Mailing List Archive: 49091 messages

## [REBOL] My Statistical Thoughts on Monty Hall Problem (non-REBOL)

### From: reboler:programmer at: 20-Dec-2001 2:25

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Okay you guys brought this up.

Here is my speculation on the Monty Hall Problem.

Let me see what you guys think about the switch/win probability for the following circumstances.

Monty picks a door randomly from [123].
You pick a door, and tell Monty.
Monty removes the door number from his list that is not yours, and not his.

You are given a choice: stay with your door number -or- pick the remaining door number.
*** You flip a fair coin: 	heads - you pick the first remaining number
tails - you pick the second remaining number
Either heads or tails could be keeper or switcher depending on your first choice

You win if your door = Monty's door.

*** Important part follows ***

Your coin flip is COMPLETELY INDEPENDENT (in the true Probability sense) of the first
choice from [123]: the coin flip can have no influence on which remaining door is the
winner, and which remaining door is the winner cannot influence the coin toss. Further,
the coin flip is completely independent of your first choice: your first choice cannot
influence the coin, and the coin had nothing to do with your first choice.

*** Therefore: Heads or Tails both have a 1/2 chance of coming up, and 1/2 chance of
matching Monty's number (winning).
That is the very definition of statistical independence.

Note: In this scenario both you and Monty have the same information as in the "traditional"
scenario (without a coin flip).

Analysis of this suggests that:

- if the Monty Hall Problem gives the switch/win probability of 2/3
- then -

A) choosing a door reduces the chances that the door is a winner (keeper win = 1/3, switcher
win =2/3)
or

B) Monty removing a non-picked door reduces the chance that a picked door is a winner.

*** If I call heads before a coin is flipped, does that reduce the chances that heads
comes up?***

*** If I call a number on a die, and you then scratch off one of those numbers (not mine)
does that reduce the chance that my number will come up? ***

(A) clearly cannot be the case: choosing a door does not change what is behind it.

(B) clearly cannot be the case: removing a door does not change what is behind the other
doors.

Seems to me that switch/win probability = 2/3 is the same as saying that a fair coin
will come up with heads or tails more often depending on which one is the winner.

My explanation: there are TWO INDEPENDENT CHOICES.

Switch and keep each have a 1/2 half probability because Monty's choice of the door to
open is not RANDOM (i.e. independent of your choice and the winning door).

*** You have essentially made TWO independent choices, whether you switch or keep. The
keeper simply chooses the same door again, the switcher chooses a different door.

Consider: Monty opens the door before you choose:

Monty shows you three doors, and opens one of them (ALWAYS A NON-WINNER no matter what
you choose)
You pick a door.
Monty asks you to switch or keep.
Since your choice was between two doors, if doesn't matter if you switch or keep
- all probabilities are 1/2

MONTY IS ALWAYS GOING TO PICK A NON-WINNER, YOUR FIRST CHOICE IS TOTALLY IRRELEVANT.
SWITCHING AND KEEPING ARE NON-REAL CHOICES, ONLY YOUR SECOND CHOICE COUNTS.
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