Dice and probability

[H5N1]

There is a formula for figuring (for instance)
the odds of getting five of a kind, all at once, in
one of three shakes of five dice.

It might be some kind of binomial distribution
formula but I cannot seem to find a simple,
spelled out version on the net.

[CBGLuthier]

1. Yahtzee probabilities here

Found some stuff here sounds like what you want.

http://www.mathsci.appstate.edu/~sjg/class/2240/finals04/winklerf.html

[kick-ass-bob]

2. YES!

Good to see my alma mater is doing some good out there!

[H5N1]

4. thanks, actually I am looking for the formula

a formula that lists and explains
the variables, perhaps (n) = the number of dice or cards,
perhaps (p) = the probability of a particular outcome
per dice or card... things like that.
I knew the formula years ago... but I forgot.
Seems it may involve some factorials as well.

[CBGLuthier]

6. see if this is it

top of page 21 on this pdf file

http://www.madandmoonly.com/doctormatt/mathematics/dice1.pdf


Looks like a continuation of the same stuff from the other page but concludes with a formula.

[H5N1]

8. must be in there somewhere, thanks

[hfojvt]

7. except they did not do it all at once

also, I have never done it with matrices, but that is probably the way to formalize all of the adding that I would do.

For the straight probability of a Yahtzee it would be the sum of the odds of getting a Yahtzee on the first roll added to the odds of getting one dice that you save and then matching it to the other four in your next two turns, although you have to track the various choices made by the player and the matrix does that very well. For example if you roll 4,4,1,3,5 on your first turn, save the fours, but then roll 2,2,2 on your second turn, your odds of a yahtzee are better if you save the twos and you have a 1/36 chance of a yahtzee on your last turn.

Unless you are my little brother - otherwise known as Mr. Dice. Once when we were playing triple Yahtzee and we both needed two Yahtzees in our last three turns. He got them. I did not.

Another time we were playing risk and he rolled two ones on defense as I was attacking him. I then proceeded to roll three ones on offense, losing two armies. What are the odds of that? 1 in 216! No problem, when you combine his good luck with my bad luck.

[hfojvt]

3. I think it is just

1/6 to the fifth power times 18 (or .0023148).
But I got my BA in math from Minnesota a long, long time ago in a galaxy far, far away.

[H5N1]

5. How about an all-purpose formula?

How did you come up with a name like hfojvt?

[hfojvt]

10. same way the creators of 2001 came up with HAL

HAL = IBM - 111. I only wish I had gone with dmpxo.

I do not believe there is an all-purpose formula for these things.

You just have to add up the possibilities. Since there are six different possible yahtzees I multiplied by 6. Since there are three turns I muliplied by 3 (only I skipped a step and just multiplied by 18). There are 6 to the fifth power possible combinations of five six-sided dice - 11111, 11112, 11113, 11114, 11115, 11116, 11121, 11122, and so on up to 66666.

But the odds of a Yahtzee in three turns are different than getting it in one turn (an also much harder to calculate). For instance if your first roll is 22225, your odds of a Yahtzee in your next two rolls are 11/36 (a 1/6 chance of getting a two on the 2nd turn added to a 5/6 chance of a non-two multiplied by a 1/6 chance of a two on the third turn or 1/6 + 5/36).

[redqueen]

13. Finally!

I get it now. :dunce:

[Ready4Change]

9. My way of figuring it:

Compare possible results against possible successes.

For example: If you roll 2 dice, you might get 1 and 1, or 1 and 2, ..., 6 and 5, or 6 and 6. That's 36 possible results. Out of those, only 6 are successes. (1 and 1, or 2 and 2, or...)

So you have:

Chance of 2 of a kind: 6 of 36 = 1 in 6

So, roll a pair of dice 6 times and odds are you'll get a pair once.

Continue that method outwards:

Chance of 3 of a kind: 6 of 216 = 1 in 36
Chance of 4 of a kind: 6 of 1296 = 1 in 216
Chance of 5 of a kind: 6 of 7,776 = 1 in 1296

Now, if you roll 5 dice 3 times you have:

3 for 1296 = 1 in 432.

I'm assuming ANY 5 of a kind will do. (all 1s, or all 2s, etc...) If you insist on a SPECIFIC 5 of a kind (all 6s, for example) you'd have

3 tries at 1 in 7,776 = 3 in 7,776 = 1 in 2592

And if that's wrong, you now know why I don't gamble. :)

[hfojvt]

12. same answer I got in post #3

only you explicated more.

[Theres-a]

11. Welcome to DU!

Hope you're not here to mutate into something that's easily transferable between humans.:scared:

[H5N1]

15. I didn't see any mention of H5N1 here

Suprisingly. So I thought I would add a little
H5N1 to the mix.

[Theres-a]

17. There's a user who posts in LBN

Pandemic 1918,who links to recombinomics articles,but they usually get moved to the science forum.

[H5N1]

18. Oh... that way

Oh well, I suppose that may change
if the wrath of H5N1 starts clobbering the hell out of us.
Most people (americans)never heard of it.

[redqueen]

14. Hi H5N1!

:hi:

Welcome to DU!


on edit: creepy username... love it. :)

[H5N1]

16. I might become very popular over the next several years

:)

Hi there!!!!!

[redqueen]

19. Well, if you mean YOU, then great!

If you mean what your name means... please don't! :scared:

[48pan]

20. The calculation is very long. Use an existing table.

The chance of getting 5 of a kind on a single roll is 6 to the 4th power.