Misclick leads to tough descison.....

[JMACK007]

Thanks for spending the time benny the cunt, but I am stuck with your "we win 1/3 and lose 2/3". It is not that black and white. I think you are missing the point now. Your 1/3-2/3 is correct, but not over the whole amount of tries. You are 1/3-2/3 against, EVERY time you try. So you cannot say after 1000 tries you will win 333 times!! That just does not happen, and therefore, your formula for generating that data is flawed.....

[JMACK007]

Reminds me of this:

http://en.wikipedia.org/wiki/Monty_Hall_problem

Haha, just had a chance to look at this. Perfect example of over thinking IMO....

(the Wiki page, not you Aaron!)...

this is definitely not over thinking, its statistics!

(by your definition, the last 7 years i have spent at uni have been overthinking

It is overthinking to say that once you see that one of the 2/3 is wrong that the other is twice as likely to hold your prize as the one you are on. Absolutely flawed logic right there....

Both the one you are on and the other one now have a 50/50 chance....

[AceLosesKing]

I'm nowhere near benny the cunt's level of thinking (level of nerdom?) but even I can understand the Monty Hall Problem when I first thought about it, and I found benny the cunt's statistical post there pretty easy to understand.

Give it some time, it gets there eventually.

[JMACK007]

Sorry Aaron? Can you explain what you "understand"??

[JMACK007]

I'm sorry, but I do not know why you guys can't see what I am getting at here!!...

[JMACK007]

we win 1/3, and lose 2/3 1/3 of the time we win $25 (getting our buyin back just counts as breakeven, this is why this number is $25 and not $35)
2/3 of the time we lose 10

if we do the maths, we get 1/3 * 25 - 2/3 * 10 = +$1.6667. i.e in the long run, it will be profitable for us to play this game, because even though we lose more than we win, the payoff makes it worth it. our expectation value for each game is +$1.67

The bold part of that statement is flawed, and is what I am getting at...

[AceLosesKing]

This is the best way I can explain it. From the wiki article:

Why the probability is not 1/2
The contestant has a 1 in 3 chance of selecting the car door in the first round. Then, from the set of two unselected doors, the host non-randomly removes a door that he knows is a goat door. If the contestant originally chose the car door (1/3 of the time) then the remaining door will contain a goat. If the contestant chose a goat door (the other 2/3 of the time) then the remaining door will contain the car.

The critical fact is that the host does not randomly choose a door. Instead, he always chooses a door that he knows contains a goat after the contestant has made their choice. This means that the host's choice does not affect the original probability that the car is behind the contestant's door. When the contestant is asked if the contestant wants to switch, there is still a 1 in 3 chance that the original choice contains a car and a 2 in 3 chance that the original choice contains a goat. But now, the host has removed one of the other doors and the door he removed cannot have the car, so the 2 in 3 chance of the contestant's door containing a goat is the same as a 2 in 3 chance of the remaining door having the car.

Another way of looking at the situation is to consider that if the contestant chooses to switch then they are effectively getting to see what is behind 2 of the 3 doors, and will win if either one of them has the car. In this situation one of the unchosen doors will have the car 2/3 of the time and the other will have a goat 100% of the time. The fact that the host shows one of the doors has a goat before the contestant makes the switch is irrelevant, because one of the doors will always have a goat and the host has chosen it deliberately. The contestant still gets to look behind 2 doors and win if either has the car, it is just confirmed that one of doors will have a goat first.

The image with the three doors explaining what would happen if you picked door 1, 2 or 3 (right under this link) helped me the most.

I can see what you're saying Jmack, I think. There is no "given" that we will pick a door with $25 behind it. We could pick 1 million doors and all pick the door with $0 money - is this right?

[JMACK007]

There is a difference between the two sentences:

you are a 66% dog.

you will win 1 out of every 3 times..

You can't run a senario based on winning 1 out of every 3 times, it just doesn't work that way.

Basically, it comes down to the fact that you know you are behind, and are gambling. You might as well put your $100 on a line or a 3rd on Roulette... It is the same thing.....

[JMACK007]

Yes Aaron, you are right. That is exactly my point. benny the cunt's examples all rely on the fact that we hit 1/3 of the time. This of course might not happen, and in fact, the odds are against it happening...

[JMACK007]

I am not a mathematician, but I can assume that the problem would be something like this:

What are the odds of hitting something 1/3 of the time when you only have 1/3 a chance of hitting it each time....