Prof. Templeton
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Everything posted by Prof. Templeton
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After following this thread for some time, the answer is obviously
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Do you have to put an operation between every digit? Otherwise you would need at most 4 digits.
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I'm only referring to B's viewpoint. 2 shots from A are worse than 1 shot from C. If B shoots C first A gets 2 shots, but if B shoots A first C get 1 shot. Provided B is still around. I see the error in my first post so all this is moot. Although the "answer" was correct if B shot A first he would still have weathered 1 shot from A and 1 shot from C before he could win, which is worse than 2 shots from A. So my logic was incorrect.
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Probability of hitting with either of his 30% chances. Isn't the probability that either occurs P(A1 ∪ A2) = P(A1) + P(A2) - P(A1) x P(A2) Where A1 is the first shot and A2 is the second? It's been a long time.
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Maybe I'm missing something. If A shoots at B and misses and B shoots at C, than A shoots again. If he hits 30% of the time wouldn't his chance of hitting B after 2 shots be .3+.3 - (.3*.3) or 51%. Greater then C's one 50% chance( if nobody takes out B he only gets one chance because he goes last). But if A intentionally shoots away, C's 50% would be greater than his remaining 30% chance(and he only gets one more chance if nobody takes out B). A and C would only get into a shootout if A were to shoot at B and score a hit on his first shot. Otherwise B will take out one of them.
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How so? You already said this is the correct solution, so why?
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You are right.
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A should shoot at the ground. If A shoots at B and misses, B would shoot at A instead of C, because A would get 2 30% chances and C would get 1 50% chance. So A has a 30% chance of winning, B has a 70% chance and C has no chance.
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This is a good example of "Bertrand's Box Paradox", where a person's intuition may tell them that they are picking cards so the probablity is 50/50, but you are picking card faces. There are 3 white faces. Two of the white faces are on 1 card. The chance of getting 1 of those 2 is 2/3, therefore the chance of flipping the card over to find another white face is 2/3. If you have drawn a white faced card, it is twice as likely that the face belongs to the all white card as opposed to the mixed card.
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Regardless of what happens before, when you get to the last decision you have 2/3 chance to win if you switch.
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The probability of seeing a White face is 3/4, and the probability of drawing the White card is 1/2. Therfore
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The first person in the room would be 365/365 and the second person would be 364/365 so the answer The first time I saw the answer it seemed incredible that so few people were needed to get at least a 50/50 outcome
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This is VERY close to the answer I have. It doesn't seem possible does it?
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Yes and yes.
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Nope. Reality and the answer are less. Nope And as a note, we'll disregard leap years
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There has been much discussion on this forum about some suprising probability questions, but I haven't seen this one yet. How many people do we need in room, so that there is at least a 50:50 chance of two people sharing the same birthday( day and month, not year ).
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I care about my partner, but more about myself, which is why I become indifferent about confessing when his jail time equals twice mine.
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Cliffs or crocodiles.
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This is the "accepted" answer.
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It was originally written as WillyWu and I missed one when changing it for this post.
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The wind is blowing the fire toward him
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BillyBob is chilling out on a heavily forested island that's very narrow: it's a narrow strip of land that's five miles long. Let's label one end of the Island East, and the other end West. A fire has started at the East end, and the fire is moving toward the West end at the rate of 2 mph. At the same time, there's a 4 mph wind blowing in the direction from East to West. What can BillyBob do to save himself from burning to death?! Assume that BillyBob can't swim and there are no boats, jetcopters, teleportation devices, etc.. (if he does nothing, willywu will be toast after at most 2.5 hours, since 5 miles / 2 mph = 2.5 hours). Let's also assume that he would not chose to off himself in some other fashion.
