Author Topic: Final Decision (Read 7357 times)

vorticon · « **Reply #60 on:** November 28, 2004, 03:07:51 PM »

he had a right handed colour blind ground crew (ever meet a left handed colour blind dude?). therefore they loaded left to right. they loaded the training bombs first, and realising there mistake, but not having time to fix it, they put on a real one, intending to warn the pilot before takeoff.

so naturally, he should set the lever to the RIGHT BOMB.

ccvi · « **Reply #61 on:** November 28, 2004, 03:15:54 PM »

Quote

Originally posted by MANDO
1 - The lost bomb is always training, being real one the pillot will abort the mission and case is over.

The lost bomb can also be the real one. Just not in your simulation. In case of mechanical failure on the pylon of the real bomb a wizzard helps and drops the dud. The pilot in the simulation isn't lucky not to lose the real bomb.

You are modifying the case of early rtb to match the case where switching is required instead of ignoring it.

edit: You're manipulating a "does not happen" to "cannot happen".

Raider179 · « **Reply #62 on:** November 28, 2004, 03:23:34 PM »

50%. With no knowledge of in what order the bombs were loaded this pilot has a 50% chance to drop the correct bomb. He can either select to switch or stay with the one he has. That is 2 choices and with no idea of which bomb is next that is a 50% chance of success/failure. You are confused because you keep interjecting the bomb that was jettissioned into the chances that you will get the right bomb. This bomb no longer exists as it relates to the problem therefore it is irrelevent.

1 : 2 thats 50% you either move it or you dont. You stand no better/worse chance because you have no idea where it is.

BlauK · « **Reply #63 on:** November 28, 2004, 03:24:23 PM »

ccvi,

the case was not about a random mechanical failure... it was: "you lost one training bomb... what now?" .. what are the chances in this case?

.. or should we also consider the possibility that the "mechanical failure" would drop more than one bomb, or that it would concern the engine.. etc. Just disregard it. The dud WAS dropped already

BlauK · « **Reply #64 on:** November 28, 2004, 03:26:53 PM »

Raider,
read them all. That is what I also thought in the beginning. But the case is not about selecting between 2 bombs... it is about keeping the alredy selected one or switching the selection. That is what makes the unbeliavable difference :\

ccvi · « **Reply #65 on:** November 28, 2004, 03:54:11 PM »

Quote

Originally posted by BlauK
the case was not about a random mechanical failure.../B]

MANDO in his initial post: "He was lucky, it was one of the two training bombs."

This clearly indicates a random mechanical failure, and the pilot was just lucky.

Otherwise it should have read something like: A bomb was lost. By more fragile design of the pylons for training bombs the pilot knew that only a training bomb could drop.

But it didn't. The pilot was lucky.

MANDO · « **Reply #66 on:** November 28, 2004, 04:56:45 PM »

Quote

Originally posted by ccvi
MANDO in his initial post: "He was lucky, it was one of the two training bombs."

ccvi, you are missing completely the point. Being a real bomb does not change the % of destroying the target once you start the real attack, because you are not going to perform that attack.

The question was "WHAT TO DO ONCE YOU ARE GOING TO DIVE AND ATTACK?", not just what were the chances of the plane since the very beginning once airborne.

ccvi · « **Reply #67 on:** November 28, 2004, 06:18:12 PM »

Please replace

253 for (la_index = 0; la_index != 3; la_index++)
254 {
255 if ((la_index != la_realbomb) && (la_index != la_lever))
256 {
257 la_lost = la_index;
258 }
259 }

with

la_lost = rand() % 3;
if ((la_lost == la_realbomb) || (la_lost == la_lever))
return (FALSE);

This IGNORES cases where the real one drops, not changes them to match the scenario described. IOW: The pilot is luck not to lose the real one, he doesn't keep the real one by design.

edit: replaced wrong 2 with 3.

bozon · « **Reply #68 on:** November 28, 2004, 06:32:57 PM »

JNOV has a good point about the selection of the fallen bomb which I assumed given.

If it is truely random, and by repeating the experiment we allow for any one of the 3 to fall even if it is the live (or even the selected one!), than the probability relevant is the one for an entirly random process P(A|B) = 1/2. In this case the initial selection is meaningless since it does not affect the next possible options.

If only a dud can fall, then the probability of the groups is conserved (1/3 for one group and 2/3 for the other group) since the 1st selection restricts the following options. So it's a question whether you think the one that fell was due to divine intervention...

Bozon

eskimo2 · « **Reply #69 on:** November 28, 2004, 06:55:22 PM »

At this point we all would have augered; too much of a brain teaser to comprehend in so little time. If the mission really is this dangerous, then this was the best choice:

Quote

Originally posted by 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

eskimo

Blue Mako · « **Reply #70 on:** November 28, 2004, 08:09:52 PM »

Man this is the dumbest thread ever but I'll bite.

There is only one choice that the pilot needs to make. Choose between one of two bombs. Each have equal probability of being correct.

The fact that a third bomb fell off the aircraft earlier is absolutely irrelevant to the final choice between two bombs. Your set of probability has been reduced to one chance in two.

50:50

Equal chance.

Any other ways to say it?

In this scenario he is just going to guess left or right. Moving the lever does not change the odds at all.

Purzel · « **Reply #71 on:** November 29, 2004, 02:05:15 AM »

I think we cann agree that the described problem suggests being different from what it really is.

Arriving at the target, the pilot has to decide which bomb he drops. There are only two hanging from the plane, and he knows one is live, the other dud. What he does with the lever (how he flips the coin) is not important.

How he got there is also unimportant. If he had a plane with only 2 pylons an one was loaded with a training bomb he would end up at the same situation. With the same probabilities.

You can write a program doing that too, and you will see, the chance of selecting the right one (or wrong one) will be 50%.

The interesting part about what happened here is the the initial descriptioin seemed to be what I said last (the 2-bomb-problem). But it isnt. Thats why it is so surprising.

MANDO · « **Reply #72 on:** November 29, 2004, 02:22:23 AM »

ccvi, by doing so you are going to "ruin" the "good luck" of the pilot for the described problem. That luck is mainly based on the selection of the lost bomb based on two other random selections. If we look for a pure random combinations where the lost bomb is red, the real bomb is green and the lever marks a present bomb, then there is no more "good luck" initially present.

The original case indicates that the pilot see a red dot falling. If you repeat that case 1000 times, he will see the red dot 1000 times because it is part of the description. He may ask himself, ok, what if the ground crew painted the real bomb red by mistake? But what he see is red, not green.

dedalos · « **Reply #73 on:** November 29, 2004, 09:29:40 AM »

Quote

Originally posted by 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.
[/B]

You missed the three door example in the probability class

midnight Target · « **Reply #74 on:** November 29, 2004, 03:25:19 PM »

The left one is already selected... drop it now, then select and pickle off the right one. Duh!