Unformatted text preview: T = 1
=4
⇡ (T T ) However, this time if we start with a T at time 1 and a T at time 0, then a
T at time 1 will give us a T T and a return to T T at time 1; while if we get a
H at time 1, then we have wasted 1 turn and we have nothing that can help us
later, so
1
1
4 = ET T TT T = · 1 + · (1 + ETT T )
2
2
Solving gives ETT T = 6, so it takes longer to observe T T . The reason for this,
which can be seen in the last equation, is that once we have one T T , we will
get another one with probability 1/2, while occurrences of HT cannot overlap.
In the Exercise 1.59 we will consider waiting times for three coin patterns.
The most interesting of these is ETHT H = ETT HT .
Example 1.49. Duration of fair games. Consider the gambler’s ruin chain
in which p(i, i + 1) = p(i, i 1) = 1/2. Let ⌧ = min{n : Xn 62 (0, N )}. We
claim that
Ex ⌧ = x(N x)
(1.26)
To see what formula (1.26) says, consider matching pennies. There N = 25 and
x = 15, so the game will take 15 · 10 = 150 ﬂips on the average. If there are
twice as many coins, N = 50 and x = 30, then the game takes 30 · 20 = 600
ﬂips on the average, or four times as long.
There ar...
View
Full
Document
This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell.
 Spring '10
 DURRETT
 The Land

Click to edit the document details