[OT] mathematical query

[CRGreathouse]

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

The correct answer to this portion of the question (and thus the entire problem) depends on interpretation. Here are the two:

Sigil's: "Is the gender of either of your childern male?" "Yes"
Mine: "What is the gender of a random child of yours?" "Male"

Deciding this decides the problem.

Ahem... why do you place MM(M) and FF(F) twice?

The first is the chance that nthe older of the two boys is pictured; the second is the chance that the younger of the two boys is pictured.

And why not MM(F) and FF(M)?

It's not possible to have a picture of the daughter if there is no daughter.

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.

I'll use your reasoning with the same problem, except that the picture is hidden.

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)
MF(F)
FM(M)
FM(F)
FF(F)

Thus, the chance of two boys (with no further infomation) is 1/6.

Since this is an obvious contradiction, your reasoning is flawed.

 

[The Sigil]

*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?

Yes.

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.

*sigh* I repeat...

The King HAS chosen something. The probability of him being correct is set based on the information when he makes his choice - subsequent revelation of information does not give him a greater probability of being correct (unless sufficient information is revealed so as to make the probability of all other solutions zero).

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?

No. Not okay. The remaining two dispositions do NOT have the same probability. Why not?

If Possibility B is true, means that the King chose poison, thus the wizard has a 100% chance of throwing out potion 3. IOW, we have a 100% chance of arriving at the point of "King has potion 1, potion 3 is discarded" if Possibility B is the correct answer.

Possibility C means that the King chose elixir, BUT the wizard had only a 50% chance of throwing out potion 3 (he had a 50% chance of throwing out potion 2). Thus, we have only a 50% chance of arriving at the point of "King has potion 1, potion 3 is thrown out" if Possibility C is the correct answer.

IOW, possibility C (where the king is correct) is only 50% as likely as possibility B. Hence, we have the 2:1 ratio.

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%.

Wrong, for the reasons shown above.

Or if you prefer, think of it in reverse...

"I'm the court wizard. I make three potions for the king - one of which grants immortality and the other two of which are poison."

1/3 of the time, the King guesses right - and I get to pick one poison potion to dump and try to trick him into choosing the other one.

2/3 of the time, the King guesses wrong. I am forced to dump the other poison vial and offer the King the chance to switch to the "right" one.

--The Sigil

 

[CRGreathouse]

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.

This assumes that the teller is truthful and will never select your choice.

 

[The Sigil]

Sigil's: "Is the gender of either of your childern male?" "Yes"
Mine: "What is the gender of a random child of yours?" "Male"

Still don't think I can agree with you on that one... should be:
"What is the gender of the nth child of yours?"

Otherwise, they are equivalent as in your suggestion, I still don't know what "Random Number" you have selected and thus for me the effective quote is the same.

If they are not equivalent, please let me know how.

--The Sigil

 

[The Sigil]

This assumes that the teller is truthful and will never select your choice.

True.

Altogether too much assuming going on today.

--The Sigil

 

[Xeriar]

It's easier to break down the problem this way.

You have three doors. Behind one lies the true prize. Behind the others, lies a goat (no prize).

You pick a door. Doesn't matter which door, but you pick one. The host then reveals one of the doors, showing you the goat.

You then get the opportunity to switch.

If you're smart, you'll switch. Think about it - your chances of picking the right one initially was one third. That means, you have a two-thirds chance of sitting on a goat. The announcer revealing the other one being a goat -doesn't matter- because there will always be one to spare.

---

If you don't believe me, let's play a game. It costs $5 to play. Under one of three cups lies $10, under the others, a grain of rice. Once you pick a cup, I will reveal one of the grains of rice, and give you the opportunity to switch.

Try it with yourself if you have to, seriously.

The original question was plagued with a different problem, however. Knowing the gender of one child does not predict the other. You only know that he has a son, his other child is either a son or a daughter.

 

[CRGreathouse]

You have three doors. Behind one lies the true prize. Behind the others, lies a goat (no prize).

You pick a door. Doesn't matter which door, but you pick one. The host then reveals one of the doors, showing you the goat.

You then get the opportunity to switch.

If you're smart, you'll switch. Think about it - your chances of picking the right one initially was one third. That means, you have a two-thirds chance of sitting on a goat. The announcer revealing the other one being a goat -doesn't matter- because there will always be one to spare.

Let's do it another way. There are two people, P1 and P2. They both get to pick a door (A, B, or C). One door has a prize, and the other two have nothing.

P1 picks A.
P2 picks A.
The announcer opens C and shows that there's nothing there.
P1 switches to B.


They're both choosing one of the two nonrevealed doors. Why should they have different odds unless the announcer has a discernable pattern in choosing doors?

 

[Blacksad]

Oops! Sigil is right (I thought eldest son, when I read son)

 

[The Sigil]

Let's do it another way. There are two people, P1 and P2. They both get to pick a door (A, B, or C). One door has a prize, and the other two have nothing.

P1 picks A.
P2 picks A.
The announcer opens C and shows that there's nothing there.
P1 switches to B.


They're both choosing one of the two nonrevealed doors. Why should they have different odds unless the announcer has a discernable pattern in choosing doors?

Because the announcer DOES have a discernable pattern in choosing doors.

*Rolls eyes*

If it helps, think of it this way... when you choose Door A, you separate the probability into A and "Not A". Door A has a 33% chance of being right. "Not A" has a 67% chance of being right. Even when you later open Door C, "Not A" STILL has a 67% chance of being right because you have not eliminated "Not A" from consideration.

For the last time, you have a 1:3 chance of being right when you make your initial pick. THIS DOES NOT BLEEPING CHANGE WHEN ONE OF THE OTHER TWO DOORS IS OPENED!!!

Play the following game and DON'T disagree with 67/33% until you've actually tried it!

Go get two six-sided dice of different colors. Let us say "red" and "blue" for sake of argument here. Cut die rolls in half, rounding up, to get a range of 1-3 on each die.

Roll both dice. The "red" die indicates the door the prize is behind (1-3). The "blue" die indicates the door person P picks.

Now, eliminate one of the wrong answers. Mark how often "staying" gets you right and how often "switching" makes you right.

For example, I roll 2, 1. That means the right answer is "2" and I picked "1". I now eliminate #3. That means if I stay, I'm wrong. If I switch I'm right. One tally for "switching." And so on...

Do this 120 times.

Come back when you notice the result is MUCH closer to 80/40 than it is to 60/60.

--The Sigil

 

[Zappo]

Possibility C means that the King chose elixir, BUT the wizard had only a 50% chance of throwing out potion 3 (he had a 50% chance of throwing out potion 2).

Nope, problem states that the wizard throws out potion 3.