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

[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.

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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...

[Kamog]

This math problem is a cumulative binomial distribution.

Let's look at binomial distribution.

Each die toss is an event with two mutually exclusive possible outcomes: either you roll a 6 (success), or you roll 1 -5 (failure). If we have N events, then the binomial distribution can be used to determine the probability of getting exactly r successes in N outcomes. The formula is:

P(r) = (N! / (r!(N - r)!)) ( p^r) (1 - p)^(N-r)

where N = number of trials,
P(r) = probability of getting exactly r successes in the N outcomes
p = probability of success on any one trial.

So, in our situation, N = 1000 dice rolls, p = 1/6.

David got 182 successes out of 1000 trials and Claude got 326 out of 1000.

So, the probability of David getting exactly 182 out of 1000, by using the above formula, is
P(182) = (1000! / (182!(1000-182)!))((1/6)^182)(1-1/6)^(1000-182)

And the probability of Claude getting 326 out of 100 is:
P(326) = (1000! / (326!(1000-326)!))((1/6)^326)(1-1/6)^(1000-326)

But so far we are talking about the probability of getting exactly the number we got. I think we're more interested in the cumulative case, where we ask, what is the probability of getting at least 182 successes out of 1000, or at least 326 out of 1000?

Well, then we have to add them up, i.e.

P(182 or more) = P(182) + P(183) + P(184) + .... + P(1000)

and
P(326 or more) = P(326) + P(327) + P(328) + ... + P(1000)

I thought, no problem, I'll do this on Excel, but I immediately found that Excel can't handle 1000! because it's too big a number.

[spoon]

Originally posted by David E. Gervais:
(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.)

I plugged those numbers into a RISK calculator... odds of you winning were 0.99999978. That's some streak of luck!

[Kamog]

OK, I did some more searching around and discovered that Excel can, in fact, do this calculation.

There is a handy pre-built function called BINOMDIST(number_s, trials, probability, cumulative).
Now this is easy.

The probability of getting 182 or more 6's in 1000 rolls is:
P = 1 - BINOMDIST(181, 1000, 1/6, TRUE)
P = 0.105008

I needed to subtract from 1 because the cumulative function adds from 0 to 181. To get the portion for 182 and above, we subtract from the whole (1) the part we don't want, the stuff at 181 and below.

OK, so there's about a 10.5% chance of getting 182 or more.

The probability of getting 326 or more 6's in 1000 rolls is:

P = 1 - BINOMDIST(325, 1000, 1/6, TRUE)
P = -0.5.08482E-14

Obviously there's something wrong. The answer can't be negative. I think Excel still can't handle this kind of big / small number.

Does anybody have a better program and/or calculation method that can do this? Or am I doing something wrong?