Re: Deal or no Deal
- From: "zcx" <sdadsa@xxxxxxxxxxxx>
- Date: Sat, 25 Feb 2006 16:17:59 GMT
"John Dean" <john-dean@xxxxxxxxxxxxxxxx> wrote in message
news:dtppt4$53t$1@xxxxxxxxxxxxxxxxxxxxxxx
zcx wrote:
"John Dean" <john-dean@xxxxxxxxxxxxxxxx> wrote in message
news:dto92d$727$1@xxxxxxxxxxxxxxxxxxxxxx
Yeah. Weird, innit? You have to distinguish between the probability
(which is set by the way the boxes are originally allocated) and the
reality (about which you may get more information as the boxes are
opened). Try it with a set of playing cards - 11 red and 11 black. Pick
one at
random (face down). The probability it is black is 0.5
Yes, at that point it is.
Draw cards from the remaining 21 one at a time and examine them.
What, in this process, could change the probability that you have a
black card?
Probability is not the same value for all time.
The more blacks you remove, the less likely that the card you have is
black. If you remove all 11 black cards, then the probability that
your card is black = 0.
The reason for this is that the events are *not* independent and
therefore information about one event gives you more information
about what is likely to happen in the others.
How do you think card counting works?
Think about it like this.
I remove 11 blacks from the pack and am left with 2 cards, which I
now know are both red.
What if I leave those 2 cards sitting about for a few days, then give
them to someone else. I tell them they are both red and they claim
the prob. of the facedown card being red is 1.
I say "No, they came from a pack with equal reds and blacks and
therefore the probability of the facedown card being black is 0.5".
They'd suggest medication...
No, because you're into a new event which begins with 2 red cards. Deal or
No deal begins with 22 boxes. That sets the probability. You're mixing
what you know about the probabilities with what information you have
discovered.
Sorry, but information you discover along the way shapes probability. You
really do need to try and understand this and also *sampling without
replacement* (which is what DOND is).
*look it up
I really don't know how make it any simpler for you (though I will try).
Say you have 22 boxes and you pick them randomly.
What are the odds that the 250k is in the last box to be chosen? (Obviously
it is 1 in 22. )
You can calculate this by the following:
On the first pick there is a 21/22 chance of you not picking the 250k.
On the 2nd pick there is a 20/21 chance of you not picking the 250k (because
you removed one of the non-250k boxes)
.............
............
On the 21st pick there is a 1/2 chance of you not picking the 250k.
So the probability of picking the 250k last = 21/22 * 20/21 * 19/20 * .... *
2/3 * 1/2 which (unsurprisngly) = 1 in 22 after cancelling.
By your logic :
the chance of not picking 250k on first pick = 21/22
the chance of not picking 250k on the second pick = 21/22 (because that's
what it was at the start and can "never change")
.................
................
And when there's only two boxes left the probability of you picking 250k
GIVEN that you have not picked it in any of the previous 20 picks) is also
21/22.
The above is obviously ridiculous, yet it's what you are claiming.
.
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