
Stanton D. answered 12/16/19
Tutor to Pique Your Sciences Interest
Hi Amy A.,
Sometimes these probability questions might seem to involve endless calculations -- but this one is just "ripe for the picking", as you will discover.
So what is the exact situation? Karpov has won 5 games, only needs 1 more game win for the championship. Kasparov has 3 wins, and would need 3 more, BEFORE Karpov gets any, to get the championship.
The first thing to do, is to disregard the non-essentials. Draw games are useless for determining a victor; therefore, consider only the universe in which the is a game winner to each of the next 3 games. Why only 3? B/c that's the minimum number required for Kasparov to win. You don't need to worry about what Karpov is doing meanwhile!
So, if the probabilities remained random, the ratio of Kasparov winning vs. Karpov winning is 3:5 or 3/8 as a probability. For 3 successive times that becomes: (You do the math!).
Of course, seldom are contests just random! Karpov had won games 3,6,7,9, and 27 (what does he favor, multiples of 3?) but Kasparov was on a roll with games 32, 47, and 48 (multiples of 2?); the Chess Federation terminated the match ("deep-sixed it", appropriately enough?) even though both players wanted to continue it, citing ostensible health concerns for the players. Really? With 48 games over 6 months?
Myself, I think the fix was in -- Kasparov went on to (narrow) victory in 1985, 1986, and 1987 subsequently.
--- Cheers, -- Mr. d.