If the probability is 4*11*6*9/132^4, then we have a hit.
... yes, but in my mind once I've picked up a letter, I must erase that letter from the sentence, sorry (how could that figure in the "puzzle-sentence" without using too many words? In French it is a "Tirage sans remise") In that case, aren't the odds 4*11*6*9/132*131*130*129? I've almost a hit in French, by reducing the set of possibilities with the "puzzle-sentence" ending in: "[The probability of...] is one on K" -- with K = 132*131*130*129/4*11*6*9. (I didn't count the digits after the comma, though, which is neither correct, nor very elegant! But my error is below 1/10000... Best, É.
Le 11 décembre 2019 à 13:16, Hans Havermann <gladhobo@bell.net> a écrit :
The odds of picking at random the four letters F, O, U, R in this sentence and in that order are Y out of Z.
EA: "The 'FOUR' letter challenge is still open!"
If Y = "one" and Z = "one hundred twenty-seven thousand, seven hundred seventy-six", I make the total letter count of the sentence to be 132, with 4 of 'f', 11 of 'o', 6 of 'u', and 9 of 'r'. If the probability is 4*11*6*9/132^4, then we have a hit. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun