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Tip: Randomly selecting alternating patterns
#4
Hello grib,

your adjusted probabilities are correct, but unless I misunderstood your description of your assignment of the probabilities the attribution to the pattern is wrong. The way I underrstood your description the correct attribution is:
P(00) = 31,6% (that you had right)
P(01) = p(01) = 25% (first decision, 25%, 75% invers event p(!01) )
P(02) = 75% * 25% = 18,8% (second decision, p(!01) * p(02))
P(03) = 75% * 75% * 25% = 14% (third decision, p(!01) * p(!02) * p(03))
P(04) = 75% * 75% * 75% * 25% = 10,5% (forth decision, p(!01) * p(!02) * p(!03) * p(04))
And for completeness sake:
P(00) = 75% * 75% * 75% * 75% = 31,6% (inveres of forth decision, p(!01) * p(!02) * p(!03) * p(!04))

The sum of these is 100% (ignoring rounding errors), so it is complete, as expected.

If you want 01, 02, 03 and 04 with 25% chance each and no pattern 00 after pattern 00 just played then you'd have to adjust the probabilities as follows:
p(01) = 25% (remains unchanged)
p(02) = 33,33333333%
p(03) = 50%
p(04) = 100%

That will also remove your previous "other 00" step.

With the formulas above you could also compute any other set of resulting probabilities Smile

Kind regards,
Michael
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RE: Tip: Randomly selecting alternating patterns - by mgd - 01-22-2021, 05:20 PM

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