Zilch (dice probability)?

Everyone is WRONG WRONG WRONG.

You cannot simply do 1 - ∑[each scoring probability]

because the scoring probabilities are NOT MUTUALLY EXCLUSIVE!!

http://en.wikipedia.org/wiki/Probability#Mathemati...

Let each event be:

A - any die 1

B - any die 5

C - 3 or more of a kind

D - 3 pair

E - straight

You must apply the principle of inclusion/exclusion, which is:

1 - P[A] - P[B] - ... - P[E] + P[ A∩B ] + P[A∩C] - etc.

Think about it like a venn diagram. To count the probability of avoiding A∪B (A or B or both), you have to count P[A]+P[B]-P[A∩B]. The other answers are DEAD WRONG because they forget this! And it's even worse, because with five events, there are many many extra terms that they ignored.

You can simplify it somewhat. The following events are mutually exclusive:

E and D

E and C

So any time we see them together we can make those terms 0.

And the following events are subsets of each other:

A ⊆ E

B ⊆ E

So that when we see A∪E, we can just write E (same with B∪E).

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So let's start computations with the one-events.

P[A] = 1 - (5/6)^6 = 0.665102

P[B] = 1 - (5/6)^6 = 0.665102

P[E] = 6! / 6^6 = 0.015432

P[C] and P[D] are substantially harder to compute than the rest. Just read on (there is no need for me to compute all these).

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Now the double events:

P[A∪B] = 1 - (4/6)^6 =

P[A∪E] = P[E] = 0.015432

P[B∪E] = P[E] = 0.015432

etc... for the rest of the events and sets of events.

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You can attempt to calculate this directly (without the 1-∑[stuff] approach).

It is actually much much simpler, and gets you the correct answer. This is why I've omitted the rest of the above method - I use it only to show that the other people are WRONG because they have serious flaws in their logic (and may have computed the probabilities incorrectly too!).

Say you have no 1s and no 5s. Then you have no straight for sure. This avoids A,B, and E.

Then you have four possible numbers: 2, 3, 4, 6.

You can use at most two of each, to avoid D.

But you can't use more than two pairs either, to avoid C.

Well then, you can use two pairs, and two singles.

That's EXACTLY enough dice.

Then the question is how many ways can we do this?

Just count them:

Pick two for the pairs, and two for the singles.

(4C2)(2C2) = 6

223346

224436

226634

334426

336624

446623

And how many possible rolls are there?

It's easier to think of these as ordered, so that the total number is simply 6^6. But the the total number of dice throws that work for us is not 6, but rather 6 times:

6! / (2! 2!)

That's the number of ways to rearrange 2pairs and 2singles. Then you get the probability:

(6 6! / (2! 2! ) ) / 6^6 = 6! / ( 4 6^5 ) = 5 / 216 = 0.0231481...

That's approximately 2.31% (fairly small).

I even wrote a short script that experimentally verifies this value to 8 decimal places in a few minutes.

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What does this tell us about other numbers of dice? If you use 7 (or more) dice, you MUST score (since you cannot avoid them with more than 6 dice).

If you use 5 dice, you still count:

[ # possible (ordered) throws of 2,3,4,6 with at most 2 pairs and no 3 of a kind ]

which simplifies to simply:

[ # possible (ordered) throws of 2,3,4,6 with no 3 of a kind ]

divided by:

[ # possible (ordered) throws ] = 6^5

The first quantity changes slightly. As the number of dice becomes 4, 3, 2, 1 you get quantities that are easier to compute for the numerator (and the denominator is always just plain 6^[number of dice]).

Zilch Dice Game

It is very simple. The answer is 1 minus the sum of the probabilities of each of the scoring throws.

It is exactly the same calculation as with poker hands - and I think you are asking too many questions.

What is the probability of getting a poker hand with one pair? And then 3 of a kind? The hands overlap in that one pair in included in 3 of a kind, but if you just calculate the odds of 3 of a kind (and ignore one pair), you get the correct results.

Another idea is to think about all the possible different hands you can get in poker. There are only a finite number. How many of those hands will be a royal flush. You subtract 4 possibilities from the total. Total different hands = 52! / 47! (52x51x50x49x48)

the chance of 1 from any die is 1/6

same as 5

so far we have a probability of 2/6

three or more of a kind is

1/6^3+1/6^4 +1/6^5 +1/6^6 (for all numbers)

1/36 + 1/216 + 1/1296 +1/7776 = (1+6+36 +216)/7776

= 259/7776

three pair :

6^3 *( (1/36)*(1/36)*(1/36))

1/216

PROBABILITY OF STRAIGHT 1-6

is

1/6^6 = 1/46656

add it all to get

1/46656 +1/216 + 2/6 + 259/7776

1- all of these = probability of getting zilch ( i think)

= 0.62870799