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Everything posted by bonanova

  1. bonanova

    Spy story

    For all practical purposes. Here's why that makes sense. On your first fold, you're folding a single sheet On your second fold, 2 sheets On your third fold, 4 sheets ----- ----- On your seventh fold, it's 64 sheets. If you want to try for an eighth fold, it would be 128 sheets. Try folding pages 1-256 of a book [numbers on both sides remember].
  2. Alex wins. By a lot. Here's why:
  3. bonanova


    *slaps forehead* thanks ... !
  4. Absolutely correct. [1] If no cup is removed, a particular pair of remaining cups match colors 12 of 26 times. [2] If a red cup is removed, a particular pair of remaining cups match colors 7 of 15 times. [3] If a green cup is removed, a particular pair of remaining cups match colors 5 of 11 times. [4] If we remove a cup 26 times [say cup 1] a particular pair of remaining cups match colors 12 of 26 times. Choose your interpretation of UR's 2nd question and you have your answer. In the Monty Hall Problem you are able to take guidance from the added knowledge. You can deduce that you'll win 2/3 of the time if you swap and 1/3 if you don't. If you think UR's 2nd question is like the MH problem, then let's give it a try: You pick a cup, say it's Cup 1, and I tell you it's green. Based on your knowledge that Cup 1 was green, which of your 6 options are you guided to choose? Cups 2 and 3 Cups 2 and 4 Cups 2 and 5 Cups 3 and 4 Cups 3 and 5 Cups 4 and 5
  5. bonanova

    who's who??

    Assuming that two halves make a whole ...
  6. Right. When two elastic bodies collide, both energy and momentum are conserved. The fly was anything but elastic, becoming and remaining a simple grease spot, so in this case only momentum is conserved. Simply put, [Mass of fly] x [velocity change of fly] = [mass of train] x [velocity change of train] Let's say for argument's sake the train is ten times more massive than the fly. [That would be one big, m......f........ fly, by the way...] That gives you ... canceling mass of fly from both sides ... velocity change of fly = 10 x velocity change of train. the fly's velocity changed from +10 to -60 = 70 mph that means the train's velocity changed by 70/10 = 7 mph. The train continues moving forward, slowing from 60 mph to 53 mph. Even a fly weighing 1/10 the weight of the train won't stop it. Here's an interesting point. A detailed analysis of what happened to the fly reveals that it never actually came to rest! The first miniscule portion of the fly to contact the train instantaneously assumed the train's speed, producing a progressive collapse and deformation of the fly into a grease spot. The rear portions of the fly were still moving forward while its forward parts were traveling backwards. As a whole, the fly never had zero velocity, and thus technically never stopped.
  7. Being it's an island, that actually might be thought of as fortunate?
  8. bonanova


    Typo check: In hex, A-F represent 10-15. UR, I've never heard of base seven being used. Being curious [actually, I've been called worse, and with some justfication] what's base 7 used for?
  9. Not sure. Alex isn't answering his phone since the story broke.
  10. bonanova

    Spy story

    [softly] ... poor writersblock - first he omits the punctuation, then he omits to mention the steamroller.
  11. I think we are agreeing loudly on this point. The positions of the colors are random [unknown to us] and the combinations of so many reds and so many greens -- it doesn't matter that we used different numbers -- permit an enumeration of all possible, equally likely distributions. What's not 50-50 is the chance that any one particular cup is red or green. Count the number of times in my 26 equally likely color distributions that the first cup [for example] is red. It's 15. Not 13. -- or is green. It's 11. Not 13. Next, you see that in 7 of the 15 red-removed cases any 2 of the remaining 4 cups have the same color. In 5 of the 11 green-removed cases any 2 of the remaining 4 cups have the same color. If all you are saying is that 7/15 is different from 5/11, then the discussion is over and we can get some sleep. [Perhaps erroneously] I took your Case 2 question to mean in all cases when a cup is picked and removed from the group [not a particular case, like a red cup is picked first] does your chance of picking two matching colors differ from your chance of picking two matching colors when all 5 cups are there. My answer to that question is that the chances do not change. This follows from the observation you can make by inspecting the 26 equally likely distributions of 2 colors among 5 cups. Namely, that whatever two cups you inspect, 12/26 of the time they will be the same color. That statement is true whether any or all of the other three cups were removed initially or not. That is, there are no circumstances that force you to pick 2 particular cups that do not have a 12/26 chance of having the same color, simply because every pair of cups in the 26 equally likely cases has a 12/26 chance of a color match. Finally, it appears that if you eliminate the 5-red case [requiring at least 1 green] you will decrease the total distributions from 26 to 25, and the colors-will-match distributions from 12 to 11. That would change the answer for both Case 1 and Case 2 to 11/25 = 0.44.
  12. This is one for the over 65 crowd. I turn up my Miracle Ear hearing aid and say, "What?"
  13. bonanova

    kids puzzle

    I can think of at least one other common oochie
  14. Alex didn't realize that; so to be fair, he should get his 18% and you should get the other 6% -- about $670 a day ... that sound alright?
  15. In Case 2, you pick a cup -- and without knowledge this is a random pick -- and then eliminate that cup. Since there are more red than green cups [in the sample space I used, where 5 reds were possible], the odds of picking a red cup first are different from the odds of picking a green cup first, and the odds of getting a pair after picking a red cup differ from the odds of getting a pair after picking a green Not that that really matters, I just made a note of it. In the space of all possible outcomes for Case 2, [that includes 15 times a red cup is picked first and 11 times that a green is picked first] the probability of getting a pair after one cup is eliminated is the same as for Case 1. In either case, it's simply the odds that any two cups are the same color: 12/26 = .461 You could take away three cups at first then ask the odds that the remaining cups are a match. Same result.
  16. What Matt proposed was not that Alex had to get aces on the first two cuts, but only that two successive cuts be aces as he continued thru the deck. The first 30 cards could be non-aces, for example, followed by aces on the 32nd and 33rd cuts. I think that's what you calculated. And now I must confess I haven't done the calculation, which I will take care of, and then post my result. I think the calculation is made easier by considering the probability that a particular ace [hearts, say] is followed immediately by one of the other aces in a well-shuffled deck, then multiply that by four. [not followed or preceded; I think that leads to double counting.]
  17. bonanova

    kids puzzle

    I agree with one thousand. We don't say, for example, twenty and three for 23, nor one hundred and forty and seven for 147. Sorry, I should say, maybe some do say that; I don't.
  18. Hmmm...? Interesting. No more than 3 green means at least TWO red. Anyway, that aside, this gives 26 distributions of chips: 1 - 0 green, 5 red 5 - 1 green, 4 red 10 - 2 green, 3 red 10 - 3 green, 2 red - but stop here: no more than 3 green are present. Specifically, the distributions are Case 1: "In a row" suggests that perhaps all five are picked in sequence and you hope that two consecutive picks are the same color. [With only two picks, they are always in a row.] But the rest of the wording suggests you have only two picks. I will assume you get only two picks. Case 2: Here you pick one cup and are told what color it is. Then you must pick two other cups. Again since all outcomes are equally likely, we can assume we pick and are told the color of the first cup, and then see in how many of the 26 outcomes cups 2 and 3 [or any other pair of cups] are the same color. We already know that two specific cups have a 12/26 chance of being the same color. Probability again is 12/26 = 0.461
  19. Sorry, my first answer was that lacking fingers he could not open his canteen. So I'm wondering how he could die of a full bladder. Good one. Add his "male"ness to the puzzle and you have a winner.
  20. It was also in Die Hard with a Vengeance.
  21. The words are special because no other English word rhymes with them.
  22. "I bet you can't cut two aces out of a shuffled deck," said Davey to Alex down at Morty's last night. Davey was still sore that Alex had flipped 10 Tails on Monday. "I'm sure I wouldn't take that bet," Alex replied, "I happen to know that my chances are [4/52]*[3/51], and I'll tell ya right now that my dear momma didn't raise no fools!" Matt the mathematician overheard the two boys talkin' and pushed Alex a little farther: "But what if you could continue cuttin' the deck until the cards were all gone? Do you know what yer odds would be to cut consecutive aces, then?" And he sipped his brew while Alex thought about it. After a moment, Alex replied, "Give me a dollar against my penny and yer on!" Was Alex correct about his chances on the first bet? What odds should he have demand to make the second bet fair? Edited for clarity.
  23. Great analysis! But ... Take it easy on Matt ... it took me 5 tries to develop his character.
  24. We assume the troll knows which road is safe. Ask: "Which road would you have told me yesterday was safe?" He'll either lie about telling you the truth or tell the truth about lying to you. Either way, he'll point to the dangerous road. Then take the other road.
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