Math Is Fun Forum

It's a good one isn't it!

Yes, the games are both independent and random.
And yes, the odds of your game are affected by information about the other contestant's choice.

If I'm right, here's the mysterious answer:

I suspect this will be considered impossible!

I simply mean: you are told whether the other contestant picked the same as you or not.

I'm asking if this information changes the odds of your game. In other words:

If he picked the same as you, are the odds for your game affected?
If he picked the opposite, are the odds for your game affected?

Hope that clarifies...

Ok I have found a solution and (unless I've miscalculated) it works perfectly.

With this method, you are asking a chief one question while inserting it within another. The result is a guaranteed truthful reply to the question you actually want answered, but as far as the Chief is concerned he has answered the other question.

At the same time, for the purposes of your question, DA and JA can mean YES and NO respectively,, irrespective of what it means to the chiefs.

So in this solution, you could have following situation:
You ask a chief, using the implicit question "Are you the Liar?" inserted within another question.
The chief replies "DA", which will truthfully answer your inserted question reliably as YES. But the Chief is actually responding falsely to the other question and for him DA might mean NO.

Baffled?

Here's the solution.

My brother, who says he has the seen the orignal puzzle, confounded me with the following information:

a) the solution does not require that you know the meaning of 'da' and 'ja'

b) the solution doesn't require that the hybrid's responses be anything except random sounds - i.e not deliberately true or false

That throws everything into disarray and my gut response is: impossible. My brother is lying.

So would I but I don't think there is one. At least, not a question that establishes the meaning of 'da' and 'ja' beyond all doubt from just one chief.

Making the question to one chief not self-referential would tell you the most likely synonyms of 'da' and 'ja'.

If he's allowed to do that, then I'm stuck.

As for your version Chilli that you got from Wikipedia:

What was your way of determining 'da' and 'ja', Chilli?

As for the rest of puzzle, I'm thinking on the same lines you put in hidden text. (I'm now doing the same with all my past and future conclusions).

This is a twister alright and I'm still working on it.

The first priority for me has been to find a question that establishes the correct meanings of 'da' and 'ja'.

I beleive I've succeeded in formulating the first question to ask a chief.

Ok, so we have three chiefs. Call them "Truthteller", "Liar", and "X" - and we don't know which is which. I'm also assuming that, although we have no idea whether "X" lies or tells the truth in his answer, he fully understands the question and knowingly must do one or the other. In other words, if "X" says 'da' or 'ja', it will definitely be a conciously true or false statement, not a random utterance.

Alright here goes:

Do others agree with this analysis?

But this presents an apparent contradiction. The probability of your selection was originally 4 Jacks and 2 Queens. There are subsequently six pairs of cards remaining. You already know there is at least one Jack in each pair. As it happens, the host reveals a Jack from each pair. Apart from not cancelling any row,  the host has simply confirmed what you already knew.

What information have you got that could change the 4-Jack-2-Queen probability of your selection, which was made before the host entered?

Therein lies the point. In those six rows, it was not impossible for the host to have revealed a Queen. It simply turned out that he didn't do so. The way the problem is posed is to consider what happens in that event. This does not rule out other events.

Let's clarify the rules. Whenever the host reveals a Queen, that particular row is null and void, but you consider the rows that haven't been voided. The most probable scenario is that the ignorant host will randomly reveal 2 Queens. In that event, there is no reason not to stand by your initial guess that you selected 2 Queens and 4 Jacks. You know that two rows where you chose a Jack have been cancelled. This means that in the four uncancelled rows, you probably selected 2 Queens and 2 Jacks. For those rows, this does indeed confirm the odds as 50/50 whether you stick or swap.

So what happens when, inevitably, you get a game in which by chance the host revealed six Jacks?

Is the intial guess for your selection - 4 Jacks and 2 Queens - still the right one? Do you in fact have a 2/3 advantage in swapping?

As far as it affects the probability of your selection, what you're saying is:

The host revealing six Jacks by chance = The host revealing six Jacks on purpose.

If so, might it not follow that in a single game:

The host revealing one Jack by chance = The host revealing one Jack on purpose.

But it has been well established in Monty Hall that what the host reveals must be based on certain knowledge. It is only then that it is impossible for the host to reveal a Queen. This is what is said to give you the 2/3 advantage.

I appreciate your position. On average, your initial selection is always likely to be 4 Jacks and 2 Queens. You're saying, therefore, that in each game, the probability of that initial selection must remain the same, regardless of what the host randomly or knowingly reveals afterwards.

If, as expected, the ignorant host cancels two rows by showing a couple of Queens, the 50/50 odds do apply, but only to the four remaining rows, from which you probably selected 2 Jacks and 2 Queens. This would still translate to your original selection having been 4 Jacks and 2 Queens.

If the host happens not to cancel any of the six rows, nothing is removed from your initial selection. A reasonable assumption is that this selection, made before the host revealed anything, probably contains exactly what was originally likely - 4 Jacks and 2 Queens.

But doesn't this contradicts the importance of the host's state of knowledge, so crucial to the original Monty Hall solution.

If, in a single game, the host was ignorant but didn't reveal the prize - there's no end of websites that explain why this reduces the odds to 50/50.

But if true, shouldn't those odds apply to each of the six rows in a game where the host was ignorant but didn't reveal a Queen? Surely each individualcard you selected should now have a revised 50/50 chance of being a Jack or Queen - because the host had no knowledge when he revealed a Jack from each row? Wouldn't your entire six card selection, therefore, now most likely contain 3 Jacks and 3 Queens,

This would mean that the 4-Jack-2-Queen probability of your initial selection can be retrospectively changed by what a host happens to do randomly, after the fact.

A strange but true conclusion?

Simon

So far, Mathsyperson has come pretty close in his solution to the six card version, so I'll start with that.

It does indeed matter who revealed what. Yes, the odds are 50/50 if the host revealed all four cards. Also if you participated and revealed one of each kind, the odds are still 50/50.

However, if you revealed two of a kind - say Kings - and the host revealed two Queens, then no matter what order they were removed, your card now has a 3/4 probabilty of being a Queen. (If you revealed just one King, and the host removed another King and two Queens, the odds would be 3/5 your card is a Queen.)

With the four card version the answer very simple but quite unbeleivable.

Lets take those two possible games where the same card is revealed.

a) You choose one of four cards. From the remaining three cards a Queen is revealed. You turned it over yourself. There are definitely two Kings and one Queen left. Result? The odds are 1/3 that your card is a Queen. Identical odds apply to the other cards.

Most people agree with the above but dispute the following:

b) You choose one of four cards. From the remaining three cards a Queen is revealed. The only difference is you instructed the host to reveal it. There are two Kings and one Queen left. Result? The odds are 50/50 that your card is a Queen. For each of the other two, the odds are 1/4.

For the record, I beleive the there are two numbers to eliminate - 67 & 34.

All the rest are divisable - or can be divided by - one or more of the others.