trazor said:
I consider the 2 statements I made, as complimentary not contradictory.
Yes, but you didn't make only those two statements, you made eight discrete points, and I agree with all except one of of them, viz:
The Monty Hall scenario is quite simple, and the odds can easily be followed......
I agree.
1) Their are 3 doors, your chance of picking the correct door with your first choice is...1 in 3 Therefore you will only win the car 33 times out of 99 attempts, if you stick with your original choice
I agree.
( if you think carefully about the above sentence, its all you need to know )
I agree.
2) So what is the advantage in switching, well we already know that in 2 out of 3 Attempts, the car will be behind the other 2 doors, that you did not choose.
I agree.
Monty now removes the goat for you.
This is true, but irrelevant.
Thus the remaining door contains the car 66 times out of 99.
This is true, but not because a goat was revealed.
Yes.
...thanks to Monty removing the goat
No!
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But again we are side stepping the original issue, which was whether any analogy could be made between the Monty Hall scenario, and DornoD.
I agree. Yours is a succinct summary of the unresolved issue.
And obviously no comparison can be made as in Montys game the Host has prior knowledge and uses it.
This statement is odd. You yourself stated "if you think carefully about the above sentence, its all you need to know" in reference to the 1/3 odds of the first choice being the door with the prize behind it. This means that Monty's prior knowledge has no bearing on the outcome, regardless of whether or not he uses that knowledge.
In the Monty Hall game, the only way Monty can influence the odds is by
not offering the swap. He is known to always show a goat instead of a car, therefore showing the goat does not provide extra information. If Monty offers the swap, then the contestant should always take it.
For any one who believes DornoD is anything other than 50-50 in the swap scenario....Give me the maths please....Not intuition.
To me it's very simple:
Let the first box chosen be Box A.
Let the last of the other n-1 boxes be Box B.
At the time of choosing Box A, there is a 1/n chance that it contains the top prize. If n-2 boxes are then opened, all showing the monetary equivalent of a goat, then the probability of the top prize being in Box B is n-1/n.
This is the same probability
calculation (not value) as the three doors in Monty Hall : n-1/n == 2/3.
For you to believe that it isn't a Monty Hall scenario, you would have to believe that Monty's knowledge, in the Monty Hall game, has an influence. And your post, trazor, contains the implicit statement that his knowledge has
no influence.
QED.
No intuition involved whatsoever. No bluff. No bluster. No insults. No coins. No trick statements. No insisting that the odds are 50:50 and then inventing new scenario variants to attempt to prove a point.
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The anomaly that people are struggling with, I believe, is that the DoND scenario of £1 and £250K being the only two prizes left with only two boxes unopened is itself an artificial one. This confuses people into comparing Monty Hall with the 'real life' DoND scenario. This comparison is a fallacy.
If you wanted to compute the probability of that artificial scenario arising, then it would be possible, but it's irrelevant because that isn't the question that Kes first posed.