If you have an expected loss of 7.07 cents on each play, the total is NOT a geometric progression. It is just -7.07-7.07-7.07...= -7.07n cents where n is the number of times you played.
you bet $5 in a game and there is a 251/ 495 probability that you wil lose that $5 and 244/495 probabilty taht you will win that you will win $5. given that your expected value is -7.07 cents, how much would you lose in the long run?
this is my working:
1st round: win -7.07 cents
2nd rounnd: win -7.07 cents
nth round: win -7.07 cents
thus the answer should be a GP of 7.07 /( 1- 7.07) ...
but my answer in my notes states that it is 1.4 cents.
may i know where i have gone wrong? thanks!
I think the book is giving an intermediate answer.
The problem seems unrealistic, though
(in terms of either losing the entire 5 dollars or not).
Round1 E(loss) =
Round2 E(loss) =
Round3 E(loss) =
E(loss) long term =
The two geometric sequences are
The sum to infinity of the first is
The sum to infinity of the second is
Hence the expected loss is 5 dollars multiplied by
The initial E(loss) is the "round 1" loss from $5.
This value shrinks from round to round, while the probability of loss remains constant.
Then in round 2 (being purely mathematical, considering that 7.07 cents had been lost),
there is 5 dollars minus 7.07 cents to bet with, to put it that way!
The expected loss is not 7.07 cents now, it's
of the "balance".
Continuing on in this way develops the pair of geometric series.