Wierd on lone dice rolls

[shadroch]

I'm playing Risk on-line with a friend when I roll 3 sixes. My friend rolls and gets two sixes. Five Dice, five sixes. Pretty rare. Less than three rounds later, same thing only he starts with the three sixes and I match him with two. Chances of it happening once are about 1 in 7600. it happening twice in three or four rolls are thru the roof.

 

[actuary]

Hey, I think its actually 1 in 1,555. Isn't it just a binomial 6C5 * (1/6)^5 * (5/6) ? That equation simply represents the probability of rolling 5 sixes in 6 rolls of a dice.

How many turns did you take in your risk game? If it was a combined 100, then the probability of this event happening twice in a game would be a binomial with n=50, p=1/1,555.2 which works out to about 1 in 2,000.

I'd say its pretty rare at once in 2,000 games, but not impossibly rare.

 

[shadroch]

Math is not my strongpoint, but from playing Sic Bo, I know the odds of three sixes with three dice would be one in two hundred and sixteen. So I figured that you'd go 216x6 as with the fourth dice, you only have a one in six chance and that would be
a one in twelvehundred and nintysix chance. Them I added the fifth die and again multiplied by six for a result of seventy thousand seven hundred and seventy six.
Did I err?

 

[actuary]

Math is not my strongpoint, but from playing Sic Bo, I know the odds of three sixes with three dice would be one in two hundred and sixteen. So I figured that you'd go 216x6 as with the fourth dice, you only have a one in six chance and that would be
a one in twelvehundred and nintysix chance. Them I added the fifth die and again multiplied by six for a result of seventy thousand seven hundred and seventy six.
Did I err?

Yes, your error lays in the fact that you did not account for all combinations that would produce 666 and 66X (X not 6). There are 6 possible ways to roll 3 sixes and 2 sixes. The possible combinations are:

666-66X
666-6X6
666-X66
66X-666
6X6-666
X66-666

You also need to multiply by 5/6 for the non-6 roll, and 6 for the combinations. So this gives us:

1/6 * 1/6 * 1/6 * 1/6 * 1/6 * 5/6 * 6 = 1/1555.2 (or 1 in 1555.2)

It's easier to think of this using the binomial distribution I used in my above post. Then you just plug it quickly into the nCr * p^r * (1-p)^(n-r) formula for the answer.

 

[shadroch]

There are only five dice being rolled. In Risk, the attacker rolls three, the defender two. Your formula is based on their being six, no?

 

[callipygian]

The probability of rolling five 6's with five dice is simply (1/6)^5, because the order of the dice doesn't matter.

However, in Risk, it's important to note that for the attacker, 66x are all equivalent because only the two highest dice are used. So 661 is no worse than 666. Once you make that assumption, the probability of attacker getting two 6's and losing to defender's two 6's is (1/6)^4.

In terms of online RNG's, they're not absolutely infallible. QFIT made some comments about the quality of the Excel RNG in this post. They start breaking down (repeating or otherwise becoming predictable) after a while, the crappier algorithms quicker than the better algorithms.

 

[actuary]

There are only five dice being rolled. In Risk, the attacker rolls three, the defender two. Your formula is based on their being six, no?

Oh, is that right? Yeah, I assumed both players rolled three dice. That means that my above numbers are wrong. If its five dice being rolled and all turn up 6, then the probability is (1/6)^5 which is 1 in 7,776.

The probability of having exactly two such events in a game with 50 turns would then be 50C2 (1/7776)^2(7775/7776)^48 which works out to about 1 in 49,700 games. Very rare indeed!