no im not

im claiming the probability of the outcome will be equal
not the outcome itself

it does and it does not

the 33% has 50% chance of winning on one try even though it is only 33%

We are talking about the statistical probabilities here damnit, and not the outcome of an event when it happens once. Trying an experiment once is not going to give you the correct insights on the probabilities of the events, does that not go into your fuckin brain?

We have now proven that you are wrong, the chance is better to switch and that's the end of the story, no matter how ignorant you are towards the subject.

I proved it to you running 10 million iterations of the experiment and you still don't get it?

The fact is that you don't know what probability means...

when you flip a coin, the probability is 50/50...... whatever you flip it 100 times or 1 times...does this mean you will ALWAYS get tail on the first shot? NO...but the odd is still 50/50


Same thing with this problem!
the odds are 66/33

does it mean you will ALWAYS win by switching? no...but you have a greater chance....whatever you play 1 or 10000000000 times....the probability is still THE SAME

the thread was about one try :) I am basing all my claims on one try and one try only

hahahahahahaha

sorry to dissapoint you but it has been proven that because both sides of a coin are not equal there is no 50% chance with coins :1orglaugh :1orglaugh

and it's no 33% becoming 50%!

by switching, you actually take 2 boxes out of 3..... the 33% becomes 66%

you toss a coin once and you will claim that the probability of getting heads is 0.0?

nigga please. your logic is faulty.

no its not statistisc win by doing it multiple times


66% = 1

33% = 0


it will be either 1 or 0 if you try once >> 50 % says 1 50 % says 0

ok....

that's it.....



i'm out of here......

no

it has a 33% chance

im not pulling 33 out of my ass here
no one is

Methinks Fake Nick is trolling, and is in fact doing this merely for entertainment. Nobody could seriously claim that "33% has 50% chance of winning" :2 cents:

I am saying it is not equal with coins , there have been studies that because both sides are not equally heavy there is no 50% chance with coins :)

again...if you have 0.00005% chances of winning at lottery...

0.000005% = 1
99.00005% = 0

it will be eighter 1 or 0, 50/50??

ok you just suck at math... nothing we can do about it! :1orglaugh

yeah no shit... this is getting ridiculous

dont have to believe me but its true

1 out of 3 is not equal to 1 out of 1 000000


and you claim I am bad at math :1orglaugh :1orglaugh :1orglaugh

later folks, have fun explaining principles of simple probability to this braindead monkey. :)

word...he can't be serious....

saying 33% chance becomes 50% becuase it's only 1 attempt is really clueless......


think of a fucking dice LOL....if you make only one throw, you have 50% chances of getting a 6??? come on LOL.... now you'll say that the dots on each face do not interfer with the wind equally or some bullshit like that? come on admit that you are kidding LOL

i'm just following your logic!

you're saying 1/3 = 1/2

which is just as stupid

This is, like, the oldest math riddle known to mankind. They dug this shit out of the pyramids in Egypt.

If you think you have a 50/50 chance of being right after they open one of the doors, press your ear to the monitor: THE CHANCE THAT YOU PICKED THE CORRECT DOOR THE FIRST TIME DOES NOT GO UP WHEN THEY TAKE THE THIRD DOOR AWAY. IT DOESN'T MATTER IF THEY REMOVE THE DOOR WITH A TACTICAL NUCLEAR STRIKE, YOU ARE STILL BETTER OFF CHANGING YOUR CHOICE.

If there are 1,000,000,000 doors and you select one then they take away 999,999,998 do you really think there is now a 50/50 chance that you picked the right door the first time? Of course not. There is still a one-in-a-billion chance that you picked the right door. There is also a 999,999,999-in-a-billion chance that the other door is the correct door.

SpaceAce

300 ignorant bitches

that would be you :Graucho

THAT IS WHERE YOU ARE WRONG

the point everyone wants to make only makes sense with more than 3 doors it never does with 3


the maths behind this base the anwser on the fact that it also has to be right with 100 doors and 100 tries


with one try and three doors changing or not changing does not matter

i'm quoting it again just for fun LOL


how to magically change 66/33 into 50/50 and call other people ignorant bitch LOL

I just hope you are kidding...

I wil write a thesis on it and try and explain the world that 1 out of 3 is not equal to 1 out of 1 000 000

i dont see how anyone could be this bad

i seriously invision you laughing at us trying to explain this to you when you already know the truth

if that is the case i feel punk'd and u rocked me

Let's take a look at the 3 cases again LOL
1 = prize, 0 = no prize

CASE 1:
A B C
1 0 0

You pick A, the host open eighter B or C...you stay with your choice, you win....you switch, you lose... ONLY CASE WHERE YOU WILL WIN

CASE 2:
A B C
0 1 0

You pick A, the host open C....you switch,
YOU WIN...you stay with your choice YOU LOSE

CASE 3:
A B C
0 0 1

You pick A, the host open B....you switch,
YOU WIN...you stay with your choice YOU LOSE


Fuck....it's easy to understand! there's no 50/50 here damnit!



tell me what part y ou don't understand

A /
B /
C /


I say A / it isn't B / therefore there is 66% chance it is C/

is what you folks claim ?

I liked it better when the equation involved invisible men. It all seemed so much simpler then. :Graucho

you don't understand the problem...

if you say A...for your first guess, and you switch, you will have it, whater the prize is behind B or C...only case you will MISS by switching is if you got it right the first time

yeah and there is 33% chance you get it right


and 66% chance you get it wrong


but changing will change everything and that is not taking in the equation

To put it simply:

If your first pick is:

A:
By switching:
You will win if it's behind door B or C
By staying:
You will win if it's behind door A


If your first pick is:

B:
By switching:
You will win if it's behind door A or C
By staying:
You will win if it's behind door B

If your first pick is:

C:
By switching:
You will win if it's behind door A or B
By staying:
You will win if it's behind door C

try it on a piece of paper LOL nothing more I can tell you....

it's been explained every possible way....


even tested on a million iterations....


if the probabiliy is 66/33 after a million attempt, it is also 66/33 after 1000 attempt

and believe it or not...ont your FIRST attempt...the probability is still 66/33

there's no way a 50/50 possibility is being transformed into a 66/33 probability on a million attempt...sorry....it's just impossible....

yeah I understand how you put it silvertab no need to do that again over and over , not good for your heart

indeed...this is obviously going nowhere!
:winkwink:

Now that does make sense. What I didn't catch the first time I read the problem is that this is NOT random. What I was trying to simulate is a set selection practice BY THE HOST to show the empty box, i.e. the host always opens B, regardless of your selection. The end result would give you 50/50 odds if you stayed or switched.

But after rereading the original problem, it's NOT random, because the host purposely avoids showing C and and always opens another box (in the fixed set practice, nothing happens if you select B). That throws off the randomness of the problem enough to change the odds. The host is effectively giving more information in round 2, because he ONLY eliminates empty boxes.

It's also interesting to note that your odds are also 66.7/33.3 if the host RANDOMLY selects which box to open out of the 2 left. Again, this is because he gives more information to you and thus throws off the randomness. i.e. The difference comes in when he randomly opens C to reveal the cash...you obviously will choose this one.

so how many people here are still not convinced?:1orglaugh

:glugglug