Was Hael
I'm here, after a long absence spent digging myself out from under piles of email. I'm sorry to have missed the action, particularly some question of probability to which I had been invited to contribute (Gene, "You Have Been Summoned," 090313).
In honor of probability, I shall offer here a delightful shortcut when calculating odds due to John Scarne. Suppose you repeatedly try for an unlikely outcome--let us say, the same birthday as mine (for simplicity, we'll ignore leap year, Scarne did). How many independent tries suffice to lower the odds of at least one success to 1-1 (even odds)? Here the odds are 364-1 against, and the exact calculation is tedious. For the only time that I noticed in his monumental work On Gambling, Scarne appealed to mathematics beyond (exceedingly elaborate) arithmetic and came up with a delightful shortcut: Take the 364 and multiply by 0.693 (the natural logarithm of two). In general, for odds of (n-1) to 1 out of n, (n-1) log2 repetitions will bring the odds of at least one success to even. But Scarne is very slightly wrong! Comparing the approximations (n-1) log2 and n log2, the proportional errors are second order, and the third order difference between them is minutely in favor of n log2. (n-1) log2 is low; n log2 is high; the correct value is about halfway between; n log2 is slightly closer; for large n, (n-1/2) log2 is very, very close.
In honor of probability, I shall offer here a delightful shortcut when calculating odds due to John Scarne. Suppose you repeatedly try for an unlikely outcome--let us say, the same birthday as mine (for simplicity, we'll ignore leap year, Scarne did). How many independent tries suffice to lower the odds of at least one success to 1-1 (even odds)? Here the odds are 364-1 against, and the exact calculation is tedious. For the only time that I noticed in his monumental work On Gambling, Scarne appealed to mathematics beyond (exceedingly elaborate) arithmetic and came up with a delightful shortcut: Take the 364 and multiply by 0.693 (the natural logarithm of two). In general, for odds of (n-1) to 1 out of n, (n-1) log2 repetitions will bring the odds of at least one success to even. But Scarne is very slightly wrong! Comparing the approximations (n-1) log2 and n log2, the proportional errors are second order, and the third order difference between them is minutely in favor of n log2. (n-1) log2 is low; n log2 is high; the correct value is about halfway between; n log2 is slightly closer; for large n, (n-1/2) log2 is very, very close.
We've missed you, Wabulon -- welcome back!
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