| Title: | Mathematics at DEC |
| Moderator: | RUSURE::EDP |
| Created: | Mon Feb 03 1986 |
| Last Modified: | Fri Jun 06 1997 |
| Last Successful Update: | Fri Jun 06 1997 |
| Number of topics: | 2083 |
| Total number of notes: | 14613 |
An easier coin-flipping problem: Flip a fair coin until you get heads. What's the average number of times you have to flip ? /Eric
| T.R | Title | User | Personal Name | Date | Lines |
|---|---|---|---|---|---|
| 641.1 | 2 | COMET::ROBERTS | Dwayne Roberts | Mon Jan 05 1987 18:30 | 7 |
(1/2)*1 + (1/4)*2 + (1/8)*3 + (1/16)*4 + ... + (2^-i)*i + ...
= (1/2)/((1-1/2)^2)
= 2
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| 641.2 | a more general solution | ESTORE::ROOS | Thu Jan 15 1987 15:31 | 24 | |
I agree wih the solution of:
(1/2)*1 + (1/4)*2 + (1/8)*3 + ... + (2^-i)*i + ...
The sum does approach 2.
As a matter of fact the sum of the first i terms = 2 - ((i + 2)/(2^i))
Thus, as i becomes large ((i + 2)/(2^i)) converges to zero
and the sum converges to 2.
Got any more interesting problems!!! The one in Note 622 looks
nice.
Note: first two terms add to 1
starting with the 3 rd term
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