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Strategies & Market Trends : ahhaha's ahs -- Ignore unavailable to you. Want to Upgrade?

To: Bilow who wrote (560)	12/8/2000 11:56:07 PM
From: Bilow	Read Replies (2) \| Respond to of 24758

Hi all; Given the challenge of finding a way of getting odds of 263 to 1 against choosing HAL by random chance, I cannot resist, so here are some rules for a game that will give those (exact) odds of winning. I don't mean to suggest that Clarke used this technique to compute his odds (my guess is that he has a list of 264 3-letter words, only one of which is "HAL"), only to show that odds of 263 to 1 are possible to compute strictly mathematically from the alphabet alone. We fill an urn with 13 checkers. Each checker has two letters on it, one on each side. The two letters are consecutive in the alphabet. The set of checkers can be listed as {(A,B),(C,D),(E,F),...,(Y,Z)}. We remove three checkers from the urn at random in order, (i.e. ordered removal without replacement), and we will turn a random side of each checker up. The resulting three-letter word will be a win in two ways. The first is if it matches "IBM", or is an offset from IBM by constant values for the three letters, with offsets wrapping around A to Z. I.e., "HAL" is a win, as is "JCN" or "GZK". The second way is if it matches the reverse of IBM, i.e. "MBI", or is an offset of that, again wrapping around A to Z. The second rule allows wins of "LAH", "KZG", JYF", ... What are the odds of a win? First note that not all letter sequences are possible. For instance, "ZZZ" or "ABC" are impossible, as the AB or YZ checker can only be removed once rather than 3 or 2 times, respectively. Also note that the various letter sequences that are possible in this game, are equally probable, and each can be obtained in only one way. The probability of getting a particular (possible) sequence is 1/26 for the first letter, 1/24 for the second, and 1/22 for the third. The probability of getting any particular sequence is therefore the product: 1/(262422) = 1/13728. The set of winning sequences is simply {HAL, IBM, JCN, ..., FYJ, GZK} + {LAH, MBI, NCJ, ... JYF, KZG}. Note that each subset has 26 elements, and that the two subsets share no common element. There are therefore a total of 26+26 = 52 winning letter sequences. The probability of winning is therefore 52/13728 = 1/264. The odds of winning are therefore 263 to 1, as required. -- Carl