⧼exchistory⧽
BBy Bot
Jun 09'24
[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]

Prove that if

[[math]] P(A|C) \geq P(B|C) \mbox{\,\,and\,\,} P(A|\tilde C) \geq P(B|\tilde C)\ , [[/math]]

then [math]P(A) \geq P(B)[/math].

BBy Bot
Jun 09'24

The probability that it is in the [math]i[/math]th box is [math]p_i[/math]. If you search in the [math]i[/math]th box and it is there, you find it with probability [math]a_i[/math]. Show that the probability [math]p[/math] that the coin is in the [math]j[/math]th box, given that you have looked in the [math]i[/math]th box and not found it, is

[[math]] p = \left \{ \matrix{ p_j/(1-a_ip_i),&\,\,\, \mbox{if} \,\,\, j \ne i,\cr (1 - a_i)p_i/(1 - a_ip_i),&\,\,\,\mbox{if} \,\, j = i.\cr}\right. [[/math]]

BBy Bot
Jun 09'24

George Wolford has suggested the following variation on the Linda problem (see Exercise). The registrar is carrying John and Mary's registration cards and drops them in a puddle. When he pickes them up he cannot read the names but on the first card he picked up he can make out Mathematics 23 and Government 35, and on the second card he can make out only Mathematics 23. He asks you if you can help him decide which card belongs to Mary. You know that Mary likes government but does not like mathematics. You know nothing about John and assume that he is just a typical Dartmouth student. From this you estimate:

[[math]] \begin{array}{ll} P(\mbox {Mary\ takes\ Government\ 35}) &= .5\ , \\ P(\mbox {Mary\ takes\ Mathematics\ 23}) &= .1\ , \\ P(\mbox {John\ takes\ Government\ 35}) &= .3\ , \\ P(\mbox {John\ takes\ Mathematics\ 23}) &= .2\ . \end{array} [[/math]]

Assume that their choices for courses are independent events. Show that the card with Mathematics 23 and Government 35 showing is more likely to be Mary's than John's. The conjunction fallacy referred to in the Linda problem would be to assume that the event “Mary takes Mathematics 23 and Government 35” is more likely than the event “Mary takes Mathematics 23.” Why are we not making this fallacy here?

BBy Bot
Jun 09'24
[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]

(Suggested by Eisenberg and Ghosh[Notes 1]) A deck of playing cards can be described as a Cartesian product

[[math]] \mbox{Deck} = \mbox{Suit} \times \mbox{Rank}\ , [[/math]]

where [math]\mbox{Suit} = \{\clubsuit,\diamondsuit,\heartsuit,\spadesuit\}[/math] and [math]\mbox{Rank} = \{2,3,\dots,10,{\mbox J},{\mbox Q},{\mbox K},{\mbox A}\}[/math]. This just means that every card may be thought of as an ordered pair like [math](\diamondsuit,2)[/math]. By a suit event we mean any event [math]A[/math] contained in Deck which is described in terms of Suit alone. For instance, if [math]A[/math] is “the suit is red,” then

[[math]] A = \{\diamondsuit,\heartsuit\} \times \mbox{Rank}\ , [[/math]]

so that [math]A[/math] consists of all cards of the form [math](\diamondsuit,r)[/math] or [math](\heartsuit,r)[/math] where [math]r[/math] is any rank. Similarly, a rank event is any event described in terms of rank alone.

  • Show that if [math]A[/math] is any suit event and [math]B[/math] any rank event, then [math]A[/math] and [math]B[/math] are independent. (We can express this briefly by saying that suit and rank are independent.)
  • Throw away the ace of spades. Show that now no nontrivial (i.e., neither empty nor the whole space) suit event [math]A[/math] is independent of any nontrivial rank event [math]B[/math]. Hint: Here independence comes down to
    [[math]] c/51 = (a/51) \cdot (b/51)\ , [[/math]]
    where [math]a[/math], [math]b[/math], [math]c[/math] are the respective sizes of [math]A[/math], [math]B[/math] and [math]A \cap B[/math]. It follows that 51 must divide [math]ab[/math], hence that 3 must divide one of [math]a[/math] and [math]b[/math], and 17 the other. But the possible sizes for suit and rank events preclude this.
  • Show that the deck in (b) nevertheless does have pairs [math]A[/math], [math]B[/math] of nontrivial independent events. Hint: Find 2 events [math]A[/math] and [math]B[/math] of sizes 3 and 17, respectively, which intersect in a single point.
  • Add a joker to a full deck. Show that now there is no pair [math]A[/math], [math]B[/math] of nontrivial independent events. Hint: See the hint in (b); 53 is prime.

Notes

  1. B. Eisenberg and B. K. Ghosh, “Independent Events in a Discrete Uniform Probability Space,” The American Statistician, vol. 41, no. 1 (1987), pp. 52--56.
BBy Bot
Jun 09'24

Let [math]R_i[/math] be the event that the [math]i[/math]th player in a poker game has a royal flush. Show that a royal flush (A,K,Q,J,10 of one suit) attracts another royal flush, that is [math]P(R_2|R_1) \gt P(R_2)[/math]. Show that a royal flush repels full houses.

BBy Bot
Jun 09'24

Prove that [math]A[/math] attracts [math]B[/math] if and only if [math]B[/math] attracts [math]A[/math]. Hence we can say that [math]A[/math] and [math]B[/math] are mutually attractive if [math]A[/math] attracts [math]B[/math].

BBy Bot
Jun 09'24

Prove that [math]A[/math] neither attracts nor repels [math]B[/math] if and only if [math]A[/math] and [math]B[/math] are independent.

BBy Bot
Jun 09'24
[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]

Prove that [math]A[/math] and [math]B[/math] are mutually attractive if and only if [math]P(B|A) \gt P(B|\tilde A)[/math].

BBy Bot
Jun 09'24
[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]

Prove that if [math]A[/math] attracts [math]B[/math], then [math]A[/math] repels [math]\tilde B[/math].

BBy Bot
Jun 09'24

Prove that if [math]A[/math] attracts both [math]B[/math] and [math]C[/math], and [math]A[/math] repels [math]B \cap C[/math], then [math]A[/math] attracts [math]B \cup C[/math]. Is there any example in which [math]A[/math] attracts both [math]B[/math] and [math]C[/math] and repels [math]B \cup C[/math]?