Monty Hall Dilemma - Winning a GTI on a Game Show

[Timbo]

Maybe in your universe buddy!

50:50 is the right answer

I take it the cat was a little bit pregnant, too

[team_v]

I take it the cat was a little bit pregnant, too

Jeez, if the cat was pregnant that just adds a whole other level of complexity.

You might as well just divide by 0.

[Spilledprawn]

This is awesome

[Timbo]

You might as well just divide by 0.

Make it a number approaching 0, and it's game on

[team_v]

Make it a number approaching 0, and it's game on

The number is inifniti, just to screw with you

[schoona]

[QUOTE=Flighter;478729][QUOTE=schoona;478697]

I think that statement says a lot about your mathematical ability.

Thankyou

It says more about my ability to open my mouth before thinking at times.
I did google the function Scott. Will have a better read later.

What would the function and graph look like then?

If it's said that "technically" you wont reach the end, surely realistically (forgoing the maths) you will get there though.

50, 75, 87.5, 93.75, 96.875, 98.4375, 99.21875, 99.609375, 99.7990625

I can't be bothered doing the rest. But once he reaces 100m, is he not "there" ?

[cme2c]

[QUOTE=schoona;479243][QUOTE=Flighter;478729]

Thankyou

It says more about my ability to open my mouth before thinking at times.
I did google the function Scott. Will have a better read later.

What would the function and graph look like then?

If it's said that "technically" you wont reach the end, surely realistically (forgoing the maths) you will get there though.

50, 75, 87.5, 93.75, 96.875, 98.4375, 99.21875, 99.609375, 99.7990625

I can't be bothered doing the rest. But once he reaces 100m, is he not "there" ?

Point is he will never get there . Google "asymptote"

He will be close enough however.

[Rocket36]

50, 75, 87.5, 93.75, 96.875, 98.4375, 99.21875, 99.609375, 99.7990625

I can't be bothered doing the rest. But once he reaces 100m, is he not "there" ?

On the basis of that number sequence, you will NEVER get to 100.

[schoona]

100 = 100 and the sequence is always increasing ??
I'll read the asymptote/hyperbola notes later though

What is your function that "technically" proves you can't get there though? I'm asking because I'm happy to re-learn, not being a smart arse

[Rocket36]

It's the same principle as starting with any number and halving it forever, you will never get to 0.

[schoona]

It's the same principle as starting with any number and halving it forever, you will never get to 0.

Lol fark me. That's obvious. Lucky I want to engineer then eh?

[Rocket36]

Well the increase sequence is only ever increasing by half the previous increase value. So if the halving a number never gets to 0 is obvious to you, then so should the increasing sequence be - same principle.

[Buller_Scott]

what rocket said!

and the graph is in the link:

http://bo.kalipedia.com/kalipediamedia/matematicas/media/200709/26/funciones/20070926klpmatfnc_81.Ges.SCO.png

[Dubya]

Regardless of what ANYONE says, the second choice is ALWAYS going to be 50/50. That is mathematical FACT and anyone who disagrees with that is simply wrong. When choosing between two things, whatever they are, it's ALWAYS 50/50, or 1 in 2.

No more conversions I see . . . does this mean the 50/50ers are set in their thinking?

If so, one last try:

If you had the choice of sticking with the first choice or switching to both of the other doors, would you switch?

I'm presuming the reader would rather have two chances of winning (67%) instead of just one (33%) and so switch from their original selection to the two other doors. . .

Now consider:

What is the difference between:

- switching to the two closed doors; and

- switching from one door to one of the two other doors when you know for certain which of the two hides a goat?

Does revealing which of the two other doors conceals a goat alter the probability that one of the two door doors will conceal the GTI?

So after the goat is revealed behind one door, there is still a 67% chance that one of these two doors conceals the GTI. And a 33% chance the door we first chose conceals the GTI.

So we get to choose between:

A door with 0% chance of concealing the GTI (the one the host has already opened);

A door with 67% chance (the door we did not first choose);

A door with a 33% chance (the door we chose first).

We already knew one of the doors we did not choose would contain a goat.

Switching to the only other closed door after the first goat is revealed is the same as being allowed to open two doors.

Whether you open both doors at once or Monty helps by first opening one for you makes no difference to the 67% probability that one of these two doors conceals a GTI.

The only question is, are you going to act on this information and have the benefit of effectively opening two doors instead of just one?

Or do you just go with your instinct and stick with your gut feel despite it offering only half the probability if you switched?

You'd kick yourself if you switched and got the goat . . . but twice as often you'd be glad you switched.

Anyone changing their position?

[Buller_Scott]

sorry dubya, i simply cant.

this is going to sound pretty claimish on my part, but no one has answered my question- if im really going to do this a thousand times with some cards, but knowing that the first one, as a rule, must be a failure, why would i bother even picking it up? i'll just deal with switching between the two remaining cards, where my chances are 1 in 2.

the 50/50ers were accused of not having the courage, bravado, nor the confidence blah blah to try this at home. i will. but i would like to know first, when im only going to be dealing with cards 2 and 3, not bothering to pick up card 1 because everyone knows it must fail- how will this prove in real life that the chances are 2/3?