wrd@tekigm2.UUCP (Bill Dippert) (01/06/86)
This are some further postings regarding dice odds for Railbaron (and other
dice games).
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Date: Tue, 17 Dec 85 10:55:20 est
Message-Id: <8512171555.AA16489@vax135.UUCP>
Subject: Re: Dice "Odds"
Organization: A Deanin c/o WE Hopkins Princeton University EECS Dept
Use permutations. In calculating probabilities by counting number of outcomes,
you want all the outcomes to be equiprobable. Imagine that the dice are
different colors and it is easy to see why 6+5 is not like 5+6.
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Message-Id: <8512181013.AA00920@unc>
Date: Tue, 17 Dec 85 03:54:24 cst
Subject: Re: Dice Probabilities
/* Written 1:14 pm Dec 12, 1985 by wrd@tekigm2.UUCP in uokvax.UUCP:net.games.board */
/* ---------- "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?
As undoubtedly hundreds of people have told you by now, in Probability
you use the permutation, NOT the combination.
That is, for 2d6, P(2)=P(12)=1/36, and P(7)=6/36.
--Carl Rigney
USENET: {ihnp4,allegra!cbosgd}!okstate!uokvax!cdrigney
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From: ihnp4!euucp
Date: Mon Dec 23 08:20:41 1985
Subject: Re: Railbaron/Dice Odds Summaries (LONG ARTICLE)
Organization: Adaptive Optics Associates, Cambridge, Mass. USA
In case you didnt read my posting, I'm recapping it here.
It turns out by experiment that dice ALWAYS act as though they
are distinguishable, even when they are the same color.
This is one of those interesting facts of nature.
This is in contrast to some other cases, such as two-photon
interactions, where the experimental outcome indicates that the photons
act as though they are indistinguishable.
In case you still doubt this, remember two things:
1) All theories must be confirmed by experiment. Thus you can't arbitrarily
decide that two white dice have different statistics than a red and a white die
merely because some net poster claims it's so.
2) What if you use a red and a green die, but you are playing with a person
who is color blind? Then , according to those who think indistinguishable
dice have different stats, you would see the "combination" stats but
your friend, who cant tell green from red, would see the "permutation"
stats....
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."
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From: ihnp4!mmm!cipher
Date: Mon, 30 Dec 85 20:06:00 cst
Message-Id: <8512310206.AA04108@mmm.ARPA>
Subject: Re: Dice "Odds"
Organization: 3M Company, St. Paul, Minn.
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 combinations. Reason as follows: use two dice, and roll them
separately. Obviously this will not affect the odds: it's the same as
rolling them together (assuming the dice roll is really random).
Suppose you want to roll a 12. There is only one way to do it... roll
a six with the first die, and then a six with the second. One chance
in 36. Now suppose you wanted to roll an eleven. There are two ways
to do it... roll a five and then a six (1 in 36) or a six and then a
five (1 in 36) for a total of 1 in 18.
If the person who says the probability is determined by permutations
is still unconvinced, suggest the following: get a pair of dice, and
roll them over and over. If the roll is 12, or 2, you give him three
dollars. If it's 11 or 3, he gives you two dollars. Keep this up
until he admits you were right, then give him his money back. He never
had a chance.
It would also be very easy to write a computer simulation of dice
rolls and get a rough empirical estimate of the probabilities.
--
/''`\ Andre Guirard
([]-[]) De Tuss from de Tonn
\ o / ihnp4!mmm!cipher
`-'
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It is beginning to get obvious to me that not everyone agrees on which is
correct, c or p. Also, as Andre shows in his message, not everyone is clear
as to which is which. Andre says combinations, but then proceeds to use the
formula for permutations. However, the majority seem to agree that
permutations are the way to go. (Regardless of die colors.)
--Bill--
tektronix!tekigm2!wrdjeff@hpcnoe.UUCP (01/13/86)
> It is beginning to get obvious to me that not everyone agrees on which is > correct, c or p. Also, as Andre shows in his message, not everyone is clear > as to which is which. Andre says combinations, but then proceeds to use the > formula for permutations. However, the majority seem to agree that > permutations are the way to go. (Regardless of die colors.) > --Bill-- > tektronix!tekigm2!wrd The problem is that your terminology is incorrect mathmatically, that is why everyone does not agree. Also this is not an election. The most votes do not win. Its like voting on "what is the result of 1+1." -- Jeff Wu
cipher@mmm.UUCP (Andre Guirard) (01/14/86)
In article <331@tekigm2.UUCP> wrd@tekigm2.UUCP (Bill Dippert) writes: >> (from my reply to his posting)... >>The combinations... suppose you wanted to roll an eleven. There are >>two ways to do it... roll a five and then a six (1 in 36) or a six >>and then a five (1 in 36) for a total of 1 in 18. >It is beginning to get obvious to me that not everyone agrees on which is >correct, c or p. Also, as Andre shows in his message, not everyone is clear >as to which is which. I think we all agree which is correct. What we disagreed on is which name to give it. Sorry. I guess "permutations" is correct. -- /''`\ Andre Guirard ([]-[]) High Weasel \ x / speak no evil ihnp4!mmm!cipher `-'
dale@aesat.UUCP (Dale Groves) (01/17/86)
In article <37800029@hpcnoe.UUCP> jeff@hpcnoe.UUCP writes: >> It is beginning to get obvious to me that not everyone agrees on which is >> correct, c or p. Also, as Andre shows in his message, not everyone is clear >> as to which is which. Andre says combinations, but then proceeds to use the >> formula for permutations. However, the majority seem to agree that >> permutations are the way to go. (Regardless of die colors.) >> --Bill-- >> tektronix!tekigm2!wrd > >The problem is that your terminology is incorrect mathmatically, that is >why everyone does not agree. Also this is not an election. The most votes >do not win. Its like voting on "what is the result of 1+1." > >-- Jeff Wu Personally, I vote for "1+1=10" ------------------------------------------------------------------------------ ________________________ Dale R. Groves ~/ | {allegra,ihnp4,linus,decvax}!utzoo!aesat!dale \__ ________________| / /_`/ Who knows what evil lurks in the hearts of men? / / / / THE SHADOW KNOWS... /____/