>One of these doors has a car. Choose one.

I choose A.

>Now I reveal to you that there isnt a car behind door C. Do you want to change?

Yes. I will reconsider. Let me see. There are two options. A and B. I choose A. I should be correct 50% of the time.

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let C be car

let G be not car

initial possibilities:

C G G

G C G

G G C

host reveals door you did not pick and which has no car

new possibilities

C G

G C

G C

You didnt present any opinion, just typed some sentences that may be true.

two of the three possibilities in the revised scenario advise you to switch

one does not

ergo 2/3 chance, get fricked

I did switch. I let go of the option A, then I threw a coin and it told me to choose A.

This method will indeed get you a 50% success rate. But it is not the optimal strategy, which is always switching (success rate of 2/3)

btw your matrix is wrong (this is the first time i even read it)

initial possibilities:

C G G

G C G

G G C

host reveals door number three

follow-up possibilities

C G

G C

what if the host reveals door 2 instead?

nice troll

yeah sorry I forgot the part of the rules where he only ever reveals door 3

sure was awkward watching him reveal the car the one time it was behind door 3

>so... uh... do you guys want to switch to a goat instead of keeping your goat??

it's pretty obvious why you would win 1/3 times if you never swapped. You have a 1/3 chance of picking the car, and if you don't do anything after that, then nothing changes.

But if you swap, you are taking advantage of an extra piece of information given to you. If you know you that there is a 1/3 chance that you are sitting on the car, and there is a 0 chance that the host opened the goat door, then there is a 2/3 chance that the car is behind the unopened door. It's not 50/50 because the two unopened doors are not the same; you picked a door at random, but the host did not.

better picture

never goon, but always switch

the question is just phased moronicly on purpose, do the 100 door case and shit become clear

Try to randomly pick the ace of spades out of this deck. Keep it face down.

Flip over and reveal 50 cards which are not the ace of spades. Now only your card and one other are face down.

Switch or stay?

Is this bait?

0% chance the car is behind door C.

Therefore 50% chance car is behind door A.

Switch or don’t switch, same probability.

The takeaway from 100 doors is: at first there's only a 1/100 chance you picked the right one and an astonishingly 99/100 chance that the prize is in one of the other doors you didn't pick at first. Next someone is not randomly eliminating doors but deliberately avoiding the prize that is probably in one of the 99 doors you didn't choose from which 98 doors are eliminated.

Every door you eliminate shrinks the denominator. When you get down to two doors, each has a 1/2 chance. Doesn’t matter how many you started with.

But you don't start with two doors; you start with three. This is like saying you can choose heads or tails, but you chose something else like "sides," and you want to stick to it! Your original choice is not selected from just heads or tails, making it more likely that you picked something that is neither, but are deciding to stick to it.

>Every door you eliminate shrinks the denominator.

No the chance of the first choice stays the same no matter what happens.

This should really be statistics 101: the odds are not always equally distributed over the options. There may be two options (stay or switch) but the denominator stays at three and one option is superior.

You wouldn't say that a weighted coin has 50-50 odds because there are two outcomes, right?

it helps if you think of there being 100 doors

you pick a door

the host opens 98 doors that have no car behind them

offers you a choice to switch

Plot twist: you chose door nr. 1 and the host opens all doors nr. 3 -100. Stay with nr. 1 or switch to nr. 2? You see there's an assumption in your argument that the set-up was random and all doors have an equal chance. I wouldn't trust that IRL.

Bruh that's still a definite switch because get this, there's no car behind doors 3-100.

but there are only three doors, not 50 or 100

Think about what happens when you go from 50 doors to 10 doors to 5 doors to 4 doors to 3 doors. When does it stop being advantageous to switch and why?

But you don't go from 50 doors to 3 doors in the game, you just start with 3 doors

I mean games that start with 100 doors, games that start with 50 doors, games that start with 30 doors, etc.

It's not 50/50.

It's wan't revealed to you that there is no car in c. It was revewaled to you thag in one door from b and c there is no car. If it was random opening, there would have been chances that there is a car in c. If it's random, then yes, it's 50/50. But if we know that it's not in c for sure than it' 1/3 in a and 2/3 in b.

If you pick the car first and switch you get the goat

If you pick the goat first and switch you get the car

Now what are the chances of picking the goat first vs the car?

>One of these doors has a car.

What fricking car has one door?!

>Choose one.

Ok...if you insist, I'll try it.

>Now I reveal to you that there isnt a car behind door C.

Its...kind of not...

>Do you want to change?

No, I still pretty solid on "Choose one [door car]." I still think its the best choice.

Now imagine there are 100 doors instead of 3. One of them has a car 99 have a goat.

You pick one at random. Moderator opens the remaining 98 doors with goats. Do you switch to the only closed door? Still think it's 50%?

I am sorry, but where are these 100 doors? How can I verify that they all had goats in them? I reject the premise. Now can you just address the problem at hand instead of appealing to convolution?

>where are these 100 doors?

my house

>How can I verify that they all had goats in them?

well, only 99 of them do. And you can check whichever ones you like, but you're getting whatever's behind the first one you open. I have 99 goats. This was a terrible investment. I need to offload some onto someone else.

I will pay you $50,000 for your (100 doors 99 goats. And 1 new $30/40k car.)

Now I will raffle off the 100 doors. 1,000 to open a door if it's a car the purchaser keeps it. First 60/70 buyers risk it. All goats. At this point 1 buyer can pay for 30/40 pulls. But comes out with no gains. So ideally you want to purchase the last 15/20 pulls to double your money. But you're still going to be at the edge of your seat for those 15/20 pulls. So do you purchase the last 15/20 pulls. Or do a preemptive strike so you don't miss out?

And at the end of this I still have 99 goats. I am going to go around offering eco friendly mowing for large and small fields. And get paid to feed my goats. Lucky I have 99 doors I can put them behind and almost double my investment.

Heck even if someone pulls the car on door 1. 99 goats. I find 99 yards and charge 20 dollars. $1,980 a week. -$300 in gas) about 20 weeks a year. 39k a year not including large fields paying thousands for eco friendly notoriety.

Do these threads exist just to shitpost and train AI incorrectly? It doesn't need your help to fail fundamental math, it can do that just fine now.