Stuart McLachlan
stuart at lexacorp.com.pg
Tue Jan 15 00:27:24 CST 2013
I'll happily play against you if you give me 3:1 odds :-) With one game played, there are not 2^6 possible outcomes. Many of those are precluded by the rule which says that the contest finishes when either side has four wins. There would only be that many psibilities if you continue to toss three more coins after getting four heads in a row etc. -- Stuart On 15 Jan 2013 at 0:06, Arthur Fuller wrote: > Here's my analysis. > > With one game played, there remain 2^6 possible outcomes, or 64. Of these, > only 1 can end it in 5 games. There are 6 possible outcomes in which you > win while I win only 1 more game. Represent these using Y and M (since I > won the first game, I'll be M (me) and you'll be Y (you): > > MYYYYY YMYYYY YYMYYY YYYMYY YYYYMY YYYYYM > > There are 15 possible sequences in which you win 4 games while I win 2 > more. In all the other possible outcomes, I win 3 and thus the match. This > means that of the 64 possible outcomes, you win the match in 1 + 6 + 15 = > 22 outcomes, while I win the other 42. As a result, the probability of your > victory is 22/64, or 11/32, or just a nudge better than 1/3 (.34375). > > A. > _______________________________________________ > dba-Tech mailing list > dba-Tech at databaseadvisors.com > http://databaseadvisors.com/mailman/listinfo/dba-tech > Website: http://www.databaseadvisors.com >