Re: coin toss

> But OP wanted a run of 6 to be only one "event" when I asked what
> comprised a "run of five"...

In the simulation I showed a run of 6 as only one event.

Notice that this:

[1 1 0 0 1 0 0 1 0 1 1 1 1 0 1 1 0 0 0 1 1 1 1 1 1 1 0 1 1 1 0 1 0 1 1
0 0 1 0 1 0 1 1 1 0 0 0 0 0 1 1 1 0 0 0 0 0 1 1 0 0 0 1 1 0 1 0 1 0 0 1
1 1 0 0 1 1 0 0 1 1 0 0 1 0 1 0 1 1 0 0 1 1 1 1 1 0 1 1 1]

was a event because it had a run of 7 heads (ones). 7 was the maximum
number of elements in a run in this particular experiment.

7 was greater than or equal to 5. So this event was represented as a 1
in the following list. Only one of the 100 experiments didn't have a
run of 5 or more.

show :many.experiments
[1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1]


Maybe a clearer way to show the process would be to create a list
composed of the
maximum number of elements in a run of each experiment:

[... 5 7 13 6 4 8 8 7 8 5 6 7 7 8 13 7 5 11 6 6 5 9 9 6 7 7 7 ...]

and then to mark all of the ones that are greater than or equal to 5:

[... 5 7 13 6 4 8 8 7 8 5 6 7 7 8 13 7 5 11 6 6 5 9 9 6 7 7 7 ...]
[... 1 1  1 1 0 1 1 1 1 1 1 1 1 1  1 1 1  1 1 1 1 1 1 1 1 1 1 ...]


Daniel


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