exercise:72af71c8ee: Difference between revisions
(Created page with "<div class="d-none"><math> \newcommand{\NA}{{\rm NA}} \newcommand{\mat}[1]{{\bf#1}} \newcommand{\exref}[1]{\ref{##1}} \newcommand{\secstoprocess}{\all} \newcommand{\NA}{{\rm NA}} \newcommand{\mathds}{\mathbb}</math></div> We can use the gambling interpretation given in Exercise Exercise to find the expected number of tosses required to reach pattern B when we start with pattern A. To be a meaningful problem, we assume that pattern A does...") |
No edit summary |
||
| Line 5: | Line 5: | ||
\newcommand{\secstoprocess}{\all} | \newcommand{\secstoprocess}{\all} | ||
\newcommand{\NA}{{\rm NA}} | \newcommand{\NA}{{\rm NA}} | ||
\newcommand{\mathds}{\mathbb}</math></div> We can use the gambling | \newcommand{\mathds}{\mathbb}</math></div> We can use the gambling interpretation given in [[exercise:18da878882 |Exercise]] to find the expected number of tosses required to reach pattern B when we start with pattern A. To be a meaningful problem, we assume that pattern A does not have pattern B as a subpattern. Let <math>E_A(T^B)</math> be the expected time to reach pattern B starting with pattern A. We use our gambling scheme and assume that the first k coin tosses produced the pattern A. During this time, the gamblers made an amount AB. | ||
interpretation given in | The total amount the gamblers will have made when the pattern B occurs is BB. Thus, the amount that the gamblers made after | ||
number of tosses required to reach pattern B when we start with pattern | the pattern A has occurred is BB - AB. Again by the fair game argument, <math>E_A(T^B)</math> = BB-AB. | ||
A. To be a meaningful problem, | |||
we assume that pattern A does not have pattern B as a subpattern. Let | |||
<math>E_A(T^B)</math> be | |||
the expected time to reach pattern B starting with pattern A. | |||
use our gambling scheme and assume that the first k coin tosses produced | |||
the pattern A. During this time, the gamblers made an amount AB. | |||
The total amount the gamblers will have made | |||
when the pattern B occurs | |||
is BB. Thus, the amount that the gamblers made after | |||
the pattern A has occurred is BB - AB. Again by the fair game argument, | |||
<math>E_A(T^B)</math> = BB-AB. | |||
For example, suppose that we start with pattern A = HT and are trying to get the pattern B = HTH. Then we saw in [[exercise:18da878882|Exercise]] that AB = 4 and BB = 10 so <math>E_A(T^B)</math> = BB-AB=6. | |||
Verify that this gambling interpretation leads to the correct answer for all starting states in the examples that you | |||
worked in [[exercise:18da878882 |Exercise]]. | |||
Verify that this gambling interpretation | |||
leads to the correct answer for all starting states in the examples that you | |||
worked in | |||
Latest revision as of 23:43, 15 June 2024
We can use the gambling interpretation given in Exercise to find the expected number of tosses required to reach pattern B when we start with pattern A. To be a meaningful problem, we assume that pattern A does not have pattern B as a subpattern. Let [math]E_A(T^B)[/math] be the expected time to reach pattern B starting with pattern A. We use our gambling scheme and assume that the first k coin tosses produced the pattern A. During this time, the gamblers made an amount AB.
The total amount the gamblers will have made when the pattern B occurs is BB. Thus, the amount that the gamblers made after the pattern A has occurred is BB - AB. Again by the fair game argument, [math]E_A(T^B)[/math] = BB-AB.
For example, suppose that we start with pattern A = HT and are trying to get the pattern B = HTH. Then we saw in Exercise that AB = 4 and BB = 10 so [math]E_A(T^B)[/math] = BB-AB=6.
Verify that this gambling interpretation leads to the correct answer for all starting states in the examples that you worked in Exercise.