larry@utcsstat.UUCP (larry) (10/07/85)
>In trying to find the probability of a certain occurence, it is often >easier to try to find the probability that it does NOT occur. So, in >trying to find the probability that two card values X and Y occur >consecutively in a deck, assume X is at position i in the deck. The >probability that a Y does NOT occur at position i-1 is p1=47/51 (the >number of non Y's over the number of cards left). The probability >that a Y also does not occur at i+1 is p2=p1*46/50 (since we used up >a non-Y at i-1). The probability that this happens for all 8 cards >is p2**8 which equals about .267. Thus the probability that NO X >and Y will be consecutive is about 26.7%. (There is probably a slight >statistical variation at the beginning and end of the deck.) > >Ken Doyle > > No! Using p2**8 assumes that all these events are INDEPENDENT which they are not. The enumeration for all the possible arrangements of the eight cards still persists as a messy problem. Larry Wasserman -- {allegra,ihnp4,linus,decvax}!utzoo!utcsstat!larry {ihnp4|decvax|utzoo|utcsrgv}!utcs!utzoo!utcsstat!larry