A math question

[jernst]

gah my prof had a whole lecture on this in stats, he spent the entire class talking even though everyone with a laptop just wikipedia'd it and knew the answer in like 2 seconds

[tenniscrazy]

ha! i finally got it.

you will choose the door it's behind one third of the time. (call this the first door)

For the other two doors...say the car is behind the second door. The gameshow host will open the third door if it is behind the second door.

And if it's behind the third door than the gameshow host will open the second door.

So really, when you choose the door that you didn't choose before, you are choosing the second door 1/3 of the time and the third door 1/3 of the time, so this means that you are actually shosing both doors because one that doesn't have the car is eliminated.

just thought i would post that because thats how i made it make sense to myself

[Insectoid]

But my math equation works, right? Seems to to me...(not about me getting 2000-ish%, about the cups)

It seems to work for me. Getting the right percentage every time!

[tenniscrazy]

in your equation

Quote:

I.E 3 cups with 33% each. 1 cup removed
(1 * 33) / ( 2 - 1 ) = 33.
33 + 33 = 66

i don't get why you divide the first line by 2-1, because that is the same thing as dividing by 1, which does nothing.

and you do it in the second one too

Quote:

50 cups with 2% chance each, - 48
(48 * 2) / (2-1) = 96
96+2=98%

[Vermette]

Don't feel bad if you argued for the incorrect answer (keeping the door you chose); this is a classic problem that has tripped up even experienced statisticians in the past before careful analysis. It was used by one my of profs to demonstrate that in the world of CS/Mathematics, your intuition can sometimes be dangerous.

[Brightguy]

Actually the probability is 50-50 for the question as you've posed it. The answer hinges on this line:

insectoid @ Wed May 28, 2008 5:28 pm wrote:

After one has been selected by Jeremy, one cup is removed and identified as a cup without the token.

That should read one cup is identified as a cup without the token [by someone who knows the token location] and removed. You must also assume that they choose randomly if they have a choice of cups to remove.

[Insectoid]

teniscrazy wrote:

in your equation

Quote:

I.E 3 cups with 33% each. 1 cup removed
(1 * 33) / ( 2 - 1 ) = 33.
33 + 33 = 66

i don't get why you divide the first line by 2-1, because that is the same thing as dividing by 1, which does nothing.

I did that so that the equation could include a number of scenarios, such as if there were 10 cups and he was alowed to pic 3, then 2 were removed. Then the equation would look like this:

((2*10)/ (8-3))+10= 14 (I think I plugged the numbers in the right spot)

[tenniscrazy]

ahk! I was tryna make sense of your equation. it works for the exaample you just did, because each of the other doors would have a 14% chance of having the car behind them.
But i'm not sure if you are putting in the same variables for each of the three examples

could you post what the actual variables in the equation mean, just to clarify. Cause it would be a cool equation if it worked and made sense

[jernst]

Vermette @ Thu May 29, 2008 10:18 am wrote:

Don't feel bad if you argued for the incorrect answer (keeping the door you chose); this is a classic problem that has tripped up even experienced statisticians in the past before careful analysis. It was used by one my of profs to demonstrate that in the world of CS/Mathematics, your intuition can sometimes be dangerous.

Yeah in the lecture our prof gave he mentioned how there were math profs who argued that they knew what they were talking about simply because they were "math profs for like 30 years" even though they argued for the wrong side.

[Insectoid]

( ( x y ) / ( n - p ) ) + x is the equation, where:

x = the base percent chance per cup (100/number of cups)
y = number of cups removed
n = number of cups remaining
p = number of cups chosen initially

It could be adjusted to use less variables, but I'm a bit lazy.

[tenniscrazy]

ah! ok i get the equation now, thanks for clarifying

[btiffin]

Just so everyone knows; as stated this is NOT a Monty Hall challenge. The key phrase is "not been revealed". For Monty Hall to pick the cup that skews the odds from 1 in 3 to 2 in 3, he has to know which cup has the prize. As stated, no one knows. It was a fluke that the first cup removed didn't have the token (it had a 1 in 3 chance).

Quote:

One has some token inside(hereafter referred to as 'the right one'), and it has not been revealed which one.

For a proper Monty Hall problem, there has to be someone that knows the cup with the token, and a desire to not give away where the token is until the very last moment (for that game show tension effect). If there is no human intervention the odds of this problem are purely common sense statistical. After seeing that the first cup removed doesn't have a token does not influence the odds (unless, again, there is a human revealing the cups in an order to heighten the tension).

But in honour of the thread; go ahead and assume that someone in Jeremy's house is acting as a game show host.

More info from the old guy...

Monty Hall was the host of a game show called "Let's Make a Deal". The show was famous for having "Door number 1, door number 2, door number 3 ..." One of the doors always had a "zonker", being a goat or somesuch. (Once it was an Osterich but the player that got that zonker actually won a spot in the finals of the show as this zonker was actually pretty expensive). It always ended with Monty walking through the audience and asking people "ok, if you have a raw egg in your pocket I'll give you $500. $200 if it's hardboiled". So the crowd always had huge bags of random stuff with them. He'd ask for hair pins, magazines, clothespins, walnuts, just about anything small and peculiar to carry around.

Monty Hall was born in Manitoba by the way - one of the first of the plethora of famous Canadians on mainstream US television.

Cheers

[Insectoid]

Well, I meant it has not been revealed to Jeremy which one contains the token. Can you really expect me to get the wording right on the first try? I didn't know at first that is was a famous problem, my brother just came home from school one day, asked my mom the question and spawned the whole argument.

I had never even heard of monty hall until someone mentioned it and posted the link. So, in order to clarify, this is now about the mony hall problem (though you can still word your answer according to my story.)

[Brightguy]

insectoid @ Sat May 31, 2008 7:48 am wrote:

Well, I meant it has not been revealed to Jeremy which one contains the token.

Of course. But you never specified if someone else knows which one contains the token or not. So which question were you having the argument over: one where someone ("Monty") knows which contains the token, or one where no one knows which contains the token?

[Insectoid]

Whichever one makes more sence.

Okay, monty know where the token is, but jeremy doesn't. Happy? Let's not talk about the story, but the solution!