You are still evading the issue. You get a further choice after choosing one door. Do you change or stick. If you do the latter then most of the time you will lose - as demonstrated in tests. You can't understand this?There are no “odds” involved with the three door option. No matter what door you choose it is 100% certain that Monty will open one of the three doors and there will be a goat Behind it, thus eliminating it as an option. So including this ‘one of three door choice‘ as part of the mathematical odds calculation for the final choice is the flaw in that calculation. The one-in-three odds never existed, and therefor cannot possibly effect the odds of the final ACTUAL choice and result.
if you set up a flawed mathematical model it will give you a flawed result … every time. And the fact that it does so every time, repeatedly and predictably, does not make it not flawed. And when it‘s flawed results confirm your flawed presumptions, it is still a flawed model giving you flawed results that you are using to “prove” your flawed presumptions to be correct, to yourself.
And this is what is happening here. Monty’s theatrics are being wrongly interpreted as being an actual option, when there is no option at all. So it is wrongly being assigned odds when the only odds are 100% that Monty will eliminate one door with one goat behind it. Leaving you with the real, actual options of two doors behind which are two different possible results. These are the only real options we are ever given. There were no previous odds to be calculated.
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