lab@qubix.UUCP (Larry Bickford) (10/12/83)
Let's make this simple by enumeration. Possibilities:
(> == I pick; M == what MH shows)
> B M B G -> Gain by switching
M B > B G -> Gain by switching
M B B > G -> Lose by switching
B M B > G -> Lose by switching
What is being overlooked (cabinet/drawer/coin problem again) is that I
can choose from TWO bad doors, which are not identical!!!
This move the odds from 2/1 (p = 2/3) to 2/2 (p = 1/2).
Can I make it any simpler?
Larry Bickford,
{ihnp4,ucbvax,decvax}!decwrl!qubix!labmcewan@uiucdcs.UUCP (10/17/83)
#R:qubix:-59200:uiucdcs:28200027:000:708
uiucdcs!mcewan Oct 16 11:09:00 1983
Let's make this simple by enumeration. Possibilities:
(> == I pick; M == what MH shows)
> B M B G -> Gain by switching
M B > B G -> Gain by switching
M B B > G -> Lose by switching
B M B > G -> Lose by switching
What is being overlooked (cabinet/drawer/coin problem again) is that I
can choose from TWO bad doors, which are not identical!!!
This move the odds from 2/1 (p = 2/3) to 2/2 (p = 1/2).
Can I make it any simpler?
/* ---------- */
You've already made it too simple. The problem with this is that the four
cases shown are NOT equally probable. The probabilities are 1/3, 1/3, 1/6
and 1/6 respectively => P(gain by switching) = 2/3.
Scott McEwan
pur-ee!uiucdcs!mcewanCSvax:Pucc-H:ags@pur-ee.UUCP (10/17/83)
Let's make this simple by enumeration. Possibilities:
(> == I pick; M == what MH shows)
> B M B G -> Gain by switching
M B > B G -> Gain by switching
M B B > G -> Lose by switching
B M B > G -> Lose by switching
What is being overlooked (cabinet/drawer/coin problem again) is that I
can choose from TWO bad doors, which are not identical!!!
This move the odds from 2/1 (p = 2/3) to 2/2 (p = 1/2).
Can I make it any simpler?
Larry Bickford,
{ihnp4,ucbvax,decvax}!decwrl!qubix!lab
No, you can't make it any simpler. It is already perfectly obvious where you
went wrong. You should have learned from the coins-in-cabinets problem that
simple enumeration gives wrong answers if you insist on always assigning
equal probabilities to different events. Your third and fourth events are each
half as likely as the other two. I suggest you go back and read the previous
discussion, since many correct answers have already been given. If you are
still not convinced, then I would very much like to play you the following
game for money:
(1) We each contribute $10 to the pot. An impartial referee (call him Monty)
puts the $20 in one of three boxes.
(2) You choose one of the three boxes.
(3) Monty displays an empty box.
(4) I take the one remaining box.
(5) We each keep the contents of whichever box we chose.
You will choose the right box 1/3 of the time. Whenever you don't get the
$20, I will. Therefore I win 2/3 of the time. I simulated this in LISP
and won 673 times out of 1000. Here is the program:
(defun makedeal (n)
(do ((trials 1 (add1 trials)) (hits 0))
((equal trials n) (list trials hits (quotient (float hits) trials)))
(and (hitp (random 3) (random 3)) (setq hits (add1 hits)))))
(defun hitp (goodprize firstchoice)
(eq goodprize (lastchoice firstchoice (badprize goodprize firstchoice))))
(defun badprize (goodprize firstchoice)
(cond
((eq goodprize firstchoice) (random-other firstchoice))
(t (lastchoice goodprize firstchoice))))
(defun random-other (n) (mod (add n (random 2) 1) 3))
(defun lastchoice (a b) (difference 3 a b))
Dave Seaman
..!pur-ee!pucc-h:ags