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

[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...

[David E. Gervais]

Who said there is no such thing as luck?

Yesterday, I had my friend Claude over and told him about this discussion. I then asked if he would like to join me in a test, he said sure, it'd be fun.

This is the test we preformed. We each rolled 5 dice 20 times, we took note of how many 6's turned up. We then repeated this test 10 times to see what kind of averages we came up with.

code:

---

David / Claude

  14  /   34

  12  /   32

  17  /   35

  20  /   37

  17  /   28

  16  /   31

  21  /   36

  32  /   27 (wow I beat him!  )

  18  /   33

  15  /   33

---------------

 18.2 /  32.6

---

Now if those numbers don't show some degree of luck I don't know how else to explain the results.

Would someone here like to roll this test and see how their results compare? (It would be nice to see how an outside 'neutral' person did with this simple test.)

Oh, now you can understand why I hate playing RISK against him, he always wins. (his best battle win against me in RISK was when I attacked him with 37 armies and he only had 5. He won. (and still had 3 of his 5 armies left after the battle.)

Cheers!

[geoschmo]

It's not luck David, but probability you have demonstrated there. Your sample is too small to be meaningfull. If you and your friend rolled 100 times, or a thousand, or ten thousand, the scores would get much closer together. Eventually within percentage points of one another. On the other hand if you and your friend sat down and rolled 10 times each again he would be just as likely to have a better score then you the second time around.

[Krsqk]

I'd start checking your friend's dice, David.

Actually, Geo is absolutely right, with one clarification. The difference in percentage points will likely decrease, eventually dropping near the break even point; the numerical difference will likely increase. There is no guarantee that actual results will even out over time (the usual misapplication of the law of averages).

Put another way, the odds of rolling a six with a single die are 1/6 every time, whether the "score" is 75 to 5 or 40 to 40. Previous results do not affect future results.

=========

In a separate thought, I found one page (granted, one which wasn't a strictly mathematical approach) which said probabilities which are mutually independent (such as rolling two separate dice) should be added; i.e., the odds of rolling at least one six with two dice is 1/6+1/6, or 1/3. If that's true, though, the odds of rolling at least one six with twelve dice would be 12/6, or 200%. That obviously isn't correct.

As I'm having much trouble finding any helpful information on the web, such as an intro to probability site or the like, I'd appreciate any links y'all can pass along.

[David E. Gervais]

geo, as I said, we rolled 5 dice 20 times (100 die rolls) and did this test 10 times. (10x100 = 1000 die rolls)

It took us about a half hour to make all the rolls. If you mean that we should have rolled this full test 10 more times to get 10,000 die rolls, that is a bit much. It wasn't my aim to prove the law of averages, but to see how in one quick session Claude always seems to out-six me.

Besides rolling 10,000 or 100,000 dice is unrealistic and does not reflect real life applications. Life does not work like that. Averages and probabilities are at best a global "best guess" and no matter what the 'odds' are or how many million tests are used to achieve those 'averages' it does not guarantee your imediate results.

Think of it this way, according to 'averages' if you roll a dice you have a 1/6 chance to hit a 6, but if you miss, your odds on the second throw don't improve, they stay exactly the same.

Cheers!

P.S...
so, anyone want to try my test and share the results?

[Kamog]

Wow, Claude should go to Las Vegas and get rich gambling!

Out of 100 dice rolls, the average number of 6's that should appear is 100/6 = 16.667 times. David's results are close to this. Claude's getting almost twice as much.

I want to calculate what the odds are of getting Claude's results, but all my statistics and probability textbooks are at work so I can't refer to them...