1 in x question

trying to understand the "1 in x" logic a little better when talking about craps...

Now I know something may have a low probability of happening 5 times before something else but there is no doubt for that same thing you will see 15 occurences before the other more often than you think.

So if we take lets say 6 vs 7. If I understand how to calculate it ( 6/11 ^ x ) then the probability of seeing five 7s before one 6 is 4.828% or 1 in 20.7.

Does that mean for any random set of 20 6/7s you should see a streak of 5 one time?
Or does it mean everytime a string of 4 happens a 5th will happen 1 in 20?
Or something else?

Thanks

I think you are wanting the dice to remember what has happened and make adjustments, afraid that it will look funny and they will get fired if there isn't that streak of 5, or 1 in 20 times a 5th happens after a 4th or there will be hell to pay [g]

Anomalies in small samples are corrected by weight of numbers in larger samples.

i fully understand that.. just trying to understand what to look at when looking for a 1 in x.. which logic is used?

1 in x means: "run x experiments. On average, 1 time you will see the result you expect."
So, ~20 times, roll the dice until you get a 7. When you get a 7, stop, and start a new experiment.
On average, one of those 20 experiments will see five 6s before the 7.

Quote: dwheatley

1 in x means: "run x experiments. On average, 1 time you will see the result you expect."
So, ~20 times, roll the dice until you get a 7. When you get a 7, stop, and start a new experiment.
On average, one of those 20 experiments will see five 6s before the 7.

five 7s before the 6
(6/11)^5
and easily expanded to the binomial probability distribution
http://stattrek.com/probability-distributions/binomial.aspx?Tutorial=Stat

For 1 in 21 experiments (20.7 is closer to)
35.3725% chance of this event NOT happening
37.6851% chance of this event happening just 1 time
26.94235% chance (about 1 in 4) of this event happening 2 times or more in 21 experiments
7.82380% chance (about 1 in 13) of this event happening 3 times or more in 21 experiments

Sally

Quote: bahdbwoy

Does that mean for any random set of 20 6/7s you should see a streak of 5 one time?

Actually, that's a good question - what is the probability of at least one streak of 5 (or more) 6s before a 7 in a set of 20 6s/7s?
Here's what I managed to calculate:
Run of 2 6s = 96.47%
Run of 3 6s = 69.32%
Run of 4 6s = 36.89%
Run of 5 6s = 17.11%
Run of 6 6s = 1 in 13.3
Run of 7 6s = 1 in 30.95
Run of 8 6s = 1 in 72.8
Run of 9 6s = 1 in 172.5
Run of 10 6s = 1 in 411
Run of 11 6s = 1 in 989
Run of 12 6s = 1 in 2397
Run of 13 6s = 1 in 5870
Run of 14 6s = 1 in 14562
Run of 15 6s = 1 in 36724
Run of 16 6s = 1 in 94643
Run of 17 6s = 1 in 251293
Run of 18 6s = 1 in 697065
Run of 19 6s = 1 in 2074793
Run of 20 6s = 1 in 7054295

Quote: bahdbwoy

trying to understand the "1 in x" logic a little better

Ever thought about that the "1 in x " is a terrible and misleading naming of a simple (actually beautiful) concept of "randomly, identically distributed, independent" events ?

"1 in 20" is the same as "100,000 in 2,000,000". The latter is a bit more realistic, but neither describes the reality of such a situation.
If you think literally about performing an experiment 20 times, and as a direct result of that you will find your event in one of them - you are already misled.

removed
silly

Sally