How about if you pick one of the doors, and I get the other two?
Now we show you one of my doors is a goat (which, of course, one of my doors had to be, so this changes nothing.)
If it's 50/50 like you say, then your odds of winning would be the same whether you switch or not. So you don't get to switch.
Do you think we'll both win half the time, or will I win twice as often?
Wrong. Look at the following slide that has 50 doors, then see if you still feel the same way.
Yeah. I think that's wrong and it only works by confusing the question.
If you have two doors, your odds can never be greater or worse than 50%. Just like if there were only two doors to begin with.
If you get two doors and I get one, then it's the same case. The revealed goat will always be one of your doors, per the rules, and still a 50/50 chance.
Same case. Doesn't matter whether I pick door 1 or door 37, I know that my door either has a goat or doesn't with equal probability. There are 49 goats, 48 of which are eliminated.
I just think this is appeasing the gambler's fallacy.
Yes.
And it doesn't matter if we're doing this 1,000 times, my answer is always going to be yes and my chance of winning is always going to be 50%.
There could be 10,000 options and the odds would always be the same. Because every time we do this exercise, you're going to show me 9,998 doors that are goats. Meaning no matter which door I pick, it's going to come down to my door or one other door.
Your logic is flawed, and your underarms smell too.
You can hate on me all you like, it won't change the right answer. Both of you should take a look at the 50 door example, but I'll put it in text.
There are 50 doors, behind them are 49 goats and 1 new car. You choose one door, then I open 48 doors with goats, leaving one unopened door. You still want to keep that first door you chose, which has a 2% chance of being the correct door?
That's wrong. Your odds were 1/3, until you were shown the door with the goat.
If you do the problem in reverse with two doors, one with a car and one with a goat, your odds are 50%. If they show you another goat, your odds are still 50%. It only works by confusing the problem.
Once you're shown the undesirable doors, they're eliminated from the equation, thus it's irrelevant how many there were.
Funny how this time the Monty Haul problem spawned the war instead of .999 = 1. It's also funny how adamantly people insist on being wrong. OK guys, put it to the test then:
Monty Haul Simulator
I wish I could find a simulator that supports more than 3 doors, because that would make it incredibly obvious that switching is better.
No, revealing the goat did not change the odds it just changed your perception.
KT
Anyone going to tell me with a straight face that the Powerball ticket you bought has a 50/50 chance of being the winner at this point?
Alright, I will restate this problem in terms of a lottery and see if it wakes anyone up. You purchase a Powerball ticket with a particular set of numbers. There's a giant bin with 175 million other Powerball tickets, each with a different number selection, so that every single number combination is in the bin, except for yours. Now the winning Powerball number is secretly drawn, and all but 1 of the tickets in the bin are destroyed. If you bought the winning ticket, the ticket left in the bin is a loser, but if you didn't buy the winning ticket, that ticket left in the bin is the winner. You have the option to trade in your ticket for the one in the bin. Do you switch?
Anyone going to tell me with a straight face that the Powerball ticket you bought has a 50/50 chance of being the winner at this point?
This is exactly the same as the Monty Haul problem, just with 175 million doors.
That's leaning on the fallacy that 1:175 million is unwinnable.
It doesn't matter which ticket I chose or how many there were, it's still 50% after the reveal.
*Taps meter*Yes.
And it doesn't matter if we're doing this 1,000 times, my answer is always going to be yes and my chance of winning is always going to be 50%.
There could be 10,000 options and the odds would always be the same. Because every time we do this exercise, you're going to show me 9,998 doors that are goats. Meaning no matter which door I pick, it's going to come down to my door or one other door.
Your logic is flawed, and your underarms smell too.
It doesn't matter which ticket I chose or how many there were, it's still 50% after the reveal.
That's assuming you had a 50% chance of getting it right with even 175 million tickets. Which is silly.
It's only 50/50 if you foolishly ignore everything that happened before the final pick. What are the odds that you picked the correct door before the 48 others were revealed? The only way you could go wrong is if you actually did pick the correct door the first time, which is highly unlikely.
That's leaning on the fallacy that 1:175 million is unwinnable.
It doesn't matter which ticket I chose or how many there were, it's still 50% after the reveal.
The problem is much simpler. You aren't 'trying to pick the right door' in the first step.I was going to answer: well, what are the odds of selecting the one goat, that he won't reveal!
Until I realized, that the revealed goats are not chosen randomly, but by exclusion - the goat you selected at first is never going to be revealed. This makes the solution evident.
Switching is only detrimental, when you ALREADY have selected the correct door, from the first batch. This chance is 1/n.
By eliminating n-2 options, with the characteristics that they are not a goat, and not selected by you, switching increases your odds of winning to (n-1)/n, as the preselected case is protected from elimination, and all other goats aren't.
Let's look at the limit case n-> infinity.
The first round of choosing merely selects one goat for protection from elimination. Then an infinite amount of goats is eliminated, and only one other case except the goat you selected is available for choice. Your winnings are behind that door with p=0,99999.... (not 1, because maybe you want to win a goat)
That's leaning on the fallacy that 1:175 million is unwinnable.
It doesn't matter which ticket I chose or how many there were, it's still 50% after the reveal.
Here's another explanation.
How does a "switcher" win. He needs to pick a wrong ticket, and then the ticket remaining in the bowl will be the winning ticket. So the probability of a "switcher" winning is 174,999,999/175,000,000 (the probability of your first pick being the wrong ticket).
How does a "non-switcher" win. He needs to pick the right ticket, because then the ticket in the bowl will be a losing ticket. So the probability of a "non-switcher" winning is 1/175,000,000 (the probability of your first pick being the right ticket)
nope.
you have one pile of tickets to begin with. yours is 1:175mill.
all other 174,999,999 tickets are placed into another pot, and you know that either your ticket or one ticket in the pot is a winner. all but one of those tickets is destroyed--again, you know that the remaining ticket, or your ticket, is the winner. This is the crux of the problem.
Do you think your original ticket now goes from 1:175mill to 50/50?
What about the remaining ticket that survived the culling of 174,999,998 other tickets?
Say you bought a lottery ticket. Then a guy from the lottery commission pulls you aside and says, "Pssst, either the ticket you bought is the winner or it's this one I have in my pocket."
50% chance.