[OT] mathematical query

The pictured son is either the oldest or the youngest.
Case 1: 50% to be oldest x 50% that the youngest is then male = 25%.
Case 2: 50% to be youngest x 50% that the oldest is then male = 25%.
Sum of probabilities: 50%.
 

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Blacksad's question is a good one.

The only real question to answer is "Is the Wizard truthful?"

If yes, then odds are better to switch.

If no, then a whole can of worms opens up.

See if the king were smart then he would choose the potion that the wizard would said was a good choice to aviod. Except that if the wizard were smart he would know that the king would do that so he would tell the king to not choose a poison. But if the king were smarter still, then he would choose the poition farthest from teh wizard because the wizard a smart man would choose to place the poison far from himself, except that the wizardknewthat and placedthepoisonnextohim, (gasp!) except the kning knew that deliberatley would choose a different potion except he knew the wizard would know so he would choose the potion closest to him.

Except that this was a Sicilian King and Sicilians know a universal truth..."you can't win a land war in Asia", but slightly lesser known than that is that you can't fool a Sicialian when it comes to matters of death. So the King would reach for the closest potion, then say "look over there" and while the wizard was looking away he would switch potions.

I did not know that the Dread Pirate Roberts was a wizard!


g!
 

In-game considerations apart, Sigil's reasoning is not correct. It begins by stating that the chance for a correct choice is 1/3 for each potion, and that's wrong already since the problem states that potion #3 is poisoned (so the chance for potion #3 isn't 1/3, it's 0), then by stating that the king chooses one of the potions with 1/3 chance, and that's wrong too because the problem states the king chooses potion #1.

The king has a 50% chance of having chosen the right potion.

Let's assume that potion 3 is poison. There are only two possibilities:

1) Potion 1 is elixir, potion 2 is poison, potion 3 is poison.
2) Potion 1 is poison, potion 2 is elixir, potion 3 is poison.

The possibilities obviously have the same chance, so the king has a 50%. :)
 

The Sigil said:
For the man:
1.) MM
2.) MF (older is male, younger is female)
3.) FM (older is female, younger is male)
4.) FF

The fact that he shows a picture of him with a son eliminates possibility four, leaving us with...

Yes, but half of 2 and half of 3 are also invalid (the pictures aren't of F, but there would be a 50% chance of that).

Try this:

First Second (pictured)
MM(M)
MM(M)
MF(M)
MF(F)
FM(F)
FM(M)
FF(F)
FF(F)

Eliminating the impossible ones:
MM(M)
MM(M)
MF(M)
FM(M)

Thus, a 50/50 chance.
 

No, Sigil's logic is correct.

If you have three potions, two losers and one winner, and...

If you (generically) select any of them and then you are told (generically) that one of the two you did not select is a loser (poison) then odds are in your favor to change your choice.

You would go from being a winner 1/3 of the time to a winner 2/3 of the time.


g!
 

Ahem... why do you place MM(M) and FF(F) twice? And why not MM(F) and FF(M)?

First Second (pictured)
MM(M)
MM(F)
MF(M)
MF(F)
FM(M)
FM(F)
FF(M)
FF(F)

Eliminating the impossible ones:
MM(M)
MF(M)
FM(M)

Thus, a 2/3 chance.
 

Heya:

My vote: (A) 50% (B) 0%.

If he had a daughter, it'd say "a picture of him with his son". Since he has two sons, the sentence reads "a picture of him with a son". Even the clarification supports this. ;)

Take care,
Dreeble
 

apsuman said:
No, Sigil's logic is correct.

If you have three potions, two losers and one winner, and...

If you (generically) select any of them and then you are told (generically) that one of the two you did not select is a loser (poison) then odds are in your favor to change your choice.

You would go from being a winner 1/3 of the time to a winner 2/3 of the time.
*sigh* ;)

I repeat:

If I select potion #1, and get told that potion #3 is poisoned, I have 50%.

There are only three dispositions of the potions:

A) 1 is poison, 2 is poison, 3 is elixir.
B) 1 is poison, 2 is elixir, 3 is poison.
C) 1 is elixir, 2 is poison, 3 is poison.

Each disposition has the same probability, for what we know. Correct until now?

King chooses potion #1 - but since he can change his mind, he hasn't really chosen anything. He just gets told, for free, that potion #3 is poisoned.

This eliminates possibility A. The remaining two dispositions must have the same probability. Since disposition A's probability is 0%, the probability of B and C is 50%. Ok?

If the king chooses potion #1, he has 50% - immortal if it was disposition C, dead if it was disposition B. If the king chooses potion #2, it's exactly the opposite, but still 50%.
 

Since we STILL seem to be having trouble with the "pictures" game, let us do a little back story.

There is a little island with four couples on it.

Each couple has 2 children.

One couple has 2 boys. The husband's name is BoB.
One couple has 2 girls. The husband's name is GaG.
One couple has a boy (older) and a girl. The husband's name is BiG.
One couple has a girl (older) and a boy. The husband's name is GuB.

One day BoB, GaG, BiG, and GuB are walking on the beach.

They see a shipwreck and a woman washed ashore.

She approaches one of them and they talk and find out that both have 2 children.

The man pulls out a picture of a son that belongs to him.

What have we learned?

Only that the man that the shipwrecked woman is talking to is not GaG. There is an equal probability that it is BoB, BiG, or GuB.

---

Or, to put it another way...

25% of men with 2 children have MM
25% have MF
25% have FM
25% have FF

When the man produces the picture of his son, you only learn that he belongs to the first 75% of people... AND NOTHING MORE!

To put it another way, 75% of men with 2 children can produce one picture of a son (assuming he has pictures of all of his children with him).

IF we assume that, as has been suggested, a man with 2 children who can produce a picture of his son has a 50-50 chance of the other child being a boy, what do we conclude?

75% of men with 2 children can produce a picture of a son times 50% chance that his other child is a son implies that 37.5% of men with two children have two sons.

IF we assume that, as I have suggested, that a man with 2 children who can produce a picture of his son has a 1/3 chance of the other child being a boy, what does that get us?

75% of men with 2 children can produce a picture of a son times 1/3 chance that the other child is a son implies that 25% of men with two children have two sons.

Which conclusion (50% or 33%) is at odds with our original statement? All things being equal, 25% of men with two children have two sons. Therefore, it seems QUITE clear that an answer of "50% chance to have another son" is wrong and "33% chance to have another son" is right.

Please LOOK at what your conclusion leads to.

--The Sigil
 

The Sigil said:
Therefore, it seems QUITE clear that an answer of "50% chance to have another son" is wrong and "33% chance to have another son" is right.
Yep. That's what I said, too. The thing with the wizard, OTOH... :D :p
 

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