Can you can guess the month another person bought their Golf?

[Buller_Scott]

lol, i had a pretty long reply already posted, but i decided i'd better reel it in.

long story short- i had questions for the 2/3 camp that no one was able to answer. e.g. my ''dub fest, 20 raffle tickets, 1 gti prize, guy gets to switch 10 times, he's one of 10 remaining at the podium come time to draw, each person has only one ticket, someone tell me just how he has a 55% chance of winning?" scenario. that was one of about 3-4 big questions that i sought explanatory answers to, yet not one peep to any of them.

so yeah- it's been done, and i applaud your enthusiasm, dubya, because if i had my way i would LOVE to see this debate/ reasoning through, with thorough explanations etc, right through to the very end. i'd also have been more than willing to oblige, and try to mimic the scenario over the course of 1000 draws using playing cards, IF my questions in the initial thread, months ago, had been addressed by those in the 66% camp.

but i think that, in light of the wave of new VWWA members, and the potential for those members too, to get dragged into this, and what must have been a headache for the mods last time, i reckon this thread might get closed soon.

cheers, though. im looking forward to the next provocative discussion in general chat.... preferably involving math of some sort

cheers,

scotty

[anthony]

In Victoria number plates help.

For example most new car registrations this month started with XS? or XT?

Although some went back as far as XMG.

Next month XU? will begin.

[The_Hawk]

Those that understand will debate it to the death, those that do not will debate to the death.

How about I repost the wiki link for anyone who wants to read through it again: Monty Hall problem - Wikipedia, the free encyclopedia

[The_Hawk]

It has been requested that this thread remain open for civil discussion. If it deviates from civil discussion it will be removed and infractions will be handed out to those who cannot play nice.

[Rocket36]

Pick one.....



There's a 50% chance you'll pick Cherry and a 50% chance you'll pick White. No more, no less. Basic maths really.

[Dubya]

lol, i had a pretty long reply already posted, but i decided i'd better reel it in.

long story short- i had questions for the 2/3 camp that no one was able to answer. e.g. my ''dub fest, 20 raffle tickets, 1 gti prize, guy gets to switch 10 times, he's one of 10 remaining at the podium come time to draw, each person has only one ticket, someone tell me just how he has a 55% chance of winning?" scenario. that was one of about 3-4 big questions that i sought explanatory answers to, yet not one peep to any of them.

so yeah- it's been done, and i applaud your enthusiasm, dubya, because if i had my way i would LOVE to see this debate/ reasoning through, with thorough explanations etc, right through to the very end. i'd also have been more than willing to oblige, and try to mimic the scenario over the course of 1000 draws using playing cards, IF my questions in the initial thread, months ago, had been addressed by those in the 66% camp.

but i think that, in light of the wave of new VWWA members, and the potential for those members too, to get dragged into this, and what must have been a headache for the mods last time, i reckon this thread might get closed soon.

cheers, though. im looking forward to the next provocative discussion in general chat.... preferably involving math of some sort

cheers,

scotty

Apologies to those who have responded in good faith with additional pointers to help determine the delivery date of the car. However this is a theoretical discussion of probability which had its genesis in the General Discussion forum here: http://www.vwwatercooled.org.au/f40/...how-42135.html).

From that discussion, two camps emerged. The 50/50s (who held it made no difference which of the two remaining options one chose), and the 67ers (who said each of the two remaining options had different probabilities, being 33% for one 67% for the other).

But in response to Scotty's post:

The 67ers explained in the original discussion that imperfect analogies were unhelpful and, not least, unnecessary.

Three cards is all that is required, or an analogous scenario such as the one in this thread where only the number of options is increased, not the method of play or rules of the game.

In the Monty Hall scenario, the contestants choose once from all available options and only two options remain (not half) when the contestant is offered the chance to switch (once, not many times).

Similarly, in the present, analogous, scenario, the contestant chooses once when theire are 12 options and once again when their are just two options.

So in both cases, number of options remaining = n-2 where n is the initial number of options.

Repsonding intuitively, most people initially conclude: "She has a 50% chance of winning now there are only two options."

However, they fail to take into account that when the contestant made her first choice, there were 3 options (or 12 in the present scenario) and so it is the number of options when the choice was made that dictates the chances of the first choice being right.

So the chances of being right the first time = 1/n, where n = the initial number of options (when the choice was made).

The host or house then discounts all but your first choice and one other option as possible winning choices.

Your first choice is only included as an option because you chose it (and/or you chose right!).

The chances you chose right in the first instance when there were n options are 1/n and this probability continues when only two options remain, your choice being no more likely to be correct then the n-2 options which have now been eliminated. It is just that the host will never do you the favour of eliminating your choice - if they did eliminate your first choice (as a dud) then the two remaining options would be a 50/50 chance. But he doesn't, so they're not!

This probability of 1/n even continues when the game is over: even if the contestant wins, they still only ever had a 1/n chance of doing so (if they did not switch) and have merely beaten the odds by winning. So the theory that a random choice from n different options can only provide a 1/n chance of picking correctly, holds true.

So given the contestant's choice has a 1/n chance and all other options an n-1/n chance, you switch because it is more likely (by a factor of n-1) that you picked wrong in the first place.

The two remaining options only present a 50:50 chance for someone who enters the game when only two options remain. If they make a choice from the two options without knowing:

- which option is the contestant's first choice (that was only included because it was chosen randomly from n options and only has a 1/n chance of being right); and

- which was the host's choice (which was included in the last two options because there is a (greater) n-1/n chance of it being right),

then that person has a 50:50 chance of picking right.

This is faithful to the theory, above, that the new entrant, given a random, uninformed choice between the two remaining options has a or 1/n chance of picking right, because they are choosing from 2 options, not 12 (or 3) like the contestant.

As for the contestant, they still have a 1/n chance of having picked right, so only 8.5% in the present game, and there is an inverse, or 91.5% chance the other remaining option is right (try viewing the other option and all of the eliminated options collectively as "the second choice" as collectively they contain the n-1/n chance of winning whereas your first choice contains the remaining 1/n chance of winning).

So hopefully you can see the two remaining options are not the same: the host deliberately chose one and n-1 times out of n the host deliberately chose the other. In the 1 out of n times the contestant guesses right the first time, the host chooses randomly from the remaining dud choices and the contestant loses if they switch. But this only happens 1 in every n times.

Now, Scotty, I guarantee, if you sat down with a friend and faithfully tried the Monty Hall game with cards you would find that you had a 67% chance of winning where n is 3 (ie n-1/n).

The 67ers (or n-1ers) could never understand why the more vociferous opponents of the theory (the 50/50s) refused to test the theory as a conclusive way of discounting it!

I'll offer the right odds if you're still not convinced!

[Dubya]

Pick one.....



There's a 50% chance you'll pick Cherry and a 50% chance you'll pick White. No more, no less. Basic maths really.

But this is not the problem.

The problem is:

Pick one month:

Jan - - Feb - - Mar - - Apr - - May - - June - - July - - Aug - - Sept - - Oct - - Nov - - Dec

You choose May, let's say.

Owner now tells you 10 months when the car was not delivered leaving your choice (May) and August as the only two options.

Now he says:

Make a choice between May and August.

Rocket36 thinks:

Two choices, May and August, hmm, makes no difference which I pick as each offers a 50% chance of being right!

Not so!

I am sure Rocket36 now realises why.

[dylan8]

Pick one.....



There's a 50% chance you'll pick Cherry and a 50% chance you'll pick White. No more, no less. Basic maths really.

Cherry!! was correct? that was my straight up choice, i DIDNT change when i had the chance

[PassatB6]

So given the contestant's choice has a 1/n chance and all other options an n-1/n chance, you switch because it is more likely (by a factor of n-1) that you picked wrong in the first place.
......

As for the contestant, they still have a 1/n chance of having picked right, so only 8.5% in the present game, and there is an inverse, or 91.5% chance the other remaining option is right (try viewing the other option and all of the eliminated options collectively as "the second choice" as collectively they contain the n-1/n chance of winning whereas your first choice contains the remaining 1/n chance of winning).

As my earlier post stated the chances were 11 times greater by switching. Your formula suggests it is 10.76 times more likely so I'll claim 'close enough'.

[Dubya]

As my earlier post stated the chances were 11 times greater by switching. Your formula suggests it is 10.76 times more likely so I'll claim 'close enough'.

The formula was right but my percentages should have been 8.33% and 91.67%, sorry.

But well done!