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

[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!

[elephino]

The answer is - they're all correct!

This is the "joy" of statistics and probabilities. Take Lotto as an example. If you play each week with the same numbers, the law of averages suggests that one day you'll win. The problem is that each time the draw is done, the probability of getting the correct numbers is the same each time.

So in the original example, yes you have a 1 in 12 chance of being correct at the first guess (assuming you cannot see the car and peek at the rego label) but once you are down to 2 choices, you have a 1 in 12 chance of still being correct and a 1 in 2 chance of choosing to correctly change or not change. The 11/12 is slightly dodgy but still correct - the dodgy part is that it's not actually the probability of the other remaining month being correct but rather that your original choice was incorrect. Of course if you know in advance that no matter what month you choose, 10 of the months will be removed from the list, then you have a 1 in 2 chance of being correct because you'd put less effort into the original choice.

The other problem with this is the mathematical assumption that the choice is totally random. By the choice not being random, as in this case as someone is attempting to make what they believe to be an educated guess, the odds are different again.

[Dubya]

The answer is - they're all correct!

Thanks for making it past all of the distractions (it is like an ADD test!). However, the answers are mutually exclusive so only one answer can be correct under the random scenario applying.

This is the "joy" of statistics and probabilities. Take Lotto as an example. If you play each week with the same numbers, the law of averages suggests that one day you'll win.

If you live to be well over 1,000 years old the laws of probability might be in your favour.

So in the original example, yes you have a 1 in 12 chance of being correct at the first guess (assuming you cannot see the car and peek at the rego label) but once you are down to 2 choices, you have a 1 in 12 chance of still being correct and a 1 in 2 chance of choosing to correctly change or not change.

If you understand the probabilities of each option and act accordingly, there is a 100% chance you will switch, and if you do so, a 91.67% chance you will win by doing so.

There is a poll on this forum showing 80% of people who responded would switch. But success on switching is determined by the underlying probabilities of 1/12 and 11/12.

The 11/12 is slightly dodgy but still correct - the dodgy part is that it's not actually the probability of the other remaining month being correct but rather that your original choice was incorrect.

Your choice being incorrect is the same outcome as the other option being correct, so the 11/12 is not dodgy at all as it refers to the same outcome.

The other possibility is that your initial choice was correct and the other choice is incorrect, of which there is a 1/12 probability, not a 50% probability as Rocket36 has suggested.

So there is a 1/12 chance that switching will end in tears (not 50% per the view that because there are two options it is 50/50.).

Of course if you know in advance that no matter what month you choose, 10 of the months will be removed from the list, then you have a 1 in 2 chance of being correct because you'd put less effort into the original choice.

All but two options are always eliminated.

No more or less effort is required in any event.

The probabilities are calculated on the basis that the game always involves only a random choice (guess) made with no extrinsic information or factors as a guide.

The other problem with this is the mathematical assumption that the choice is totally random. By the choice not being random, as in this case as someone is attempting to make what they believe to be an educated guess, the odds are different again.

This scenario is intended to be totally random and analogous to the Monty Hall Dilemma discussed elsewhere in this forum.

[Rybky]

OK, so maths was never my strong point but logic is.

It's simple, you have a 50/50 chance.

The two guesses are unrelated. All the first guess did was help pick one of the two options for the second guess. The odds related to being right on the first guess have nothing to do with the odds of the second guess.

If the guy had said from the start that you had to guess whether his car was built in May or August then you have a 50/50 chance of being right, right?

So how is the situation any different now??

You still have two choices, 1) stick with your first choice, or 2) pick the other option. That is a choice of two options to me, either of which could be right.

Adam

[maca]

OK, so maths was never my strong point but logic is.

It's simple, you have a 50/50 chance.

The two guesses are unrelated. All the first guess did was help pick one of the two options for the second guess. The odds related to being right on the first guess have nothing to do with the odds of the second guess.

If the guy had said from the start that you had to guess whether his car was built in May or August then you have a 50/50 chance of being right, right?

So how is the situation any different now??

You still have two choices, 1) stick with your first choice, or 2) pick the other option. That is a choice of two options to me, either of which could be right.

Adam

You see, that's what I thought as well. But it turns out it's a lot more complicated than that for some reason. At the end of the day, isn't it still a 50/50 choice with a 50% chance of being incorrect?

[Rybky]

You see, that's what I thought as well. But it turns out it's a lot more complicated than that for some reason. At the end of the day, isn't it still a 50/50 choice with a 50% chance of being incorrect?

I think everyone else is over thinking it.

Adam

[Buller_Scott]

You see, that's what I thought as well. But it turns out it's a lot more complicated than that for some reason. At the end of the day, isn't it still a 50/50 choice with a 50% chance of being incorrect?

yes. at the end of the day, that's all that matters. you can choose either may, or august- one in two, right?

ohhhh no wait. it's in the rules that even though you 1. can choose may, or 2. can choose august, it says in the rules that you have to count all the other months of the year in your odds, because you know that it WASNT built in january, feb, march, april, june, july, september, october, november, december.

so, we KNOW that it wasnt built in those months, right? apparently, when choosing either 1. may, or 2. august, the fact that you know that it wasnt built in all those other months, means that your chances are now 11/12.

they are now 11/12, when faced with a choice of either may, or august. 11/12. two choices, you get one pick.

one pick. two choices. 11/12. two to choose from, one chance. 11/12.

so. may or august. pick one. i would have thought that the chances of it being may, are 1/2. and the chance that it's august, is the other 1/2.

wrong. the chance that its may, is 11/12. AND, the chance that it's august, is 11/12. apparently.

and with that indisputable logic, i'm out of this debate. unless there's sheilas involved.

[Rybky]

Since when were there rules??

Two choices. May or August. That is what you have to work with. Only one is right. So you have a 1 in 2 chance of picking the right one.

Sorry, everything you else you said is BS designed purely to invoke an argument.

Adam

[Buller_Scott]

Since when were there rules??

Two choices. May or August. That is what you have to work with. Only one is right. So you have a 1 in 2 chance of picking the right one.

Sorry, everything you else you said is BS designed purely to invoke an argument.

Adam

1. there are very clear rules in the ''monty hall dilemma'' thread, upon which this thread has a strong basis.

2. if you read that thread, you'll note that i am one of the biggest proponents for the 1/2 line of logic, which might also have been somewhat evident in my literary style in my post just above yours.

3. i wont respond/ provide feedback, but please feel free to outline just what, in my post, was ''provocatively argumentative".....

cheers,

scotty

[Rybky]

Sorry scotty, didn't mean you specifically, I realise you were quoting other people there. I meant the "rules" and resulting theories are designed to provocate an argument. There wouldn't be any point posting it on a forum otherwise!

I am not going to read the other thread ... I have a life!

Adam

[Buller_Scott]

I am not going to read the other thread ... I have a life!

yes, well....... there were a few of us who proved that we dont, with the shenanigans in the monty hall thread, haha!

and no need to apologise- no matter what the debate/ argument [even heated], there's never any bad blood with me - it's all about Veedub love, after all!

cheers,

scotty