Author Topic: Final Decision (Read 8687 times)

MANDO · « **Reply #120 on:** December 06, 2004, 06:41:37 AM »

lasersailor,
now imagine your plane has 1000 bombs, you lost 998 training bombs, what is the probability that the lever was selecting the real bomb from the very beginning? 1/1000? But what is the probability for the other rack? 999/1000?

dedalos · « **Reply #121 on:** December 06, 2004, 09:41:38 AM »

Quote

Originally posted by BlauK
dedalos,
ok... show me the formula all the way from the beginning with an accidentally lost training bomb. If you lose a real bomb, you can disregard those cases, because they do not fulfill the stated case.

How will you end up with 50:50 for switching case?

What is your bases for the statement that this does not apply for an accidentally lost bomb (which WAS already lost and therefore happens in 100% of the possible relevant cases!!!)

I don't think I can because of this line:
The probability that Monty Hall opens door B if the prize were behind C,
P(Monty opens B|C) = 1

If the location of the real bomb/prize is not known, then P(Monty opens B|C) = .5 and therefore the result of the the formula will not be 2/3.

The base of this prob problem is the fact the who ever opens the door or drops the bomb, knows where the real one/prize is. If not, then it is just random acts and at the end you have 2 choices to make a random decison.

Lets try it anyway, without the knowlege of where the prize is:
The probability that Monty Hall opens door B if the prize were behind A,
P(Monty opens B|A) = 1/3

The probability that Monty Hall opens door B if the prize were behind B,
P(Monty opens B|B) = 1/3

The probability that Monty Hall opens door B if the prize were behind C,
P(Monty opens B|C) = 1/3

The probability that Month Hall opens door B is then
p(Monty opens B) = p(A)*p(M.o. B|A) + p(B)*p(M.o. B|B) + p(C)*p(M.o. B|C)
= 1/9 + 1/9 + 1/9 = 3/9 = 1/3

Then, by Bayes' Theorem,

P(A|Monty opens B) = p(A)*p(Monty opens B|A)/p(Monty opens B)
= (1/3)/(1/2)
= 2/3
and
P(C|Monty opens B) = p(C)*p(Monty opens B|C)/p(Monty opens B)
= (1/3)/(1/2)
= 2/3

All choices are equal and therefore the selction will be random.

BlauK · « **Reply #122 on:** December 06, 2004, 11:31:46 AM »

Dedalos,
why are you loking at the situation BEFORE opening the door or losing the bomb. Try to look at the situation when the final decision has to be made. Th edoor is already opened P(goat was shown)=1 just like P(training bomb was lost)=1

Actually one does not even have to know the LOCATION of the real bomb or the price to be able to open an unselected door with a goat to lose an unselected training bomb. One only need to know that a goat was shown or that a training bomb was lost. That knowledge is what counts in the final P(switch selection)=2/3 and P(dont switch)=1/3.

The accident of losing the real bomb is not included in the question or the case of probability. If you are asked about how many bananas you can eat, why would you start calculating also some bananas?

Consider the following case: You had 2 coins, lost one accidentally or someone took one from you intentionally. Now you flip all your coins once. Is the P(heads) at this particular moment somehow different in these 2 cases only because in one case the the prior happening imcluded intention and the other was accident???

dedalos · « **Reply #123 on:** December 06, 2004, 11:45:54 AM »

Blauk

The reason I look at the situation before opening the door (look at the formula above) is that for it to work and give you the desired 2/3 result, the person openning the door or dropping the bomb, must have knowlege of where the prize or the real bomb is.

If not, and the door/bomb is opened/droped randomly then your probability is the following:
3 bombs --> prob = 1/3,1/3,1/3 for any of the bombs to be droped.

Once one bomb is droped, the prob is 1/2 for the one on the left and 1/2 for the one on the right. Real or fake does not make a difference at this point. Because you did not select the droped bomb but it was droped randomly, you have a 50/50 situation.

Appologies if I cannot explain it any better (Einglish not my first language), but the key to the problem is knowing that the dropped bomb was selected to not be the real one for a reason. If it was an accident then it is a random event resulting in the 50/50 situation.

BlauK · « **Reply #124 on:** December 06, 2004, 12:13:56 PM »

Same here dedalos, English is not my 1st language. Still I wonder why you insist that there is any probability included in the case of losing the first training bomb. The description of the problem already states that IT WAS LOST!!!! Therefore P(training bomb was lost)=1... it is a pre-condition for the whole case. Just as it is for the Monty Hall scenario. When the competitor is asked to switch or not, a goat is already shown.

This question is about switching after training bomb is lost, not about probability of losing a training bomb. If you want to include the latter, you have to find out somewhere the probability for accidentally losing a bomb.

ps. I just find this phenomenon very captivating and always enjoy good logical arguments

That is why I have sticked with this thread. Nothing personal is meant towards anyone here

MANDO · « **Reply #125 on:** December 06, 2004, 01:02:53 PM »

Quote

Originally posted by dedalos
3 bombs --> prob = 1/3,1/3,1/3 for any of the bombs to be droped.

Sorry, but wrong: one of them has 0% to be lost. You know that training bombs fail, real bombs dont.

lasersailor184 · « **Reply #126 on:** December 06, 2004, 01:22:57 PM »

If you lost 998 training bombs, you suddenly have 1 real bomb and 1 training bomb.

Thus you have a 50% chance of selecting the right bomb. It doesn't matter how many bombs you have before you have to make the decision. It only matters how many bombs you have to make the decision. You have 2, one real one training bomb.

So it doesn't matter whether you flip the lever or not, it's a 50% chance you have the correct bomb selected.

So my solution still stands.

dedalos · « **Reply #127 on:** December 06, 2004, 01:24:20 PM »

Quote

Originally posted by BlauK
Just as it is for the Monty Hall scenario. When the competitor is asked to switch or not, a goat is already shown.

I think this is where we disagree. It is not about the goat/bomb being shown. The difference is that in the Monty example, his door was not selected randomly. He is not going to open the door with the prize. If door A had the prize then the prob of opening door A=0. This is not the case with the bombs. The probability of a good bomb droping by accident is 1/3. Once a bomb is droped then there is a 50/50 that one of the remaining is good (since you know a training bomb was lost)

The only way the Monty example would apply to the airplane would be if tyhe story changed to:

The grownd crew knows where the good bomb is and they remotly made a traning bomb drop. Then, the pilot should switch since there is a 2/3

Quote

Still I wonder why you insist that there is any probability included in the case of losing the first training bomb. The description of the problem already states that IT WAS LOST!!!! Therefore P(training bomb was lost)=1... it is a pre-condition for the whole case. Just as it is for the Monty Hall scenario. When the competitor is asked to switch or not, a goat is already shown.

Not the same. What makes the difference is not that the door/bomb was opened/droped. The difference is on the why the door/bomb was selected. In any case, try it using prob formulas and you will see. The only way to get a 2/3 is by knowing that the who ever made the bomb drop, knew that it was not going to be a real one.

dedalos · « **Reply #128 on:** December 06, 2004, 01:26:03 PM »

Quote

Originally posted by MANDO
Sorry, but wrong: one of them has 0% to be lost. You know that training bombs fail, real bombs dont.

Right, did not think of that

dedalos · « **Reply #129 on:** December 06, 2004, 01:27:48 PM »

Quote

Originally posted by lasersailor184
If you lost 998 training bombs, you suddenly have 1 real bomb and 1 training bomb.

Thus you have a 50% chance of selecting the right bomb. It doesn't matter how many bombs you have before you have to make the decision. It only matters how many bombs you have to make the decision. You have 2, one real one training bomb.

So it doesn't matter whether you flip the lever or not, it's a 50% chance you have the correct bomb selected.

So my solution still stands.

If those 998 were lost completly randomly/accidentaly then you are right.

BlauK · « **Reply #130 on:** December 06, 2004, 01:33:59 PM »

lasersailor,
sorry, but you are wrong there

This thing has been already explained it tnis thread and in some of th elinks provided. I wont repeat it. Go find it out yourself

I repeat once more.. it is not about selecting one or another. It is about keeping the original selection or switching to a new one. If you cannot see the difference between those, it is useless to continue

dedalos,
The competitor in Monty Hall scenario does not really know how Monty picked the door, does he?

And why would it matter since it already happened. Not statistics, but probabilities in this situation for switching

BlauK · « **Reply #131 on:** December 06, 2004, 01:38:16 PM »

for those 1000 bombs there is an initial chance or 1/1000 that the real bomb is selected. In 999 of 1000 cases it is not.

Since the condition is that the real bomb cannot be dropped by accident, in 999/1000 cases the only remaining (not dropped) unselected bomb is the real one and in 1/1000 cases it is the selected one...... naturally everyone switch the lever, right?

----

If nothing was selected initially, all those dropped bombs dont mean a thing!!! One just selects between 2 bombs and the chance of getting the real bomb is 50:50.

---

Having something selected and switching or not is a completely different case from just choosing between two previously unselected options. Funny, interesting and almost unbeliavable... but true

MANDO · « **Reply #132 on:** December 06, 2004, 01:56:03 PM »

lasersailor, on the first or second page of this thread there is a link to my "game" where you can try your "luck". Download it and test it by yourself, it is very small.

lasersailor184 · « **Reply #133 on:** December 06, 2004, 02:37:44 PM »

I have seen and read everything on this post. (Just to make sure someone didn't post my idea first).

This isn't some complex statistics problem. It's something a second grader could solve easily.

There is a 50-50 chance that the person would select the real bomb.

Say you have a bag with 3 MM's in it. 2 red, one blue. Just by dumb luck, a red falls out of a hole in the bottom of the bag. You reach in and grab 1 MM. What are the chances it's going to be blue? What are the chances it's going to be red?

The problem is no different then the one I just described.

dedalos · « **Reply #134 on:** December 06, 2004, 02:39:54 PM »

Quote

Originally posted by BlauK

dedalos,
The competitor in Monty Hall scenario does not really know how Monty picked the door, does he? And why would it matter since it already happened. Not statistics, but probabilities in this situation for switching

So, we are in agreement that (Monty) knowing makes the difference?

The competitor knows unless it is the first time that game is run. He does not really need to know. What matters is how Monty is selecting the door. The contestant needs to know only so that he can realize what is going on and switch the door. His knowledge has nothing to do with what he should do. He may not do the right thing if he does now, but his knowlege does not effect the results of the problem. Clear as mud?

This has nothing to do with statistics. It is one of the first probability problems you will find in probability classes. I've seen it 4 times, 2 dif Universities, 3 diff teachers.

Any teachers here? Heeeelp, we are stuck