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(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> A perpetual craps game goes on at Charley's. Jones comes into Charley's on an evening when there have already been 100 plays. He plans to play until the next time that snake eyes (a pair of ones) are rolled. Jones wonders how many times he wil...")
 
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\newcommand{\mathds}{\mathbb}</math></div> A perpetual craps game goes on at
\newcommand{\mathds}{\mathbb}</math></div> A perpetual craps game goes on at Charley's.  Jones comes into Charley's on an evening when there have already been 100 plays.  He plans to play until the next time that snake eyes (a pair of ones) are rolled. Jones wonders how many times he will play.  On the one hand he realizes that the average time between snake eyes is 36 so he should play about 18 times as he is equally likely to have come in on either side of the halfway point between occurrences of snake eyes.  On the other hand, the dice have no memory, and so it would seem that he would have to play for 36 more times no matter what the previous outcomes have been.  Which, if either, of Jones's arguments do you believe? Using the result of [[exercise:Cf2779f01a |Exercise]], calculate the expected to reach snake eyes, in equilibrium, and see if this resolves the apparent paradox.  If you are still in doubt, simulate the experiment to decide which argument is correct.  Can you give an intuitive argument which explains this result?
Charley's.  Jones comes  
into Charley's on an evening when there have already been 100 plays.  He plans
to
play until the next time that snake eyes (a pair of ones) are rolled. Jones
wonders
how many times he will play.  On the one hand he realizes that the average time
between snake eyes is 36 so he should play about 18 times as he is equally
likely to have come in on either side of the halfway point between occurrences
of snake eyes.  On the other hand, the dice have no memory, and so it would
seem that he would have to play for 36 more times no matter what the previous
outcomes have been.  Which, if either, of Jones's arguments do you believe?  
Using the result of Exercise [[exercise:Cf2779f01a |Exercise]], calculate the expected to  
reach snake eyes, in equilibrium, and see if this resolves the apparent
paradox.   
If you are still in doubt, simulate the experiment to decide which argument is
correct.  Can you give an
intuitive argument which explains this result?

Latest revision as of 02:29, 15 June 2024

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A perpetual craps game goes on at Charley's. Jones comes into Charley's on an evening when there have already been 100 plays. He plans to play until the next time that snake eyes (a pair of ones) are rolled. Jones wonders how many times he will play. On the one hand he realizes that the average time between snake eyes is 36 so he should play about 18 times as he is equally likely to have come in on either side of the halfway point between occurrences of snake eyes. On the other hand, the dice have no memory, and so it would seem that he would have to play for 36 more times no matter what the previous outcomes have been. Which, if either, of Jones's arguments do you believe? Using the result of Exercise, calculate the expected to reach snake eyes, in equilibrium, and see if this resolves the apparent paradox. If you are still in doubt, simulate the experiment to decide which argument is correct. Can you give an intuitive argument which explains this result?