Oldsoul
Think of it this way,
suppose the host didnt throw away the empty box but said, ' do you what to keep the one you have or exchange it for these two'
steve
by stevenyc 26 Replies latest jw friends
Oldsoul
Think of it this way,
suppose the host didnt throw away the empty box but said, ' do you what to keep the one you have or exchange it for these two'
steve
always stick with your first choice es
Break it down... it is easy with this few of choices... In order to show it, we have to assume that Monty Hall knows the answer or he wouldn't be able to discard a box... that is why the odds increase greater than 50% to change... Here we go... Monty knows Box C wins...
Pick A - Discard B - Swap to C, WIN
Pick B - Discard A - Swap to C, WIN
Pick C - Discard A or B - Swap, Lose...
2 out of 3 times, you WIN if you swap...
The odds increase to 2/3 BECAUSE of the knowledge required to remove a 'losing' box. If Monty didn't know the answer and randomly removed a box... well, let's see.
Pick A - Discard B - Swap to C, Win
Pick A - Discard C - Swap to B, Lose
Pick B - Discard A - Swap to C, Win
Pick B - Discard C - Swap to A, Lose
Pick C - Discard A - Swap to B, Lose
Pick C - Discard B - Swap to A, Lose
The odds become 1/3 to WIN - this is the way street hustlers play the game... inverting the odds of a well known game that if played like Monty gives the contestant the advantage.
ugly
I can?t watch this thread any more without commenting. It is absurd for anyone to think that changing one?s selection from C to B increases the odds of winning!
When three options were available the person had a 33 percent chance of winning. Period.
When the options were reduced to two, and the person?s choice was among the two, the chance of winning just became 50 percent regardless of whether B was selected over C. Period. The mere fact that options had been reduced is what changed the odds of winning, and not what choice the individual made!
Marvin Shilmer, who is annoyed by bad statistical math
I think I have figured out why it works that way. The folks who said the odds are better are right. I fully expected to find roughly fifty/fifty chance. Here is some VB code I wrote to prove it:
Function intRnd(intLB As Integer, intUB As Integer) As Integer
intRnd = Int((intUB * Rnd) + intLB)
End Function
Sub trythis()
Dim winner As Integer
Dim remove As Integer
Dim guess As Integer
Dim score(2) As Integer
Dim intLoop As Integer
Dim intCounter As Integer
Do
winner = intRnd(1, 3)
guess = intRnd(1, 3)
' Loop insures that remove is never equal to winner or guess, but is still random among available
Do
remove = intRnd(1, 3)
Loop While remove = winner Or remove = guess
If guess = winner Then
score(1) = score(1) + 1
Else
score(2) = score(2) + 1
End If
If remove = winner Or remove = guess Then score(0) = score(0) + 1
intCounter = intCounter + 1
Loop While intCounter < 10000
MsgBox Str(score(1)) & vbCrLf & Str(score(2)) & vbCrLf & Str(score(0))
End Sub
Running through 10000 iterations the process consistently returns score(1) as just about 1/2 of score(2). score(0) is a control that always returns 0 and proves that the "remove" was not equal to "winner" or "guess". score(1) represents staying with first choice, score(2) represents changing your mind.
Here are some results:
score(1): 3344
score(2): 6656
score(1): 3303
score(2): 6697
score(1): 3330
score(2): 6670
You can test the code yourself. The odds of winning are consistently nearly double if you change your choice.
You are asked to pick a door | |||||
you pick the prize door | you pick wrong door a | you pick wrong door b | |||
Wrong door b is removed | Wrong door b is removed | Wrong door a is removed | |||
you swap | you stick | you swap | you stick | you swap | you stick |
you lose | you win | you win | you lose | you win | you lose |
out of the 3 options from swapping 2 of them will win, so the probability is 2/3 chance of winning. Out of the 3 options for sticking 1 will win so the probability is 1/3. So you are twice as likely to win if you swap.
Actually, those other two boxes had goats, garbage cans, granny's swimsuits, etc.--remember?
So your odds of winning something odd were very great.
bebu
Let the doors be called q, r and s.
Let Cq be the event that the car is behind door q and so on.
Let Hq be the event that the host opens door q and so on.
supposing that you choose door q, the possibility that you wil the car if you then switch you choice is given by the following formula:
P(Hr^Cs)+P(Hs^Cr)
=P(Cs).P(Hr|Cs)+P(Cr).P(Hs|Cr)
=(1/3 . 1)+(1/3 . 1)
=2/3
Taken from the curious incedent of the dog in the night time by mark haddon.
If you do a simulation with random numbers then it shows you are better off swapping. I think its to do with the fact that the showman can not open just any box ... he has to open one that he knows is empty.
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