A logic puzzle

lumpy

I thought that I had been convinced about the 66% by switching thing was correct, but something is still bugging me.

can someone please explain this to me. (i did not take statistics in college) How can your odds be 2/3 to win by switching if 1/3 of the equation has been removed from the game?

OR

can someone explain to me why my point of refference needs to stay at the starting point of the game (original choice) and not move to the second descision (stay or swap), since the rules do get changed after all.

tommyboy

lumpy wrote: »

I thought that I had been convinced about the 66% by switching thing was correct, but something is still bugging me.

can someone please explain this to me. (i did not take statistics in college) How can your odds be 2/3 to win by switching if 1/3 of the equation has been removed from the game?

OR

can someone explain to me why my point of refference needs to stay at the starting point of the game (original choice) and not move to the second descision (stay or swap), since the rules do get changed after all.

There's nothing that should be bugging you. We have posted good explanations and videos explaining this.

Ok, you understand that if you have the money and switch, you lose and vice versa right? We never said that was in question. Ok, make sure you understand that cause that is key...

So if you initially pick the money and switch then lose. What was the chances you picking the money 1/3 = 33%. You initially pick an empty box and switch then win(I hope you understand initally picking an empty box will win if you switch). What was the chances you initially picking an empty box, 2/3= 66%. So that means what?

bobman1235

tommyboy wrote: »

so that means what?

a witch!!!

lumpy

In my opinion the two empty boxes are a single entity, since the games host always plans to remove one of them.

if you choose the winning box first you will always loose if you switch and if you choose an empty box first you will always win if you switch, and you have an equal chance either way. this is 50/50 to me.

I see that the math works out to 66.7% from the starting point of the game by using 3 seemingly available choices, but this is an illusion based on the two empty boxes being the same choice.

at least this is how see this

PhantomOG

lumpy wrote: »

I see that the math works out to 66.7% from the starting point of the game by using 3 seemingly available choices, but this is an illusion based on the two empty boxes being the same choice.

Its not an illusion. There are 3 boxes. You can either: pick one and stick with it. Or: pick 2 and take the prize if its located in either of the 2 you picked.

The whole 'show you an empty box' and 'switch vs. not switch' is the illusion.

tommyboy

lumpy wrote: »

In my opinion the two empty boxes are a single entity, since the games host always plans to remove one of them.

if you choose the winning box first you will always loose if you switch and if you choose an empty box first you will always win if you switch, and you have an equal chance either way. this is 50/50 to me.

I see that the math works out to 66.7% from the starting point of the game by using 3 seemingly available choices, but this is an illusion based on the two empty boxes being the same choice.

at least this is how see this

Oh it all makes sense now! You know, you should write your own textbook.."Understanding Statistics" by Lumpy.

wz2p7j

There's only 3 scenarios based on your original pick:

1) You picked the car. You switch, you lose
2) You picked goat number 1. Announcer reveals goat number 2. You switch and win car
3) You picked goat number 2. Announcer reveals goat number 1. You switch and win car

By switching, 2 out of 3 times you win the car.

Chris

bobman1235

lumpy wrote: »

In my opinion the two empty boxes are a single entity, since the games host always plans to remove one of them.

if you choose the winning box first you will always loose if you switch and if you choose an empty box first you will always win if you switch, and you have an equal chance either way. this is 50/50 to me.

I see that the math works out to 66.7% from the starting point of the game by using 3 seemingly available choices, but this is an illusion based on the two empty boxes being the same choice.

at least this is how see this

This isn't an opinion, it's a little thing called "math." You're just wrong.

BeRad

The people that can't wrap their heads around this are missing the most important element. The person that takes away a box is an inside influence. Regardless of your choice, weather right or wrong, he will take away a box that is empty. This doesn't create a new, 50/50 chance, but compounds the original odds if you switch. If you do not switch, your odds remain the same.
Odds right the first pick: .33
Odds of it being one of the two left: .5
0.333/0.5=0.666 if you choose and then choose again. Remains .33 if you don't.

That's a quick cole's notes reasoning.

Tony M

wz2p7j wrote: »

There's only 3 scenarios based on your original pick:

1) You picked the car. You switch, you lose
2) You picked goat number 1. Announcer reveals goat number 2. You switch and win car
3) You picked goat number 2. Announcer reveals goat number 1. You switch and win car

By switching, 2 out of 3 times you win the car.

Chris

:rolleyes:
That's an awsome breakdown of the game and odds right there if the host is definately going to give you a second choice. Simple..2 out of 3 we win.

lumpy

tommyboy wrote: »

Oh it all makes sense now! You know, you should write your own textbook.."Understanding Statistics" by Lumpy.

the book would be titled "NOT understanding statistics and how to get someone who does understand statistics explain it"

thank you

maximillian

Lumpy, since you still don't believe the many explanations offered here then please make sure to watch the video on page 2 of this thread. Or, read up on the Internet about "The Monty Hall Problem". If after all this you still don't believe then I guess we can't help you.