling 11 with two dice is twice the probability of rolling
12 and the chance of getting 10 is 3 times the chance of getting 12.

If there is more than one way to get a certain result (e.g. odd-7 or
even-3), add together the probability of each different way.

If you still don't see why it works this way, don't worry about it.  I've
always found probability theory to be very intuitive; either it's common
sense to you or it isn't.
-------------------------------------------------------------------------
From: watmath!watrose!rtummers1
Subject: Re: Dice "Odds"

	Rolling two dice is identical to rolling dice one at a time.  Therfore,
you use the permutations aspect as order matters.

	Another way of thinking of this (if you don't like the first), is to
roll dice of different colors.

	Either way, it matters if one die is a 5, and the other a 6, or the
opposite.

	Richard Tummers (BMath).
-------------------------------------------------------------------------
From: uw-beaver!cornell!bonnie!dnc
Subject: Re: Dice "Odds"

Bill,
Your permutation column gives the correct chance of throwing a particular
number. If you think about it, it's pretty reasonable.

If you want a two, there is only one way to get it:

		1st Die		2nd Die
		1	+	1	=	2
and the probability is
		1/6	X	1/6	=	1/36

If you want a three, there are two ways to get it:

		1st Die		2nd Die
		1	+	2	=	3
with a probability of
		1/6	X	1/6	=	1/36
and
		1st Die		2nd Die
		2	+	1	=	3
with a probability of
		1/6	X	1/6	=	1/36

So the total probablity of throwing a three is 1/36 + 1/36 = 2/36 = 1/18.

					Don Corey
-------------------------------------------------------------------------
From: decvax!seismo!philabs!pwa-b!mmintl!franka
Subject: Re: Improved truly fair Rail Baron rules

In article <300@tekigm2.UUCP> you write:
>The odds of rolling an odd 3 are [(l/2)(1/6 + 1/6)];
>the odds of rolling an odd 10 are {[1/2][(1/6 + 1/6)(1/6 + 1/6)]};
>and finally the odds of rolling an even 6 are 
>	 {[1/2][(1/6 + 1/6)(1/6 + 1/6)(1/6 + 1/6)]}.

I don't know quite how you got these numbers, but they aren't anywhere
near being right.  There are two ways to roll a 3 on two dice: roll a 1
on the first die, and a 2 on the second, or roll a 2 on the first die,
and a 1 on the second.  So the chance of rolling a 3 is 2(1/6)(1/6) or
1/18; and an odd 3 is (1/2)(1/18) = 1/36.  Similarly, the chance of rolling
an odd 10 is (1/2)(3)(1/6)(1/6) = 1/24; and an even 6 is (1/2)(5)(1/6)(1/6)
= 5/72.  The total is 5/36 which is about .139.

Frank Adams                                                             
-------------------------------------------------------------------------
From: decvax!dartvax!chuck
Subject: Re: Dice "Odds"

> 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?

Use "permutations".  To see this...  You will roll an eleven if the first
die comes up six and the second die comes up five, or if the first die comes
up five and the second die comes up six.  However, the only way to get
twelve is for the first die to come up six and the second die to come up six.

If you find explanations unconvincing, you might try throwing a pair of dice
a few hundred times...
-------------------------------------------------------------------------
From: earle@tekcrl
Subject: Dice "Odds"

In response to your question:

>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 to figure out the frequency of outcomes of two
identical dice.  For non-identical dice, like a red one and a white one,
you use the combinations for each.  You will notice that they're then 36
such combinations, which gives the frequencies that appear in the
permutations with indestinguishable dice.

Earl Ecklund  (earle@tekcrl)
**************************************************************************
*Earl is the only respondent who seemed to think that you should use     *
*permutations for identical dice and combinations for non-identical dice.*
*All others said to use permutations.                                    *
**************************************************************************
-------------------------------------------------------------------------
From: allegra!sjuvax!bhuber
Subject: Re: Dice "Odds"

In article <309@tekigm2.UUCP> you write:
>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?

The outcomes are unaffected if you paint the dice two different colors, say
red and green.  There are 36 different possible outcomes, corresponding to
the six ways the red die can turn up and the six ways the green die can turn
up.  Assuming that the dice are fair, and recognizing that they are independent
of one another, we conclude that all thirty-six outcomes are equally likely.

Now you can compute any probability.  For example, if you wish to know the
probability of rolling a sum of 7, you need to count the different outcomes
among the 36 which have such a sum.  There are six such, so the probability of
this event is 6/36.

In your language, the 'permutions' are the things to look at.  (Technically,
they aren't permutations at all; they are 'ordered pairs'.)

It's always hard to be convincing with probability, so I urge you to do the
following:  spend a few minutes tossing a pair of dice, recording the outcomes,
until you have done a couple of hundred tosses (this is much easier if you
coerce your RailBaron partners to help you).  Look at the distribution of
the sums on the faces; it should be pretty close to what is predicted.  
(Warning:  if you only look at less than 100 tosses, the results can be pretty
peculiar, due to chance variation!)

				Bill Huber
-------------------------------------------------------------------------
>From: hes@ecsvax.UUCP (Henry Schaffer)
Subject: Re: Dice "Odds"

> Given that you use two dice in a game and that you are trying to figure 
> ....etc.

   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
> 
> 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.

--henry schaffer
-------------------------------------------------------------------------
>From: ark@alice.UucP (Andrew Koenig)
Subject: Re: Dice "Odds"

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.
-------------------------------------------------------------------------