[OT] mathematical query

[The Sigil]

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

*FRUSTRATED SCREAM*

Yes, but that only tells us that he happened to roll 01-50 on his percentile die. The potion we happened to label "3" could just as easily have been labelled "2."

GO DO THE ***#@)*(%&)@#*%&@#$)*(%&)@#$(% EXPERIMENT AND THEN COME BACK HERE!!!

If you know I'm correct and are just trying to read things in to be a comic and tick me off, you're succeeding. If you're not, you're being bullheaded about your ignorance. Either way, you're about to go on my "ignore" list.

--The Sigil

 

[Zappo]

*FRUSTRATED SCREAM*

Ooop, sorry about my last post. I've got what you meant with that now (hey, it's 2 AM here). I'm not fully convinced, but I'll think some more about it. G'night.

 

[The Sigil]

One last time, REAL SLOWLY...

1.) Woman with kids:

4 possibilities.
MM
MF
FM
FF

All have equal odds. 50% chance of having one child of each gender.

2. Man with picture of son:
4 possibilities
MM
MF
FM
FF

All have equal odds. When he shows a picture of his son, we do NOT then have "MX" and we do not have "XM." We have "MX or XM" which is much different. All we learn is that FF is impossible. In other words, all we learn is that we are chatting with someone from one of the first three groups. That leaves us with a 67% chance of having one child of each gender.

3. Pick, reveal, ask to switch:
9 possibilities ("choice picked" and "right choice"):
AA
AB
AC
BA
BB
BC
CA
CB
CC

We have a one in three chance of being right (AA, BB, CC). When a choice is revealed to be wrong, provided it is NOT the choice we have chosen (bloody loophole-searchers), we get the following:
AA (B or C)
AB (C)
AC (B)
BA (C)
BB (A or C)
BC (A)
CA (B)
CB (A)
CC (A or B)

The above all have equal probability - and we are still right only 1/3 of the time. It does NOT look like this for equal probability:

AA (B)
AA (C)
AB (C)
AC (B)
BA (C)
BB (A)
BB (C)
BC (A)
CA (B)
CB (A)
CC (A)
CC (B)

It DOES look like this with the following unequal probabilities:
AA (B) - 1/18
AA (C) - 1/18
AB (C) - 1/9
AC (B) - 1/9
BA (C) - 1/9
BB (A) - 1/18
BB (C) - 1/18
BC (A) - 1/9
CA (B) - 1/9
CB (A) - 1/9
CC (A) - 1/18
CC (B) - 1/18

Now let us trace those who wish to quibble the specifics of the problem ("it said #3 was taken away" or "he picks #1").

If he picks A, we are left only with....

AA (B) - 1/18
AA (C) - 1/18
AB (C) - 1/9
AC (B) - 1/9

And if we know C is taken away, we are left only with...

AA (C) - 1/18
AB (C) - 1/9

This tells us that you are right half as often as you are wrong. NOT 50-50.

Again, go do 120 experiments with dice and come back if you are still sure you are right.

--The Sigil

 

[The Sigil]

Ooop, sorry about my last post. I've got what you meant with that now (hey, it's 2 AM here). I'm not fully convinced, but I'll think some more about it. G'night.

Sorry for the scream. I've had a long day at work.

I'm going home now, hopefully to get into a better mood.

--The Sigil

 

[CRGreathouse]

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

That was my whole point. If you know he won't choose your door, that's a pattern you can use to affect your probabilities. If he chooses a bad door at complete random, you get no information.

Call the door initially chosen "A" an the other two "B" and "C".

Correct (Revealed) New choice
A(B)C
A(B)C
A(C)B
A(C)B
B(A)B
B(A)C
B(C)B
B(C)B
C(A)B
C(A)C
C(B)C
C(B)C

Randomly chosen - 1:1

Correct (Revealed) New choice
A(B)C
A(B)C
A(C)B
A(C)B
B(C)B
B(C)B
B(C)B
B(C)B
C(B)C
C(B)C
C(B)C
C(B)C

Never chooses yours - 2:1

 

[CRGreathouse]

One last time, REAL SLOWLY...

1.) Woman with kids:

You did read my post on this, right? It's the only logical objection to your position (that I've seen, anyway).

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.

You said, above:

Yes, but that only tells us that he happened to roll 01-50 on his percentile die. The potion we happened to label "3" could just as easily have been labelled "2."

That's essentially what I'm saying here. Look, let me put this another way:

The guy has two marbles, one in each hand, which could be blue or pink. You ask if he has any blue marbles, and he procedes to open one of his hands, revealing a blue marble. What's the chance they're different colors? 2/3.

BB
BP
PB
PP - not possible

(BP and PB)/(BB, BP, and PB) = 2/3

The guy has two marbles, one in each hand, which could be blue or pink. You ask him to open his left hand. He responds by opening his left hand, revealing a blue marble. What's the chance they're different colors? 1/2.

BB
BP
PB - not possible
PP - not possible

(BP)/(BB and BP) = 1/2

Edit: added numbers to the marble example.

 

[2WS-Steve]

Alright, since using Bayes’ Theorem (a theorem entailed by the probability axioms and used by economists, game theorists, decision theorists, mathematicians, and statisticians…) wasn’t convincing try the following:

You walk along the beach and meet eight men; each man has two children and shows you a picture of one of those children chosen at random. The table attached below shows the probability space.

Every time you see a photo of a male child you guess that the other kid is also male. How many times did you guess and how many of those guesses were correct?


The mistake you’re making here Sigil is that you’re eliminating possibilities 7 and 8 but not eliminating possibilities 2 and 5.

Note also that it doesn’t matter whether or not you specify that it’s the older kid.

>>>>>>>>>
On the other hand, Sigil’s interpretation of the potion puzzle is spot on. Do a Google search for probability puzzles Monty Hall problem, and you’ll see a variety of sites with the solution.

 

[RyanL]

One last time, REAL SLOWLY...

1.) Woman with kids:

4 possibilities.
MM
MF
FM
FF

All have equal odds. 50% chance of having one child of each gender.

2. Man with picture of son:
4 possibilities
MM
MF
FM
FF

All have equal odds. When he shows a picture of his son, we do NOT then have "MX" and we do not have "XM." We have "MX or XM" which is much different. All we learn is that FF is impossible. In other words, all we learn is that we are chatting with someone from one of the first three groups. That leaves us with a 67% chance of having one child of each gender.
--The Sigil

Edit: Err, what was I thinking?

-Ryan

 

[CRGreathouse]

On the other hand, Sigil’s interpretation of the potion puzzle is spot on. Do a Google search for probability puzzles Monty Hall problem, and you’ll see a variety of sites with the solution.

Did you see my posts on these two problems? They both come down to the assumptions. Both answers to both problems can be defended, given appropriate assumptions.

 

[Blacksad]

Now I know what my math teacher meant when he said: "probability problems are too hard to correct, that's why there is none for the exam to get into engineer school"