Trying to Roll Dice Towards a Target
March 18th, 2012 at 7:17:11 AM
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If I roll "n" dice, what is the probability that at least "m"of them will land on a target value (e.g. n=60, m=14, t=1)?
March 18th, 2012 at 7:44:32 AM
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That's the Bernoulli sum for m+ successes out of n trials with p=1/6.
Generally: SUM_ {from i = m to n} (n Choose i) * p^i * (1-p)^(n-i)
If someone who cares more wants to put that into nicer formatting, please go ahead.
For your example, it's 0.115216
Generally: SUM_ {from i = m to n} (n Choose i) * p^i * (1-p)^(n-i)
If someone who cares more wants to put that into nicer formatting, please go ahead.
For your example, it's 0.115216
Wisdom is the quality that keeps you out of situations where you would otherwise need it
March 18th, 2012 at 9:54:37 AM
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Quote: dwheatleyIf someone who cares more wants to put that into nicer formatting, please go ahead.
Is this what you had in mind?
I wish there were an easier way to format such nerdy stuff on this forum.
BTW, Brunhilde, note that "t" does not appear in the formula.
Edit: I noticed a typo in the formula that I originally posted. I had the exponent as (n-1) instead of (n-i) and have now corrected it. I think the old, incorrect version will continue to show up in FleaStiff's comment below.
Re-edit: Perhaps not.
March 18th, 2012 at 10:32:34 AM
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Quote: Doc
This reminds me of the student, instructed to find "x", who circled it ... and said Here It Is.
March 18th, 2012 at 10:53:08 AM
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silly
silly
I Heart Vi Hart
March 18th, 2012 at 11:50:05 AM
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mustangsally,
I'm rusty at Excel. Do you do that entire calculation in a single cell? If so, what is your formula? I did the same calculation in Excel but used 61 cells for the (zero to m) individual terms that might have to be added together, then did the two sums over the appropriate sets of cells. The result had an error (difference between answers) of 3.69149E-15, so I would be interested in knowing your method that gives a lower error. Thanks.
I'm rusty at Excel. Do you do that entire calculation in a single cell? If so, what is your formula? I did the same calculation in Excel but used 61 cells for the (zero to m) individual terms that might have to be added together, then did the two sums over the appropriate sets of cells. The result had an error (difference between answers) of 3.69149E-15, so I would be interested in knowing your method that gives a lower error. Thanks.
March 18th, 2012 at 12:00:26 PM
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silly
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I Heart Vi Hart
March 18th, 2012 at 12:31:29 PM
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Thanks. I guess Microsoft has a more efficient algorithm for the BINOMDIST() than the individual multiplication and exponentiation steps. The individual terms (BINOMDIST vs. the equation) differ by no more than 2E-16 and mostly much smaller, with some differences being positive and some negative. Guess I don't do much work these days where that kind of error is significant. In my checkbook I don't even consider fractional cents. ;-)
March 18th, 2012 at 12:52:00 PM
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Yes, many Excel rounding errors are well known.
My BF uses xlPrecision for his Excel calculations
xlPrecision
I also have it but never use it.
Wow, I don't even have a checkbook!
added:
OP, nice question.
I hope you can now solve these kind of problems
A lot of math without a calculator or spreadsheet
My BF uses xlPrecision for his Excel calculations
xlPrecision
I also have it but never use it.
Wow, I don't even have a checkbook!
added:
OP, nice question.
I hope you can now solve these kind of problems
A lot of math without a calculator or spreadsheet
I Heart Vi Hart
