Quote:
Originally Posted by SheetWise
Are you sure?
In the case of a binary choice (win/lose each bet), yes.
Take 10 bets and these two ways to break even (of many).
+++++-----
and
+-+-+-+-+-
These have the same probability of occurring, but one has a a maximum excursion of 5, and the other a maximum excursion of 1.
Let's look at why. We start with + in both cases.
Then 50% +/- for 2 possible 2-bet sequences with equal probability
++
+-
Then 50% +/- for 4 subsequent 3-bet sequences all equal probability
+++
++-
+-+
+--
Then 50% +/- for 8 subsequent 4-bet sequences all equal probability
++++
+++-
++-+
++--
+-++
+-+-
+--+
+---
If we continue this we get two paths to the two 10-bet sequences I showed at the beginning, and both of them will be a summation of identical probabilities.
It should be obvious that all binary sequences we can construct of a given length and common endpoint always have the same probability to occur in a random walk. Substitute 1s and 0s if it helps conceptualize it.
Edit:
This thread might be helpful too. They discuss why the dice sequences (a) 11111111111111111111; or (b) 63335643331622221214.
have the exact same probability to occur.
Last edited by NewOldGuy; 01-23-2012 at 11:58 PM.