Ignore Bait: Highest IQ, Many Questions, Odds makers invited...

Now that this thread has settled, I think I'll go with my original impression which is:

Taker takes a stab at 33 percent.

Monty puts a nothing move on him - a hustle so to speak. And in a hustle the odds are off the table.

Taker is left with a coin flip.

So, you’ve decided you’re smarter than everyone over the last 30 years including the computer simulations?
 
The only hustle in this scenario is when people such as yourself pull 50% out of thin air when it’s 33% and 66%.

That’s the hustle. Letting people lie to themselves when they could get the best odds possible without knowing.
 
Why is everyone assuming this person has 100 chances? The game is played one time by one player once.
 
The only hustle in this scenario is when people such as yourself pull 50% out of thin air when it’s 33% and 66%.

That’s the hustle. Letting people lie to themselves when they could get the best odds possible without knowing.
The math you cite is based on a hypothetical situation. You are suggesting betting this that according to the equations is the correct thing.
You don't have multiple tries. You don't even have 2 tries. That the arithmetic says _switch and win_, no longer applies.

Taker guessed once, and has the chance to guess again. That's all.
 
The math you cite is based on a hypothetical situation. You are suggesting betting this that according to the equations is the correct thing.
You don't have multiple tries. You don't even have 2 tries. That the arithmetic says _switch and win_, no longer applies.

Taker guessed once, and has the chance to guess again. That's all.

Have you read this thread at all? Or read the literal hundreds of pages on Google about this?

You don’t need 100 tries for odds to be odds.


When you get in your car tomorrow, (going to make up numbers here) and the odds of dying are say 2%.

But, the two possible outcomes are that you either die or you don’t…..

Does that mean it’s a 50/50 coin toss?
 
Have you read this thread at all? Or read the literal hundreds of pages on Google about this?

You don’t need 100 tries for odds to be odds.


When you get in your car tomorrow, (going to make up numbers here) and the odds of dying are say 2%.

But, the two possible outcomes are that you either die or you don’t…..

Does that mean it’s a 50/50 coin toss?
You got a lot of rhetoric for what is forgone. Savant says switch, the math says so.

The taker played at 33%.
Monty creates the illusion of a better 66% percent chance but that's all it is; an illusion.

The taker (who is fixed on winning) has in fact, nothing to go on but one guess - stay/switch.
 
That’s not how it works. It’s 33% and 66%.

In the scenario you posted on the first post, it’s literally impossible to ever be 50%.
The scenario Is the Savant/LMAD/Anyone else. All you seem to know on the topic is 66%. Fine that's correct.

None of that applies to the one outcome the taker gets.
 
The scenario Is the Savant/LMAD/Anyone else. All you seem to know on the topic is 66%. Fine that's correct.

None of that applies to the one outcome the taker gets.

At this point, I’m not being intelligent arguing with someone who has zero idea what they are saying.

Enjoy thinking that outcomes = odds.

With your logic, everything in life is an coin flip because you either get something or you don’t.

Carry on.
 
If you’re rounding up/down, it’s 33% and 67%, not 66%. 33% + 67% = 100%

It’s actually not perfectly 67 and 33 once you figure in confidence intervals and such.

But people aren’t able to understand the non complicated explanation. So, not even going to go down that rabbit hole.
 
For example, you could run the scenario 9 times.

And anything from winning 3 times up to 9 times is “normal”.

The long term would be 66.6r and 33.3r, however, it’s almost never going to hit those numbers exactly without an exponentially large sample size.
 
Confidence intervals are the real meat and potatoes of predictable outcomes in life.

I know that by switching, I’m going to be 66.6r% odds. So it’s the correct choice.

However, I know that in 95% of situations (95% confidence interval) if I ran it 9 times, I would win anywhere from 33% of the time to 100% of the time.
 
The math you cite is based on a hypothetical situation. You are suggesting betting this that according to the equations is the correct thing.
You don't have multiple tries. You don't even have 2 tries. That the arithmetic says _switch and win_, no longer applies.

Taker guessed once, and has the chance to guess again. That's all.
After 17 pages you still have it 100% wrong. Just because there are 2 choices doesn't make it 50-50 because the 2 choices do not have an equal chance of being the winner. Your original door has a 1/3 chance of being the winner and the other door has a 2/3 chance of being a winner.
 
If you’re rounding up/down, it’s 33% and 67%, not 66%. 33% + 67% = 100%
That is why it is smarter to use 1/3 and 2/3. No rounding error. With percents it is not 33% it is 33.333 . . . %(repeating infinitely), and 66.666 . . . % (repeating infinitely) not 33/66, or even 33/67.
 
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