Final Decision

[MANDO]

The following is an interesting logic problem:

An attack pilot took off with a payload of three bombs under the fuselage. Few minutes later he received a radio call, by mistake, the ground crew loaded 2 training bombs (red) and only 1 real bomb (dark green).

The airplane has a lever in the cockpit with three positions to select and arm each bomb (left, center and right pilons), at the left of each position of the lever is a illuminated red led inditating that the ordenance is present at the corresponding pilon. This lever was at the lower position, so left pilon was selected.

To drop a bomb, the pilot needs to move the lever to the desired position, click a button on the lever to arm the bomb, wait three seconds and then press the trigger at the stick to release the bomb.

As he was aproaching the target, the center pilon led switched off, by some mechanical problem that bomb was released. The pilot inverted the plane quickly and looking at his high six saw a bright red object descending. He was lucky, it was one of the two training bombs.

Minutes later he was just over the target and due the heavy flak he will not have more than a single chance to drop one bomb and run back to home.

What should he do with the bomb selector lever?

[hornet63]

*****punt of course*****

[MANDO]

To clarity the final situtation of the pilot:
He now has only two bombs attached to left and right aircraft underbelly pilons, one of these bombs is real (dark green), the other is a practice bomb (red). Remember that he lost the center pilon bomb and it was a red practice bomb.
The bomb lever selector is still at its original position, marking left pilon.

The pilot is starting the final attack dive, he has time left only to make a final decision: move the lever to select the other pilon or not. He will not have time to do a second attack. What should he do?

[DieAz]

do like in MA arm both and be suicide dweeb

[MANDO]

Originally posted by DieAz
do like in MA arm both and be suicide dweeb

The lever only allows you to select and arm one pilon. His final decision will show how dweeb he is (hopefuly he will not select center pilon, he would be dweeb and retard at the same time )

[bozon]

switch to the other bomb. 66% that this is the live one.
that just goes to show the absurdity of statistical probability.

Bozon

[BlauK]

Just drop either one of them before the attack and check the color... then u know if it is worth to do the final attack or to just bug off home

Bozon, how did u figure out that 66%?
I would suggest 50% at that point, whereas before the failure it was only 33%... or am I totally missing something?

[bozon]

from statistics point of view, this is a problem of partitioning.
1st, the pilot chose one option out of the 3. So it means that he has got 33% to get it right, but 66% to be wrong.

Think of it as dividing his options into to groups, one with 33% to have the live bomb and the other with 66%.

The new information, that the center one was dud, left him with 2 bombs to choose from, but the partitioning to groups is still the same - one group has 33% and the other group (now containing only one bomb) has 66%. So he should switch.

The absurdity is that this assumes a cognitive first selection. Had the center bomb fallen without him looking at the selection lever position, and then he was asked to choose, he'd have 50-50% chance. In order for the new information (center is not live) to change something, a first decision had to be made prior to the info arriving. Alternatively, the random position of the lever before him choosing is also information.

Bozon

[MANDO]

If we look at the final situation, the pilot has two bombs, one of them real. So, it would seem that he has same chances selecting any of them, 50%.

But Bozon is right, knowing that the center pilon had a training bomb is the key to change the lever from left pilon to right pilon.

If the pilot does not change the lever, or simply decices randomly which pilon to select (left or right), he will have 50% chances of dropping the real bomb, but changing to the right pilon he has 66%.

[BlauK]

Yeah, right.
And if his Grandma had said no to his grandpa, the chance would be 88,9%

Looking at he lever or not does not change the chance % when he knows that one training bomb was lost. Prove me wrong, drop the bomb in similar case 100 times and prove that you get even close to 66 cases of the real bomb  

[Furball]

most boring thread EVER!

[MANDO]

BlauK, may be you will understand with a different example:

Lets suppose the plane had 1000 bombs, only one of them real. Then the plane releases 998 bombs (all of them training bombs), and only two remain: the one initially selected by the lever and another. What would you do?

If you need an even more clear example, lets suppose you play Lottery. You have your number, and only one wizard knows the winning number (this wizard does not play Lottery). The wizard ask you to see your number, and then he tells you: "ok, the winning number is yours or this other one (and shows you another number)". You may change your number now by the other. What would you do?

[MANDO]

Originally posted by BlauK
drop the bomb in similar case 100 times and prove that you get even close to 66 cases of the real bomb  

You would be really surprised ... To reach a stable 66% probably you will need to repeat the case 150 times.

[DREDIOCK]

Originally posted by MANDO

.

What should he do with the bomb selector lever?

Nothing. He returns home dumping the last two along the way in case there were any other malfunction, then beats the crap out of his mechanic

[ALF]

I musta missing something in college, if you flip a coin 10 times and it comes out heads all 10 times, what are the odds it will come out heads on try 11....50% because the next flip is a new subset, completely removed from the old set with no direct imperical relation.

You drop 2 of 3 bombs, only one of which is a live good bomb....your odds of dropping a dud are 66%, you now have 2 bombs, the odds are not 50%, because you are now dealing with a NEW probability, that is completely seperate from the initial one.  Dropping the 1st bomb has nothing to do with the other two.  

I believe that where some are making their mathmatical error is in the assertion that if we have a 33% chance of getting the 'live' one, and the first is a dud, they we can take the 33% chance we had initially and add the new 50% chance we have now, and somehow conclude that its a good bet you will get the live bomb on this second drop.  And while if you fly the plane 1000 times, in the end you will see a 66% rate of a live bomb being dropped, this is only because 33% of the time the damned live bomb will drop on the initial malfunction.  
You cannot take the information that the initial bomb is a dud and still use the probability of a three bomb subset, because youve made it a two bomb equasion.