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[sevenlayermuddle]

Maybe 7 was a 6 offender?

Nice alternative, but Kitz’ answer is the one presented when I first encountered the ‘riddle’, about 15-20 years ago.   

[kitz]

Here's another old favourite:

A farmer with a wolf, a goat, and a cabbage must cross a river by boat. The boat can carry only the farmer and a single item. If left unattended together, the wolf would eat the goat, or the goat would eat the cabbage. How can they cross the river without anything being eaten?

[sevenlayermuddle]

I think this works…?

First, he takes the goat to the other side.  He leaves the goat there and returns alone in the boat.
Then he takes the wolf to the other side.   He leaves the wolf there, but brings back the goat.
Then he takes the cabbage to the other side.  He leaves the cabbage with the wolf, and returns alone.
Then he takes the wolf goat to the other side, reuniting it with wolf and cabbage.

Apols for the edit, I made a genuine slip when composing!

[kitz]

Almost.   In fact I think you were actually on the right track, but made a typo rather than logic error.

ETA
apols, my post crossed with your edit.  You are indeed correct.  Your turn
I could see it was likely a typo rather than the wrong logic.

[sevenlayermuddle]

Almost.   In fact I think you were actually on the right track, but made a typo rather than logic error.

Yes and I’m afraid I already edited my typo which was a little bit naughty, probably crossed with your post. 

Not sure if I deserve next go for that?

[kitz]

Oops, like you say our posts must have crossed.   
It takes me a while to type anything these days. :/
I will edit my post as I can see from your post that you had realised and corrected.

[sevenlayermuddle]

I hope I’m not stealing Kitz’ thunder as this one is closely related to the three bags of marbles.  It’s my favourite probability poser.   If you want to see a discussion, find the Wikipedia page for Monty Hall problem.  But be brave and pitch your answer first.   


Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?"

Is it to your advantage to switch your choice?

[kitz]

I've been mulling this one over for some time, but usually tie myself up in a knot.  It seems simple at face value.   I have made some notes in notepad and keep thinking I will go back to it when my head is clearer.   I normally love stuff like this.   

You start off with a 1:3 chance 33%
Then offered at 1:2 chance 50% of it being door B

This is where confusion sets in as Door A was 33%
Door C has a goat
Door B has 50% of being a car/goat
Surely that should mean Door A is also 50%


Now door B is 50%, but surely when you make the decision to swap if door B is 50%, then Door A should be 50% too.  I havent looked at any of the discussions yet as I'm trying to get it straight in my own head.
I keep thinking one in three so still the same odds, but no its now 50 A or B at 50%.  There has to be a twist somewhere with the wording.
At face value it appears the offer to switch has no advantage...  but purely because this is a puzzle I know there has to be more to it.

I think I will have to admit defeat and go look up the Monty Hall solution.

[sevenlayermuddle]

Ref the wording, I quoted the exact text as originally published as it seemed dangerous to change anything.

I personally think it can be taken at face value.   But tiny tiny hint… the wikipedia article points out that the commonly accepted solution depends upon a few assumptions that, whilst arguably obvious, are not explicitly stated…..

The host must always open a door that was not picked by the contestant.
The host must always open a door to reveal a goat and never the car.
The host must always offer the chance to switch between the originally chosen door and the remaining closed door.

[kitz]

Based on the not explicitly stated statments I came to the conclusion that its Monty who has the 50% choice.

This one is quite deep. I have looked at a solution page that says the odds are better to switch.  I see that this puzzle really has caused a lot of discussion.

There are alot of other pages that discuss this puzzle that Ive yet to read and I kind of get it...  but I'm not yet fully convinced.

I'm still here.

You had a 1:3 choice
Monty had a 1:2 choice
Monty eliminates one of your 3
------  and its here that things blur

You could argue that you are getting a fresh roll of the dice.
Your new chances are between door A and door B meaning 50:50
The counter argument is that you have a choice of 3 doors but Marty eliminated door C

I think the twist on words that I was looking for is "is it to your advantage to switch"
That infers we are looking at roll 1 of the dice.
Door A,B,C were 1:3
Door C was 50%
,,, and Im back in knots again.   Goes to look for another solution that explains it better. 

[kitz]

Unfortunately some sites are failing to convince me. 
For example this one here that claims to show all the possible outcomes.
https://statisticsbyjim.com/fun/monty-hall-problem/

Yet I keep thinking it doesnt...  as it doesnt show that Monty has a choice of 2 possible outcomes if the contestant guesses the correct door.  This takes the odds to 50:1 - See my attached image showing what I think are all the possible outcomes. 

I'm glad that its supposed to have puzzled Mathematical geniuses and is said to be a paradox.

[sevenlayermuddle]

Isn’t it astonishing?   

It goes back a few decades but I first came across it quite recently, last 5-10 years.   I absolutely refused to accept it was advantageous to switch, iirc I was so convinced that I became downright hostile to anybody who disagreed.    But I was eventually persuaded, if only because some serious mathematicians brow-beat me into it.  Also, I gather it’s been proven by computer simulation.

One way I look at it that seems nearly acceptable to my remaining few brain cells, and I think the wiki page suggests this line of thought too…  the door initially chosen was a 1/3 chance, and nothing can change that.  You know it’s not the door that Monty opened, so to balance the odds, the door you can switch to must have a 2/3 chance?

[sevenlayermuddle]

Yet I keep thinking it doesnt...  as it doesnt show that Monty has a choice of 2 possible outcomes if the contestant guesses the correct door.  This takes the odds to 50:1 - See my attached image showing what I think are all the possible outcomes. 

I can see nothing wrong with your reasoning, and will ponder it, but I remain convinced (for now) that the odds improve by switching.

It seems Monty Hall is actually a reformulation of an article by Martin Gardner, in his Scientific American column.   He called his the “Three Prisoners” problem….

https://en.wikipedia.org/wiki/Three_Prisoners_problem

[sevenlayermuddle]

I can see nothing wrong with your reasoning, and will ponder it, but I remain convinced (for now) that the odds improve by switching.

I’ve thunk about this for a while now, trying to find a weakness in Kitz’ challenge.  Best I can suggest is…. isn’t it the case that the event in the list is that “Monty revealed a goat”, rather than “Monty opened some specific door”?  After all, he knows what’s behind each of the doors, so even if he had a choice of goats one goat was as good as the other.

I did a bit of maths and statistics in pursuit of an engineering degree, but that was many moons ago, I’d never have called myself a mathematician or statistician, so I’m decidedly unsure.

And I certainly don’t understand the notation let alone the logic in used in some of the analysis, such as quoting something called Bayes theorem, eg https://www.statisticshowto.com/probability-and-statistics/monty-hall-problem/

[kitz]

This is one heck of a puzzle with many mathematicians trying to prove or disprove the theory.   I'm not steadfast in either camp.  I can in part believe it might be more advantageous, it's just that I haven't seen an explanation yet that has fully managed to convince me.  I dont for one minute think Im going to be able to come up with a solution that no-one else before has.  There's a lot of highly educated mathematical and statisticians who have worked on this.  I just want to try and understand how they got there. 

Whilst I haven't spent long periods of time looking at it,  I do keep going back to it when I think I have a clear head, but then give in about 5 mins later.  I'm not yet prepared to say Ive exhausted all arguments, I've not even fully read the wiki article yet....  Ive skimmed over it but some of it seems to be too deep.

>>such as quoting something called Bayes theorem,

I saw that and my brain exploded 

Ive looked at several sites but they seems to have answers that I feel have missed some questions in my own head.  There was a site where I clicked on the door so many times..  but tbh I wasnt convinced by that either - not if the doors were coded with an amount that would favour a switch. 

>>  isn’t it the case that the event in the list is that “Monty revealed a goat”, rather than “Monty opened some specific door”?

Yes.  There's something in the wiki article that does account for it, but says it makes the odds 1/6.... so its there that I am until I brave spending some more time on it   
.

I tried a different method, but still stalled when it comes to Monty having a choice of 2 doors if the contestant has selected the door with the car behind.  At least wikipedia makes mention of this statistic, a lot of sites dont.   So Im at the stage of trying to convince myself that Monty's choice is 1/6th at this point.