Author Topic: Final Decision (Read 8104 times)

BlauK · « **Reply #165 on:** December 08, 2004, 02:25:39 AM »

laser,

The your probability of winning in a lottery with an already bought ticket does not get any better if non-winners are revealed to you before finding the winner

Consider this:
-1000 lots, only 1 winner.
-you buy one
-Someone else buys 999 and asks his secretary to open all "no win" lots. Should she find the winner, it should be closed again. If all are "no win" lots. then one lot should still be kept closed.

Now you and the other buyer both have only one lot each. Neither of you know which one is the winner.

What do you think... what are your own chances now? 50:50 ???? or 1:999 ????

With your reasoning so far you should claim 50:50. Do you?

MANDO · « **Reply #166 on:** December 08, 2004, 03:43:18 AM »

Quote

Originally posted by lasersailor184
No, I definately used yours. And since you claimed that the pilot should change the lever because of a 66% chance, I ran it on auto for roughly 100 runs a bunch of times with always change lever.

Ok, your result with 100 should be near 66%, if not, let it running 2 minutes with each option. And please, post your results.

My results for 200 attacks:

"Never change the lever" 31%
"Change the lever always" 69%
"Change the lever randomly" 54%

If you let the program to run even more attacks, 1000 for example, you will have more accurately 33, 66 and 50%.

BlauK · « **Reply #167 on:** December 08, 2004, 04:27:09 AM »

This is surely interesting and it is also interesting to see my own position switching...

It starts to look like this whole case is purely mathematical and dependent on how the pre-conditions are defined.

mando,
What if we consider th epylos instead of bombs, since they are actually selected instead the bombs, which can be on which ever pylon:

RB = pylon with real bomb
L-TB = pylon with lost training bomb
K-TB = pylon with kept training bomb

Possible cases:
A: RB selected, L-TB lost, K-TB unselected (1/2)
B: K-TB selected, L-TB lost, RB unselected (1/2)

So is the problem actually based on interpretations. When we are counting probabilities for this one (once in a lifetime) case where the TB was lost from center position, can we say that it could also have been the right pylon if there was a training bomb?

-TB was lost from an unselected pylon.
or
-TB was lost from center pylon, which was unselected.

Does it affect building the equation, that we know the location (pylon) of the lost bomb? Also we dont know if it was TB1 or TB2 that was lost... just a TB... so how can we differentiate between them? If TB1 was red and TB2 yellow, it would be a completely different case, since we would know which TB was lost.

A: RB selected, TB lost from middle, TB unselected (1/2)
B: TB selected, TB lost from middle, RB unselected (1/2)

So... thinking while writing here

What if the bomb selector was kind of automatic (semi-intelligent) in such way that it would not tell the positions, only that one could switch with it to a next available bomb. Then we would not know the position of the lost bomb or the selected bomb.

Mando's program is actually performing like above. But if we are originally told that in this particular case the left was selected and TB was lost from middle, we seem to get too much information to apply it to Monty Hall scenario.

Should I now position myself somewhere in between??

If we had this automatic bomb selector, the question would go: "one unknown bomb is selected, a TB is lost from unknown position, should we switch from originally selected unknown position to the other unknown position?" ..... Yes, with 2/3 probability.

Since Mando originally stated that left was selected and center bomb was lost, it seems that we cannot use Monty Hall scenario in this case.. or can we? Do we calculate the pylons or the bombs

More comments anyone??

Accident for single case against intelligence (with options) for repetitive cases, which kind of make the location info irrelevant seems to make the difference..... like someone stated already quite early in this thread.

OK... made my mind

Without a selector I just described higher above (lets call it Monty Hall bomb selector) the Monty scenario does not apply to the original case.

Sorry Mando...
I seem to be a weather vane, but I still have enjoyed this argument

.. and thanks to everyone for your own arguments.. sorry for possible harsh words.

MANDO · « **Reply #168 on:** December 08, 2004, 11:22:15 AM »

BlauK,
considering that the initially selected bomb may be training and lost, if we change always the lever, we'll get 50%, if we dont, 33%. Even in this case the answer is the same: TO CHANGE THE LEVER.

dedalos · « **Reply #169 on:** December 08, 2004, 12:16:00 PM »

Quote

Originally posted by BlauK
Heeelloooo dedalos

The pilot ALREADY SAW that a RED TRAINING BOMB WAS LOST! Cases E and F are not possible in this case. If you start adding things to possible cases, you can go on adding everything that is possible in the world.

If that is the only thing we find worth arguing, it is better to let it be I cannot show any better evidence than what was said about the situation. If you believe that TB was not lost, we are talking about different cases.

Ha, thats exactly the point

Without those cases, the pilot knows that he has two bombs. One good one bad and he flips a coin.

But you are right, we will never agree. That does not mean we cannot argue the point

lasersailor184 · « **Reply #170 on:** December 08, 2004, 01:21:02 PM »

From what I can understand is that the Monte Hall situation calculates the center pylon into the equation, even though it was lost.

You shouldn't do this. It's almost exactly like the pilot only upped the airplane with 2 bombs, one real and one fake. Hence it is only a coin flip.

dedalos · « **Reply #171 on:** December 08, 2004, 02:33:52 PM »

Quote

Originally posted by lasersailor184
From what I can understand is that the Monte Hall situation calculates the center pylon into the equation, even though it was lost.

You shouldn't do this. It's almost exactly like the pilot only upped the airplane with 2 bombs, one real and one fake. Hence it is only a coin flip.

What he said

Raider179 · « **Reply #172 on:** December 08, 2004, 04:18:10 PM »

blauk I understand the question just tell me if this is right.

He has two bombs left..
A lever will switch to either bomb.
He can choose A. To switch

B. Not to swtich

There are two outcomes
A. Drops training bomb
B. Drops real bomb

Is this right?
These are the only choices and only outcomes correct?
Is there something I am missing.

its 50:50 unless he knows exactly where the ordance is on his wings. Then he could just select the right ord and his chances are 100%

Raider179 · « **Reply #173 on:** December 08, 2004, 04:24:30 PM »

Quote

Originally posted by MANDO
Ok, imagine you do not have a plane with racks, but three doors. Behind two of them there are goats (instead training bombs), and the other door hides a golden doblon instead of a real bomb.

To select a door, you need to be in front of it.
Currently you are in front of LEFT door.

Suddenly, you ear a goat bleating (is that the english word?) at your door. Should you switch to other door? Yes (0% to 50%).

But, what if the goat bleats at the CENTRAL door? Should you switch to the RIGHT door? Yes also (33% to 66%). (<- This is the equivalent to the original case).

This makes no sense. On what are you basing switching is more probable. You have no idea where they put the goat. none The odds dont favor you either way. YRevealing the location of one goat just makes the odds go from 1:3 to 1:2. you have 2 curtains now instead of 3. I have to be misunderstanding this somewhere.

MANDO · « **Reply #174 on:** December 08, 2004, 05:27:32 PM »

Raider179, the idea is that first time you were in front of a door, you have 2/3 chances of missing the correct door. So, it is more probable that the correct door is any of the other two.

If you dont agree this from the beginning, do not continue.

If a goat sounds in any other door but yours, then you take out that door from the group with 2/3 chances of having the correct door. Now you have your original door and another one. Your door has the same chances as the beginning 1/3. But the group with 2/3 now has a single door left (with 2/3). So, it is better for you to change your door.

lasersailor184 · « **Reply #175 on:** December 08, 2004, 05:30:57 PM »

No mando, that's wrong.

If you're standing in front of the middle door and suddenly it bleats, then you know a goat is behind that door. Or if it bleats behind another door (but not both).

So you now have *******************TWO******************* doors to choose from.

Without any further information I.E. Another goat bleating, you have 50-50 chances of selecting the correct door.

Switching which door you are standing in front of has *****NO***** bearing on the final problem, as does which door the goat bleated behind.

Raider179 · « **Reply #176 on:** December 08, 2004, 06:08:42 PM »

Quote

Originally posted by MANDO
Raider179, the idea is that first time you were in front of a door, you have 2/3 chances of missing the correct door. So, it is more probable that the correct door is any of the other two.

yeah that is correct 2:3 but this is because each individual door has a 33% likelihood of being correct. 2 x 33 =66%

If you dont agree this from the beginning, do not continue.

If a goat sounds in any other door but yours, then you take out that door from the group with 2/3 chances of having the correct door. Now you have your original door and another one. Your door has the same chances as the beginning 1/3. But the group with 2/3 now has a single door left (with 2/3). So, it is better for you to change your door.

yeah that is correct 2:3 but this is because each individual door has a 33% likelihood of being correct. 2 x 33 =66%

Its a fundmental problem with your logic here. You take away one of the doors but you are still saying it is part of the group. It is not. each of the three has 33% chance. take one away you have to remove it from the odds.

lasersailor184 · « **Reply #177 on:** December 08, 2004, 06:14:18 PM »

Yup, just like taking marbles out of a bag.

You have 50 marbles in a bag, 10 red, 10 blue, 10 green, 10 clear, 10 black.

What are the chances you'd pick a black one? 10/50. You keep the marble out of the bag.

Now, what are the chances you'd pick a red one. It's not 10/50 because you took a marble out, it's 10/49.

MANDO · « **Reply #178 on:** December 08, 2004, 07:00:36 PM »

Ok Raider179,
now imagine you have 1000000 doors with only one golden coin, the rest are goats. Initially you select door number 710567.

Suddenly, all the goats but one piss their doors, only your door and other remain clean. Would you change the door? If you dont change your door, and you win, you are really, really, really lucky.

And yes, at the end you have two doors, one good, one bad, 50%?

Raider179 · « **Reply #179 on:** December 09, 2004, 12:28:07 AM »

yeah its a 50 50 situation you just described.