OT: Home Sick for the Holidays, or Probability and Yahtzee

[Krsqk]

As my family has all gone out shopping, I am left alone to ponder the mysteries of life. My family are Yahtzee fanatics, and I find myself struggling to calculate probabilities for this game. For basic probability, I know the formula is

Chosen Outcome
-----------------------------------
# of Possible Outcomes

For example, on a standard d6, there is a 1 out of 6 chance that any number will come up. If I have a 2,3,4,5 and am rolling for a large straight, I should have a 2/6 or 1/3 chance to get it in one roll (since either a 1 or a 6 will work).

IIRC, the odds for multiple dice with separate outcomes for each are calculated by multiplying the individual odds; i.e., the odds for rolling a Yahtzee (all 5 d6 on same number) in a single roll would be 1/6^5, or 1/7776. However, it's been quite some time since I've studied probability, so I don't remember if that's correct.

My dilemma comes when you factor in multiple dice and multiple rolls for a single outcome. Rolling 6 d6 does not guarantee a roll of any particular number. Likewise, rolling 1 d6 six times does not guarantee a roll of any particular number. Obviously, simple experimentation shows there is something more to this than a simple formula such as:

Chosen Outcomes * Rolls
--------------------------------------
Possible Outcomes

What, then is the formula? Using the large straight example, given a 2,3,4,5 and 2 rolls of a single die, what is the probability that I will roll a large straight?

I impatiently await your elucid answers to these vital and perplexing questions about life's meaning (at least while my in-laws are here, so I can finally beat them )!

[Fyron]

IIRC, the odds for multiple dice with separate outcomes for each are calculated by multiplying the individual odds; i.e., the odds for rolling a Yahtzee (all 5 d6 on same number) in a single roll would be 1/6^5, or 1/7776. However, it's been quite some time since I've studied probability, so I don't remember if that's correct.

That is correct.

Rolling 6 dice at once or 1 die at a time six times makes 0 difference.

There will never be a guarantee of getting that 1 or 6.

[Fyron]

I forget if you want to use permutations or combinations here... it will be one over one of the following formulae:

n is the total number of items, r is the number of those items you want.

Combination (order does not matter):
n! / [ ( n - r )! * r! ]

Permutation (order matters):
n! / (n - r)!

So for probability, you have either

[ ( n - r )! * r! ] / n!

or

(n - r)! / n!

Depending on which is the one you want.

[se5a]

the meaning of life?
42 isint it?

[David E. Gervais]

Yahtzee has absolutely nothing to do with 'odds' or probabilities.. It's pure luck. I know. When I play and my friend Claude is also in the group playing he regularly gets 4-5 Yahtzee's per game. I don't remember a game where he did not get at least 1 Yahtzee. (And he rolls an unusually high number of Six's)

Luck, that's the only thing that matters in Yahtzee.

So, each time you roll the dice, just say a simple prayer... "Thy will be done." and see just how well that works. (oh, ans you don't have to say this prayer out loud, so you can keep the secret.)

Cheers!

[ December 24, 2003, 19:56: Message edited by: David E. Gervais ]

[gregebowman]

There was a dice game I used to play with my D&D friends when I was in Korea. Forgot now all of the rules, but IIRC we threw 6 dice also. But that's all the similarities to Yahtzee there was. Dang, I wish I could remember how that game went now. I got 6 brass six-sided dice I got from Korea I call my "Dice of Death". They made quite a racket when I rolled them (they're quite heavy compared to regular dice).

[se5a]

there is no such thing as luck...

[Krsqk]

Originally posted by Imperator Fyron:
I forget if you want to use permutations or combinations here... it will be one over one of the following formulae:

n is the total number of items, r is the number of those items you want.

Combination (order does not matter):
n! / [ ( n - r )! * r! ]

Permutation (order matters):
n! / (n - r)!

So for probability, you have either

[ ( n - r )! * r! ] / n!

or

(n - r)! / n!

Depending on which is the one you want.

These formulae work when r=1, but don't seem to work when r>1.

code:

---

[ ( 6 - 1 )! * 1! ] / 6!

[ 5! * 1 ] / 720

[ 120 * 1 ] / 720

120 / 720

1/6, or 16 1/6%

---

but,

code:

---

[ ( 6 - 5 }! * 5! ] / 6!

[ 1! * 120 ] / 720

120 / 720

1/6, or 16 1/6% (should be 83 1/3%)

---

Obviously, the odds that I will roll a 6 and the odds that I will roll less than 6 are not both 1/6 (at least on a standard d6 ).

It's worse with the permutation formula:

code:

---

( 6 - 1 )! / 6!

5! / 720

120 / 720

1/6, or 16 1/6%

---

but,

code:

---

( 6 - 5 )! / 6!

1! / 720

1/720, or ~.1389%

---

Things are even more exciting when you factor in more dice, such as the odds to roll at least one six with five d6:

code:

---

[ ( 30 - 5 }! * 5! ] / 30!

[ 25! * 120 ] / 2.6525285981219105863630848e+32

[ 15,511,210,043,330,985,984,000,000 * 120 ] / 2.6525285981219105863630848e+32

1,861,345,205,199,718,318,080,000,000 / 2.6525285981219105863630848e+32

7.0172483965587413863275932241449e-6

.0000070172483965587413863275932241449, or ~.0007%

---

I'd have thought my chances of rolling a six would have been slightly better than that! That's pretty close to 1/6^4.

Offhand, I'd guess there's a problem with the denominator in this formula--could it be n!/r, which would make the formula:

code:

---

r [ ( n - r )! * r! ] / n!

5 [ ( 6 - 5 }! * 5! ] / 6!

5 [ 1! * 120 ] / 720

5 [ 120 ] / 720

600 / 720

5/6, or 83 1/3%

---

but, then again:

code:

---

3 [ ( 6 - 3 )! * 3! ] / 6!

3 [ 3! * 3! ] / 720

3 [ 6 * 6 ] / 720

3 * 36 / 720

108 / 720

3/20, or 15% (should be 50%)

---

Any insight here?

[ December 25, 2003, 16:48: Message edited by: Krsqk ]

[Fyron]

Those formulae are correct for combinations and permuations. In truth, I was hoping someone would come along and fill in the blanks... Try google...

[Ed Kolis]

Originally posted by Krsqk:
IIRC, the odds for multiple dice with separate outcomes for each are calculated by multiplying the individual odds; i.e., the odds for rolling a Yahtzee (all 5 d6 on same number) in a single roll would be 1/6^5, or 1/7776. However, it's been quite some time since I've studied probability, so I don't remember if that's correct.

Wouldn't it be 6/7776, or 1/1296, since there are six mutually exclusive ways to score a Yahtzee (five ones, five twos, five threes, etc.)?

If you want, you could always write a program to run through all the permutations and check whether they match any of the scoring combos... I've written two Yahtzee programs myself so I could always borrow something from one of those and just run all the permutations through a loop instead of generating them randomly...