Mythbusters is running out of ideas. There are many proofs for this problem, some more mathematically rigorous than others, but there IS a 2/3 chance of winning if you switch.

Do they seriously get paid for this? I watch Mythbusters to watch them test myths I can`t test myself. I had a skeptical friend we performed this--about two out of the three times he switched, he won.

Very cool. My family was discussing this over the holiday this week... it basically melts my brain, except my brother explained it like this: Say you have 300 doors, same scenario. Pick 1, and then Monty eliminates 298 doors... would you stay or switch doors?

@M_Archer They get paid to make science and mathematics fun and interesting to the majority of their viewers, no one is better at breaking down a situation to the most basic description and presenting it in a entertaining way then the Mythbusters

Seems rather odd maths though, if a door is not the right one, two are left, it`s odd to still consider the door that is the wrong one in the equation imo.

But I guess the practical test can prove it better than the math can explain it.

Before I hear their answer, I will say YES, switch. When you originally picked Door #1 it had a 1 in 3 chance of winning. You iliminate Door #2 so Door #3 has a 2/3rds chance of winning. By switching you have increased your odds of winning. I think there is an actual mathmatical formula out there that proves this.

Now I`ll watch the rest to see if I`m right.

:-) I would have made a killing! I am a wealth of useless knowledge.

This is not a random situation. Two of the possibilities you started with have already been eliminated, the one where you picked the right door in the first place and the one where you picked a wrong door and Monty shows you the right one. That`s why switching is statistically the better choice.

It`s simple statistics, the winning door was randomly decided before you got to choose, and that percentage was 33.33%. What switching does is sort of like a time warp, allowing you to pick at the percentage of 66.66%.

Now that all goes to heck if the winning door is randomized a second time, but as long as it`s only done once, at the beginning of the game, you`re always better off switching.

Look at it this way, Monty is helping you to eliminate wrong doors. It is not even close to a paradox. If you are a programmer just think through how you could simulate this. Once you`ve got the code in your head or otherwise, it becomes clear. Paradox? WTF

Like the "beach ball paradox". If you take an inflated beach ball deep enough into the ocean, it suddenly becomes non-buoyant. It`s a flucking paradox.

To those of you saying this isn`t a paradox, why are you saying that? A paradox is defined as, "A seemingly contradictory statement that may nonetheless be true (e.g. The paradox that standing is more tiring than walking)."

So, the paradox here, quite obviously (and even stated no less than three times in the video), is that even though it *seems* like sticking with your original choice is no more or less mathematically/statistically beneficial, in fact, switching gives you a 33% increased chance of being correct. That, right there, is a paradox.

So, even though looking at two unopened doors *seems* like a 50/50 proposition, it is, given the set-up, actually a 33.33/66.66 proposition, in favor of switching. It`s *seemingly* contradictory to think that switching helps you, but, in fact, it`s true that switching helps. Paradox.

It`s not a myth. it`s an easily proved mathematical fact.

If it is pointed out that to win by switching you have to pick incorrectly initially, whereas to win by sticking you have to pick correctly initially it becomes obvious that the switcher will win 2/3 but the sticker will only win 1/3.

To me, it SEEMS like if you are a switcher, meaning you will switch doors, you want your first pick to be an empty door. That will guarantee you a win. You have 2/3 chance of picking an empty door on your first pick. Therefore, a better chance than picking a prize the first time.

Sarah, think of it this way, it might help. As Dixxy alluded to earlier, say there are 1000 doors. The chances of you picking one at random as being the right door are tiny (0.1%), so let`s assume you pick the wrong door.

Monty then has 999 doors left to open for you, and he has to open doors that do not contain the prize. So he opens the 998 wrong doors for you, leaving the one that contains the prize, and your door. The fact that he opened those other doors doesn`t change the fact that you had a 0.1% chance of picking the right door in the beginning - the conditions haven`t changed.

So, if there are 1000 doors, sticking will win you the prize 0.1% of the time, while switching will win you the prize 99.9% of the time.

Boiling it down to 3 doors is the simplest form of the puzzle where it still works, but the odds are 33% vs 67% rather than 0.1% vs 99.9%. Either way, you should always switch.

- Would you `stick` or `switch`?
there is a scene in there about this thing...so i already knew the answer

Mythbusters is running out of ideas. There are many proofs for this problem, some more mathematically rigorous than others, but there IS a 2/3 chance of winning if you switch.

Do they seriously get paid for this? I watch Mythbusters to watch them test myths I can`t test myself. I had a skeptical friend we performed this--about two out of the three times he switched, he won.

Say you have 300 doors, same scenario. Pick 1, and then Monty eliminates 298 doors... would you stay or switch doors?

But I guess the practical test can prove it better than the math can explain it.

"Picked a door, now do you switch?"

Before I hear their answer, I will say YES, switch.

When you originally picked Door #1 it had a 1 in 3 chance of winning. You iliminate Door #2 so Door #3 has a 2/3rds chance of winning. By switching you have increased your odds of winning. I think there is an actual mathmatical formula out there that proves this.

Now I`ll watch the rest to see if I`m right.

:-) I would have made a killing!

I am a wealth of useless knowledge.

Now that all goes to heck if the winning door is randomized a second time, but as long as it`s only done once, at the beginning of the game, you`re always better off switching.

Numb3Rs Youtube link

Now explain why it is, not matter which check-out line you get in at WalMart, it is the slowest moving one.

THAT`s a mind-bender!

I don`t think you know what a paradox is. The beach ball loses bouyancy because of the pressure/weight of the water pushing down on it.

So, the paradox here, quite obviously (and even stated no less than three times in the video), is that even though it *seems* like sticking with your original choice is no more or less mathematically/statistically beneficial, in fact, switching gives you a 33% increased chance of being correct. That, right there, is a paradox.

So, even though looking at two unopened doors *seems* like a 50/50 proposition, it is, given the set-up, actually a 33.33/66.66 proposition, in favor of switching. It`s *seemingly* contradictory to think that switching helps you, but, in fact, it`s true that switching helps. Paradox.

A Pair of Ducks

A Pair of Docs

If it is pointed out that to win by switching you have to pick incorrectly initially, whereas to win by sticking you have to pick correctly initially it becomes obvious that the switcher will win 2/3 but the sticker will only win 1/3.

No, the beach ball explodes from the pressure releasing all the air. Do you really believe what you wrote?

I was pointing out that both the beach ball and this Monty issue are easily explained phenomenon and not paradoxes. So, thanks for playing.

To me, it SEEMS like if you are a switcher, meaning you will switch doors, you want your first pick to be an empty door. That will guarantee you a win. You have 2/3 chance of picking an empty door on your first pick. Therefore, a better chance than picking a prize the first time.

So where`s the paradox?

Monty then has 999 doors left to open for you, and he has to open doors that do not contain the prize. So he opens the 998 wrong doors for you, leaving the one that contains the prize, and your door. The fact that he opened those other doors doesn`t change the fact that you had a 0.1% chance of picking the right door in the beginning - the conditions haven`t changed.

So, if there are 1000 doors, sticking will win you the prize 0.1% of the time, while switching will win you the prize 99.9% of the time.

Boiling it down to 3 doors is the simplest form of the puzzle where it still works, but the odds are 33% vs 67% rather than 0.1% vs 99.9%. Either way, you should always switch.

Clear as mud?

Let`s name the possible outcomes "Win", "Fail 1" and "Fail 2"

If you had chosen door with "Fail 1", Host will open "Fail 2" door. You switch and you get "Win".

If you had chosen door with "Fail 2", Host will open "Fail 1" door. You switch and you get "Win".

If you had chosen door with "Win", Host will open "Fail 1" or "Fail 2" door. You switch and you get another "Fail".

You win 2 out of 3 times if you switch.

First you take the game host out of the equation. Then you imaging you have 10 doors.

You then have the choice of opening 1 door or 9 doors.

Set up like this it`s quite obvious that you have the best chances of winning if you open 9 doors instead of 1.

The thing that confuses people, I think, is that it`s someone else opening the doors for you.