BobbyGo
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Posts posted by BobbyGo
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Spoiler
If we map out all the possible 'day of the week' combinations on 4 separate grids of days, we get the following totals where at least one boy is born on a Tuesday:
B/G = 7
G/B = 7
B/B = 13
G/G = 0
Overall, there are a grand total of 27 possible outcomes. The probability of both of the children being boys is then simply 13/27.
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Two guesses (both of which are more fun than logical).
"In the box" guess:
SpoilerChuck is the liar.
Each brother references 3 twice in their statements except Chuck.
- Al: 3 numbers, cubes
- Bert: 3 sides, trisected
- Chuck: 3 trees
- Dick: 3 miles, 3 (living) brothers
Plus, Chuck is the third brother. Kind of poetic.
"Out of the box" guess:
SpoilerEddie is the liar.
Dick ran 3 miles and saw Eddie. When he got back to the cabin, he knew that the body had to be a fake and Eddie was still alive.
(I guess this also makes Dick a liar for saying he only has 3 living brothers, but that still leaves 3 absolute truth-tellers per the conditions of this problem.)
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Spoiler
Assuming that any positive number is valid, I'd buy one token and pick:
0.00000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001
as my number.
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Spoiler
Either, like Kuldeep Sarma stated, they start on opposite sides of the river, or they start on the same side and cross at two different times. (e.g. Person 1 crosses the river, does what they want to do, and then crosses the river back. Then, Person 2 does the same. At this point, they've both crossed the river... they just didn't stay there.)
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Spoiler
Is it some strange hotel where the room numbers are posted inside the rooms but not in the hallway/outside? If the friend is told to meet the man in room ##, he would only know if he was ringing room ## after his friend answered the door.
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Spoiler
Jack be nimble, Jack be quick. Jack wasn't wearing a seat belt and flew through the windshield.
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32
# of Segments -> X
# of Points on Circle -> Z
X = 2^(Z-1)
Unless you are only counting the segments that are made up of straight sides and not the segments that utilize part of the circle.
In that case: 26
X = 2^(Z-1) - Z
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Should we assume that x and y are always unique values?
If x=y, then there are an infinite number of solutions.
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Here is a slightly reversed approach (using geometry):
Instead of finding out how many soldiers it would take to find the poison amongst x number of bottles, here we find out how many bottles can be investigated using x number of soldiers.
Take the number of soldiers used for testing and plot that many non-linear points onto a plane. The total number of bottles that can be tested is equal to the sum of one plus every possible point, line, and shape that the plotted points are capable of creating.
For example, 3 soldiers could find the poisonous bottle from a total of 8 bottles. (3 points + 3 line segments + 1 triangle + 1)
4 soldiers could find the poison from 16 bottles. (4 points + 6 line segments + 4 triangles + 1 square + 1)
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On down to 9 soldiers testing 512 bottles. (9 points + 36 lines + 84 triangles + 126 squares + 126 pentagons + 84 hexagons + 36 heptagons + 9 octagons + 1 nonagon + 1).
And 10 soldiers testing 1024 bottles. (10 points + 45 lines + 120 triangles + 210 squares + 252 pentagons + 210 hexagons + 120 heptagons + 45 octagons + 10 nonagons + 1 decagon + 1)
So the least number of soldiers you would need to test 1000 bottles is 10.
Although...
If I were the King, I would want to minimize the number of deaths (the more soldiers that die, the harder it is to defend your kingdom). With that in mind, I would have 999 soldiers each drink from a different bottle. Maximum casualties = 1
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The smallest US coin ever minted was the half cent. So you could have 199 of these coins and still not have even change for a dollar.
Although, I wouldn't recommend it
. Coin collectors will pay from $30 to $18,000 per coin depending on the type, year, and condition. -
If you're strictly asking about dice equalities (because draws technically go to the defender in Risk) and if all dice comparisons need to equal in a given roll to count as a draw, then:
3v2 = 4.8997% (381 / 7776)
3v1 = 16.6667% (216 / 1296)
2v2 = 5.0926% (66 / 1296)
2v1 = 16.6667% (36 / 216)
1v1 = 16.6667% (6 / 36)
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Attempted it a second time. Tried to put it in the spoiler, but it wouldn't go there and now I can't move it? Not my day I guess.
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Behold the limit of my artistic abilities:
Imagine that the red lines are ladders and the black lines are tubes. The snake can get to every nest. You can get to every nest except the top middle.
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Does this have to do with the deck being brought back into its original order (assuming perfect shuffling) after so many shuffles? If you were skilled enough to "follow" a card and you knew where it started, you would have a pretty good chance of finding it.
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For all cases larger than 2 moves, the total number of unique squares that a knight can end up on after n moves is equal to 7n
2 + 4n + 1. (This equation's results for cases 1 and 2 are four larger than the actual number of squares the knight can end up at. Not sure how to balance.)The bishop problem is a bit of fun. If the bishop can move to an infinite number of spaces on the first move (infinity times 4, if you want to count all diagonals separately), then on the second move it can reach an infinite number of spaces for each of the infinite spaces from the first move. Additionally, after the second move, no additional spaces can be reached; all the possible spaces that the bishop can reach have already been counted.
So after the first move, on an infinite chess board, a bishop can uniquely reach infinity2 spaces... all while only using half of the infinite board!
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Or maybe instead of 9/16 it would be 37/64?
(Was too late to change in post.)
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Since everybody must hook up with every individual of the opposite sex, there needs to be at least 5 "rounds". Since there are 5 girls and 4 guys, each round one girl will not participate. If you are able to schedule the round that you sit out, then you would want to do so on either the 4th or 5th round. Use protection on your last active round (either the 5th or the 4th) to maximize the likelihood of not catching the std.
Using the above strategy, you should only have a 9/16 chance of catching the std.
I would think that anyone who participates in the orgy without protection would have a 100% chance of catching the std.
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Assuming there are no obstacles in the way (experimental equipment, life-support gadgets, etc), the first astronaut could just let go of the wrench without throwing and it should remain in the same position as both astronauts rotate into each other's position.
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For your third set (the one beginning in 2,0), I count 30 times instead of 60. Also, I had 300 for the last one because any digit 0-9 could be in the rightmost seconds column during any of the valid hELLO times (and should count as separate times).
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2 = 4 - 2
(or something similar)
Inverting the 2 on the left and removing the 4 makes it
-2 = -2
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5?
(10 - 6) + (10 - 9)
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60 divided by the first number equals the second number.
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Custard
Triangle probability
in New Logic/Math Puzzles
Posted · Edited by BobbyGo
A couple of questions:
Is a side of the triangle determined by a single die roll? Or do you sum up three rolls per side?
Are non-possible triangles considered valid solutions? e.g. A roll of 1,1,6 might appear isosceles, but would not be a valid triangle.