exercise:F66abbb0e8: Difference between revisions

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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 deck of ESP cards consists of 20 cards each of two types: say ten stars, ten circles (normally there are five types). The deck is shuffled and the cards turned up one at a time. You, the alleged percipient, are to name the symbol on each card...")
 
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<div class="d-none"><math>
A deck of ESP cards consists of 20 cards each of two types: say ten stars, ten circles (normally there are five types).  The deck is shuffled and
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the cards turned up one at a time.  You, the alleged percipient, are to name the symbol on each card ''before'' it is turned up.
\newcommand{\mat}[1]{{\bf#1}}
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\newcommand{\secstoprocess}{\all}
\newcommand{\NA}{{\rm NA}}
\newcommand{\mathds}{\mathbb}</math></div> A deck of ESP cards consists of 20 cards each of two types:
say ten stars, ten circles (normally there are five types).  The deck is shuffled and
the cards turned up one at a time.  You, the alleged percipient, are to name the
symbol on each card ''before'' it is turned up.


Suppose that you are really just guessing at the cards.  If you do not get to see each card after you have made your guess, then it is easy to calculate the expected number of correct guesses, namely ten.


Suppose that you are really just guessing at the cards.  If you do not get to see
If, on the other hand, you are guessing with information, that is, if you see each card after your guess, then, of course, you might expect to get a higher score.  This is indeed the case, but calculating the correct expectation is no longer easy.
each card after you have made your guess, then it is easy to calculate the expected
number of correct guesses, namely ten.


But it is easy to do a computer simulation of this guessing with information, so we can get a good idea of the expectation by simulation.  (This is similar to the way that skilled blackjack players make blackjack into a favorable game by observing the cards that have already been played.  See [[exercise:2257237708|Exercise]])


If, on the other hand, you are guessing with information, that is, if you see each
<ul style="list-style-type:lower-alpha"><li> First, do a simulation of guessing without information, repeating the
card after your guess, then, of course, you might expect to get a higher score.  This
is indeed the case, but calculating the correct expectation is no longer easy.
 
 
But it is easy to do a computer simulation of this guessing with information, so we
can get a good idea of the expectation by simulation.  (This is similar to the way
that skilled blackjack players make blackjack into a favorable game by observing the
cards that have already been played.  See Exercise \ref{exer 6.1.29}.)
<ul><li> First, do a simulation of guessing without information, repeating the
experiment at least 1000 times.  Estimate the expected number of correct answers and
experiment at least 1000 times.  Estimate the expected number of correct answers and
compare your result with the theoretical expectation.
compare your result with the theoretical expectation.
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and <math>h(0,0) = h(-1,0) = h(0,-1) = 0</math>.  Using this relation, write a program to
and <math>h(0,0) = h(-1,0) = h(0,-1) = 0</math>.  Using this relation, write a program to
compute <math>h(S,C)</math> and find <math>h(10,10)</math>.  Compare the computed value of <math>h(10,10)</math> with
compute <math>h(S,C)</math> and find <math>h(10,10)</math>.  Compare the computed value of <math>h(10,10)</math> with
the result of your simulation in (c).  For more about this exercise and
the result of your simulation in (c).  For more about this exercise and [[exercise:58957e3826|Exercise]] see Diaconis and Graham.<ref group="Notes" >P. Diaconis and R. Graham, “The Analysis of Sequential Experiments with Feedback to Subjects,” ''Annals of Statistics,'' vol. 9 (1981), pp. 3--23.</ref>
Exercise \ref{exer 6.1.26} see Diaconis and Graham.<ref group="Notes" >P. Diaconis and R. Graham, “The Analysis of Sequential Experiments with Feedback to
Subjects,” ''Annals of Statistics,'' vol. 9 (1981), pp. 3--23.</ref>
</li>
</li>
</ul>
</ul>

Latest revision as of 16:52, 14 June 2024

A deck of ESP cards consists of 20 cards each of two types: say ten stars, ten circles (normally there are five types). The deck is shuffled and the cards turned up one at a time. You, the alleged percipient, are to name the symbol on each card before it is turned up.

Suppose that you are really just guessing at the cards. If you do not get to see each card after you have made your guess, then it is easy to calculate the expected number of correct guesses, namely ten.

If, on the other hand, you are guessing with information, that is, if you see each card after your guess, then, of course, you might expect to get a higher score. This is indeed the case, but calculating the correct expectation is no longer easy.

But it is easy to do a computer simulation of this guessing with information, so we can get a good idea of the expectation by simulation. (This is similar to the way that skilled blackjack players make blackjack into a favorable game by observing the cards that have already been played. See Exercise)

  • First, do a simulation of guessing without information, repeating the experiment at least 1000 times. Estimate the expected number of correct answers and compare your result with the theoretical expectation.
  • What is the best strategy for guessing with information?
  • Do a simulation of guessing with information, using the strategy in (b). Repeat the experiment at least 1000 times, and estimate the expectation in this case.
  • Let [math]S[/math] be the number of stars and [math]C[/math] the number of circles in the deck. Let [math]h(S,C)[/math] be the expected winnings using the optimal guessing strategy in (b). Show that [math]h(S,C)[/math] satisfies the recursion relation
    [[math]] h(S,C) = \frac S{S + C} h(S - 1,C) + \frac C{S + C} h(S,C - 1) + \frac {\max(S,C)}{S + C}\ , [[/math]]
    and [math]h(0,0) = h(-1,0) = h(0,-1) = 0[/math]. Using this relation, write a program to compute [math]h(S,C)[/math] and find [math]h(10,10)[/math]. Compare the computed value of [math]h(10,10)[/math] with the result of your simulation in (c). For more about this exercise and Exercise see Diaconis and Graham.[Notes 1]

Notes

  1. P. Diaconis and R. Graham, “The Analysis of Sequential Experiments with Feedback to Subjects,” Annals of Statistics, vol. 9 (1981), pp. 3--23.