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Pickett

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Posts posted by Pickett


  1. Haha, that's kind of fun to wrap your head around...

    Spoiler

    But really, it should be fairly simple, I think:

    Each option has a 25% chance of being randomly picked. That is the correct answer: 25%.

    BUT there are two options where "25%" is the answer. This means that the answer that you pick has a 50% chance of being "25%", which is STILL the correct answer regardless of the probability of picking an answer that HAS that value. So the probability that you are correct is 50% which is answer 2...

     


  2. well...

    Spoiler

    O=0, M=1, Y=2, E=5, N=6, D=7, R=8, S=9

    Now, I'm not quite sure what you are asking in your question...as substituting in, you would get this:

    065 065+2 065+4 065+17 =?

    I don't know what you want us to do with those...I guess you could get

    65 67 69 82

    Are you wanting those multiplied (24640590)? added (283)? concatenated (65676982)? re-substituted (NE ND NS RY)?

     


  3. Depends...

    Spoiler

    If you go with purely "expected damage value" as your basis for "best", then it would be

    1. Shotgun
    2. Pistol
    3. Railgun / Grenade
    4. Rifle
    5. Knife

    However, I personally would rather have something that guarantees SOME damage over something that has a 50% chance of doing 0 damage (so Railgun > Grenade by that logic). Some people, though, would probably prefer something that CAN do the most damage (Grenade > Railgun)

    Either way, the way I see it: the shotgun is the most consistent and highest average damage weapon...so it should be #1. The pistol, railgun, and rifle all do the same MAX damage, but have different min damages...so I would say Pistol, Rail gun, and Rifle in that order...Which gives us:

    1. Shotgun
    2. Pistol
    3. Railgun
    4. Rifle

    That leaves the Knife and the Grenade and where they should go...while the grenade will do 0 damage 50% of the time, the knife will do the lowest expected value (4.5)...and while the knife CAN do 0 damage some of the time, it won't do 0 damage nearly as often as the grenade. Soooo...if I had to choose, I'd put the Grenade then Knife...so my final order would be:

    1. Shotgun
    2. Pistol
    3. Railgun
    4. Rifle
    5. Grenade
    6. Knife

    But again, that's my personal preference based on my definition of "best"...and depending on your playing style, risk level, and other unknown factors (does grenade hit multiple targets??), it may vary.

     

    • Like 1

  4. Spoiler

    Alright, this time, I'm getting the smallest number of you who drank all three is 10 (the largest is 13).

    Basically it still turns out that there are 0 people that didn't drink anything, so that didn't impact my original answer...but by maximizing the number of people that drank beer and water, I found a different answer for people that drank all three. Here's a breakdown of all 8 groups for reference:

    • NOTHING: 0
    • WINE ONLY: 3
    • BEER ONLY: 0
    • WATER ONLY: 9
    • WINE/BEER: 15
    • WINE/WATER: 11
    • BEER/WATER: 6
    • ALL: 10

    Most of my above approach still worked, except there was a "+ N" (non-drinkers) added to a couple of my equations, which led to a few additional checks by trial and error, but it wasn't too bad.

    Hopefully that matches your answer...or I should maybe just stick to my day job...

     


  5. 1 hour ago, bonanova said:

    @Pickett It may not impact your solution, but the OP did not intend to say that all of us drank something.

    "Class of drinkers" was not meant to preclude anyone from drinking nothing.
    Meaning there are eight "classes" of drinkers.

    Sorry for that. I have this thing about insisting that zero is a number^_^ rather than a denial.

    Also, I get a different answer. I'll check my analysis against yours to see why.

    I'm guessing it's because I missed this piece of the assumption: "who drank beer and water but not wine were as many as possible".

    I'll give it another go with the non-drinkers included and that piece and see what I come up with.


  6. Approach

    Spoiler

    Let's start by identifying the 7 different groups of people involved and label them

    • Total people that had only Wine = W
    • Total people that had only Beer = B
    • Total people that had only Water = H
    • Total people that had Wine and Beer = WB
    • Total people that had Wine and Water = WH
    • Total people that had Beer and Water = BH
    • Total people that had Wine, Beer, and Water = WBH

    Now, given that, we can write our system of equations based on the problem statement:

    EQUATION 1 (total cost of all drinks):
    5*(WINE) + 2*(BEER) + 1*(WATER) = 293
    5*(W+WB+WH+WBH) + 2*(B+WB+BH+WBH) + (H+WH+BH+WBH) = 293
    Simplifies to:
    5W + 7WB + 6WH + 8WBH + 2B + 3BH + H = 293

    EQUATION 2 (total glasses used):
    W + B + H + 2WB + 2WH + 2BH + 3WBH = 106

    EQUATION 3 (total non-water drinkers):
    W + B + WB = 18

    EQUATION 4 (total wine drinkers):
    W + WB + WH + WBH = 39

    EQUATION 5 (total non-alcohol drinkers):
    H = 9

    EQUATION 6 (ending assumption that wine-only was half the number of beer/water drinkers):
    BH = 2W

     

     

    So, now that we have our 6 equations, we can start doing our algebra:

    Plug equation 5 into equation 1 and 2 to simplify slightly:
    1) 5W + 7WB + 6WH + 8WBH + 2B + 3BH = 284
    2)  W + 2WB + 2WH + 3WBH +  B + 2BH =  97

    Plug equation 6 into equations 1 and 2:
    1) 11W + 2B + 7WB + 6WH + 8WBH = 284
    2)  5W +  B + 2WB + 2WH + 3WBH =  97

    Re-order equation 1 and 2:
    1) 2(W+B) + 9W + 7WB + 6WH + 8WBH = 284
    2)  (W+B) + 4W + 2WB + 2WH + 3WBH =  97

    Plug equation 3 (W+B = 18-WB) into equations 1 and 2:
    1) 9W + 5WB + 6WH + 8WBH = 248
    2) 4W +  WB + 2WH + 3WBH =  79

    Plug equation 4 (W=39-WB-WH-WBH) into equation 1 and 2:
    1) 4WB + 3WH + WBH = 103
    2) 3WB + 2WH + WBH =  77

    Solve for one of the remaining variables (WB):
    WB = (103 - 3WH - WBH) / 4

    Plug into equation 2:
    WH = 1 + WBH

    Alright, so we now have 1 equation with 2 unknowns, where one of the unknowns is what we're trying to determine! So, we can list out all of the possible values of WBH and WH...which happen to be:

    WBH = 1, WH = 2
    WBH = 2, WH = 3
    ...
    WBH = 18, WH = 19

    Anything above WBH = 18 will cause equation 4 to not work...so we can stop there with those 18 possibilities. Now it's just a matter of substituting in each of those values into our equations to find the lowest one that works. I'll spare the gory details, but WBH < 11 results in negative values for other variables at some point...so let's just show trying WBH = 11:

    WBH = 11, WH = 12:

    Substitute those values into our 4 equations and simplify:

    1.     5W + 2B + 7WB + 3BH = 124
    2.      W +  B + 2WB + 2BH = 40
    3.      W +  B +  WB       = 18
    4.      W +       WB +     = 16

    Now substitute 4 into 3 to solve for B:

    B = 2...so we know B = 2 and W = 16-WB

    Plug those two into equations 1 and 2 and simplify

    1.     2WB + 3BH = 40
    2.      WB + 2BH = 22

    Solve (WB = 22-2BH)
    You find BH = 4...so given BH = 4, WBH = 11, WH = 12, B = 2, and H = 9, we can find the other two values of WB = 14 and W = 2...at which point we can verify all of the original equations hold true. so therefore, WBH (total number of people who drank all three drinks) is at least 11 given the assumptions we stated.


  7. FWIW

    Spoiler

    There are only 83 total...and they are all under 100,000,000...

    Spoiler

    Also going with the "tree" approach that Molly Mae mentioned:

    2
    23
    233
    2333
    23333
    23339
    2339
    23399
    233993
    2339933
    23399339
    239
    2393
    2399
    23993
    239933
    2399333
    29
    293
    2939
    29399
    293999
    2939999
    29399999
    3
    31
    311
    3119
    31193
    313
    3137
    31379
    317
    37
    373
    3733
    37337
    373379
    3733799
    37337999
    37339
    373393
    3739
    37397
    379
    3793
    3797
    5
    53
    59
    593
    5939
    59393
    593933
    5939333
    59393339
    59399
    593993
    599
    7
    71
    719
    7193
    71933
    719333
    73
    733
    7331
    7333
    73331
    739
    7393
    73939
    739391
    7393913
    73939133
    739393
    7393931
    7393933
    739397
    739399
    79
    797

     

     


  8. Spoiler

    AL: Since he removes the lowest value each time, he ends up with all of the values from N+1 to 2*N in his bag. So the total can be calculated as (3N2N) / 2

    BERT: He's removing an even each time...and since he only gets one even every round, he's left with only the ODD numbers left. So the total can be calculated as simply N2

    CHARLIE: He's obviously a little bit trickier...I will spare you the ugly details of how I arrived at this, but basically you are calculating the expected value by multiplying the number by the probability that that number will be present at midnight. This calculation can be ultimately simplified down to a sum: ∑ x=1 -> N  (4x2-x) / (N+1)

    So...Here's a quick example of their payouts for increasing values of N:

    N        | AL                  |  BERT                | CHARLIE (expected)        
    1        |                   2 |                    1 |                   1.5
    4        |                  26 |                   16 |                  22
    10       |                 155 |                  100 |                 135
    250      |              93,875 |               62,500 |              83,375
    1000     |           1,500,500 |            1,000,000 |           1,333,500
    75998    |       8,663,582,005 |        5,775,696,004 |       7,700,940,671.895
    250250   |      93,937,718,875 |       62,625,062,500 |      83,500,125,041.563
    5000000  |  37,500,002,500,000 |   25,000,000,000,000 |  33,333,334,166,666.794
    10000000 | 150,000,005,000,000 |  100,000,000,000,000 | 133,333,335,000,000

     


  9. 8 hours ago, ThunderCloud said:

    I think...

      Hide contents

    Let a diagonal road connect each of the four towns to the endpoints of a vertical road that runs through the center of the square. Call the length of the vertical road x. Then the total length of all of the roads, geometrically, is

     

    T = x + 2√(x2 -2x +2)

     

    Then, dT/dx = 1 + (2x -2) / √(x2 -2x +2), which has a zero at x = 1 - (1/√3). The total road length, by the above formula, is then ~2.73205.

    Special thanks to Cygnet for suggesting there might be a minimum hidden in such a configuration. ^_^

    I assumed my answer was along the right lines, but I was too lazy to actually find the true minimum with it...thanks for doing the dirty work!


  10. My brain isn't exactly firing on all cylinders today, but...

    Spoiler

    So there are infinite number of triangles available, divided into 3 categories (or sets), each with an infinite number of elements

    1. An infinite number of right triangles
    2. An infinite number of acute triangles
    3. An infinite number of obtuse triangles

    As a quick example of that, imagine a right triangle with points at (0,0), (0,1), and (1, 0)...you can just move the point on the y-axis up and down however you want to stretch the triangle infinitely.

    So then, it depends on how you define "what fraction".

    • If you look at it from a statistical perspective and ask "if you draw 3 completely random, non-colinear points in the x-y plane, what are the odds of that triangle being obtuse?" I'd have to say it's 1/2 for the following reason:
      • While there are an infinite number of right triangles, it is statistically a zero percent chance of the 3 random points creating one. Which leaves 2 possible outcomes, each with equal probability.
    • However, if you are asking for total number of obtuse triangles divided by total number of triangles, I'd say it's indeterminate.
      • It's similar to asking "what fraction of points are inside of a circle on a plane?"
      • Technically, since there are an infinite number of points inside the circle AND outside of the circle, you could map every single point on the inside of a circle to some point outside of the circle and vice versa...however, conceptually, it's obvious that there should be more points outside of the circle...but that defies the idea that there's an infinite number inside...

    As I said, my brain is pretty scattered today, and I'm assuming that's not the answer you were looking for...

     


  11. Spoiler
    1. 43
    2. 63
    3. 135
    4. 175
    5. 518
    6. 598
    7. 1306
    8. 1676 <-- Note: v and x have the same value...I'm not sure if that's allowed by the OP.
    9. 2427 <-- Note y and @ have the same value...I'm not sure if that's allowed by the OP.

    BONUS: Not finding a solution...

          abcdefgcc = a4 + b3 + c8 + d5 + e7 + f9 + g0 + c8 + c8  OR

          abcdefgcc = a4 + b3 + 3c8 + d5 + e7 + f9 + 1

    Even with allowing duplicate values for the variables and allowing leading zeros in the number on the left, I'm not finding a solution...

     


  12. Spoiler
    1. He took note of the time on his clock when he left. We'll call this value TS '
    2. He walked x amount of time to his friends house.
    3. He took note of the time there. We'll call this value TF '
    4. As he leaves he notes the time on his friend's clock. We'll call this value TF '' 
    5. He knows what time he arrived and what time he left, so he knows he spent y amount of time at his friend's house: 
      y = (TF '' TF ')
    6. He then walks x amount of time back home and looks at his clock. This is TS ''. He knows this value would be: 
      TS '' = TS ' + y + 2x
    7. So with that information, he can determine x (The amount of time to walk home from his friend's house)  
      x = (((TS'' - TS') - (TF'' - TF')) / 2)
    8. He knows the CORRECT current time would be: 
      TF '' + x
    9. Therefore, he has all the info he needs to set the correct time:
      TF'' + (((TS'' - TS') - (TF'' - TF')) / 2)

    QUICK EXAMPLE:

    • Let's say his clock showed 12:00 when he left (TS ' = 12:00)
    • His friend's clock showed the actual current time of 17:00 when he arrived. (TF ' = 17:00)
    • His friend's clock showed the actual current time of 20:00 when he left. (TF '' = 20:00) therefore y = 3 hours
    • His clock shows 17:00 when he arrives home (TS '' = 17:00)
    • He knows he should set his clock to:
      • 20:00 + (((17:00 - 12:00) - 3 hours) / 2)
      • 20:00 + ((5 hours - 3 hours) / 2)
      • 20:00 + (2 hours / 2)
      • 21:00

     


  13. I've not heard this one...but I'll take a stab:

    Spoiler

    Let X be the amount Bill originally bought the scooter for.

    Bill's total profit (or loss) = $110 - X

    Example: If he originally bought it for $100, then:

    1. Bill buys for $100 (total profit = -$100)
    2. Bill sells to Tom for $100 (total profit = $0)
    3. Tom sells to Bill for $80 (total profit = -$80)
    4. Bill sells to Hermon for $90 (total profit = $10)

    So unless we know what he originally bought the scooter for, we can't really calculate his total profit.

     


  14. I guess I took this question to be something different. Rather than just "what's the smallest positive value you can think of?" for which I would go with something like: 

    1-(3↑↑↑↑...↑↑3) 
    or in other words: 1 over Graham's number (or simply "G")...but why stop there? why not go with
    1/G↑↑↑↑↑↑↑...↑↑↑↑↑↑↑↑G

    That interpretation of the OP isn't as interesting or fun because it just gets ridiculous... The way I took the original question was pick the smallest positive INTEGER that no one else picks...what is your strategy when up against 9 others doing the same (all allowed to pick however many tokens as they want)?

    So, if that's the case, it becomes a much more challenging problem. Obviously it doesn't make sense to purchase more than 9 tokens, as with 10 the BEST you could hope for is breaking even...I would probably purchase the following numbers (and my rationale next to them):

    • 1 - I'm assuming I will lose this $10, but maybe everyone else will think someone else picks this and therefore no one else does!
    • 2 - Might as well try this one as well, but again, most likely will lose this $10 as well...
    • 7...ish - Some relatively low number that hopefully no one else picks...I would assume I'd lose this.
    • 10 - Let's assume everyone buys their 9 tokens and picks 1-9...10 would be the first number that no one would pick...
    • 16...ish - Again, some pretty low number, that's pretty much just a shot in the dark and you hope no one else picks it. 
    • 42 - If by some crazy chance everyone buys 9 tokens and all are duplicated, this is the lowest possible number that wouldn't be duplicated...plus isn't it really the answer to everything?

    At this point, I've already bet $60 in hopes of winning $100...and really, the odds are still not in my favor...so to me, it's not worth the risk and so my ultimate strategy would be to not play and "break even" :c)


  15. Spoiler

    679: 679 --> 378 --> 168 --> 48 --> 32 --> 6

    Then for what it's worth, here are the minimums with persistence 6-9 (notice how they all build on each other):

    • 6788: 6788 --> 2688 --> 768 --> 336 --> 54 --> 20 --> 0
    • 68889: 68889 --> 27648 --> 2688 --> 768 --> 336 --> 54 --> 20 --> 0
    • 2677889: 2677889 --> 338688 --> 27648 --> 2688 --> 768 --> 336 --> 54 --> 20 --> 0
    • 26888999: 26888999 --> 4478976 --> 338688 --> 27648 --> 2688 --> 768 --> 336 --> 54 --> 20 --> 0

     


  16. Spoiler

    Agree with 681...Next up would be:

    • September: 9919
    • October: 71015
    • November: 81114
    • December: 8124
    Spoiler

    The "code" is simply:

    • {NUMBER OF LETTERS IN NAME}
    • {ORDINAL NUMBER OF MONTH IN YEAR}
    • {ORDINAL NUMBER OF FIRST LETTER IN ALPHABET}

    For example, for "January", it is:

    • 7 letters long
    • First month in the year (1)
    • begins with "J" which is the 10th letter of the alphabet

    So January = 7 1 10

    Going with that, August would be 6 8 1

     

     

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