Unusual Dice Rolls

[Jimbo123]

Is it just me or do you notice an unusual amount of rolls on the dice for 2, 3, 11 and 12?

[coyote303]

[quote:d42d18da64="Jimbo123"]Is it just me or do you notice an unusual amount of rolls on the dice for 2, 3, 11 and 12?[/quote:d42d18da64]

Think of it this way. The combined chances of 2, 3, 11, and 12 are equal to rolling a 7.

[Phenomenal]

What I notice is a highly unusual and suspect occurrence of duplicate numbers back to back. If there is 216 possible roll combinations each at bat then the odds off rolling back to back dice rolls would be something around 40,000. However, I have been paying attention now for the past month and find that it occurs at least once almost every game and sometimes more than once. Of course there are games where it doesn't happen but I can say with a high degree of confidence that it is happening more than the odds would say.

It leads me to believe the dice jgenerator is not 100% random. I'm not complaining though, I have a blast with this game on this site but it is an observation

[ROBERTLATORRE]

[quote:d252406b47="coyote303"][quote:d252406b47="Jimbo123"]Is it just me or do you notice an unusual amount of rolls on the dice for 2, 3, 11 and 12?[/quote:d252406b47]

Think of it this way. The combined chances of 2, 3, 11, and 12 are equal to rolling a 7.[/quote:d252406b47]

GREAT way to look at it

[coyote303]

[quote:000012f940="SToNe_WaLLeD2"]What I notice is a highly unusual and suspect occurrence of duplicate numbers back to back. If there is 216 possible roll combinations each at bat then the odds off rolling back to back dice rolls would be something around 40,000. However, I have been paying attention now for the past month and find that it occurs at least once almost every game and sometimes more than once. Of course there are games where it doesn't happen but I can say with a high degree of confidence that it is happening more than the odds would say.

It leads me to believe the dice jgenerator is not 100% random. I'm not complaining though, I have a blast with this game on this site but it is an observation[/quote:000012f940]

Dice--and random number generators--have no memory. If you roll a 1-7, there is exactly a 1/36 chance the next batter will roll another 1-7. I calculate that if each team has 36 batters, there is better than a 75 percent chance you will get at least two numbers back-to-back.

You can figure this out yourself with a calculator. Estimate you have about 1 chance in 50 of getting the same roll twice in a row. With a 1-7, 2-7, 3-7, etc. you would have a 1/36 chance; with a 1-12, 2-12, 3-12, etc. the chance would only be 1/216. However, you will roll more 7s than 12s, so it works out to about 1/50 chance that any roll will duplicate the next or 49/50 that you [i:000012f940]won[/i:000012f940]'t duplicate it. Now simply multiply .98 * .98 seventy-one** times (thirty-six times for each team less one for the original batter) and see what your answer is. The decimal (about .233 or just over 23%) is the chance that you don't roll any back-to-back numbers in any given game.[b:000012f940] So, only about a quarter of your games can expect zero back to back identical rolls, and many games will have more than one.[/b:000012f940]

The way I calculated was a simplification and may be off a couple of percentage points, but it should be close to the actual number.

**You only do this 71 times because the first batter cannot duplicate a previous roll. Point 98 (.98) is equal to 49/50. The odds for any two consecutive batters NOT having a duplicate roll is therefore .98 times .98. The odds that any consecutive three batters NOT having a duplicate roll would be .98 *.98 *.98. As I said before, I am simplifying rather than accounting for every permutation. Perhaps a better mathematician than myself can supply the exact figures.

I'll share another probability oddity in another post.

EDIT: I used 1/50 instead of 1/53 which would have been more precise. Using 1/53 will put you a bit over 25 percent chance (instead of 23 percent) of not rolling any back-to-back rolls in a game with 72 dice rolls. Using 1/50 seemed to be a bit easier to work with for explanation purposes. Otherwise, you are multiplying .981132 *.981132 seventy-one times instead of .98 * .98! Even .981132 is slightly off since 52/53 is still an (albeit closer) estimate.

[coyote303]

Here is something you can try next time you are in a group of about 40 people. Find out what everyone's birthday is. What do you suppose the chances are that out of 40 people, two will share a birthday? One in nine? A little better than that maybe? Believe it or not, the odds are around 90 percent. Do it with 60 people and the odds are better than 99 percent.

That's because randomness is so fickle. It doesn't just spread itself out evenly in the short run. It can also explain why numbers seem to get rolled back-to-back more often than you might otherwise expect. It's just the way randomness and probability work.

http://en.wikipedia.org/wiki/Birthday_problem

[coyote303]

[quote:2d30e830c5]If there is 216 possible roll combinations each at bat then the odds off rolling back to back dice rolls would be something around 40,000.[/quote:2d30e830c5]

I just reread Stonewall's post. You are correct that if I pick up a set of Strato dice, the odds of me rolling specifically a 1-12 twice in a row is around 1 in 40,000 (actually 1 in 46,656). However, if I already rolled a 1-12, the odds of me rolling it again are only 1 in 216.

Look at it another way: You don't roll a 1-12 in every game you play, but if you do roll a 1-12, there is a 1 in 216 chance that the next batter rolled one as well.

And as I said two posts ago, if I pick up the same three Strato dice, the odds of me rolling [i:2d30e830c5]any [/i:2d30e830c5]two combinations in a row are about 1/50.