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Posts posted by Molly Mae


A start:
Less than 99%. My first instinct is that Al only has a 1% chance of randomly choosing his assigned seat (and that's true). But Al doesn't have to move if the current person displacing another passenger finds their assigned seat empty.
My intuition now says that the chance Al has to move is actually rather low (probably below 5%). Basically, Al has to be in either (a) his assigned seat, or (b) a closed loop of passengers who have each others' seats. I can't even think of how to go about solving that mathematically.

10 hours ago, flamebirde said:I hate to bring this one back up again, but what's the rationale behind #3?
It's just a letter substitution. 1 = A, 2 = B, 3 = C, etc.

On 3/20/2018 at 10:05 PM, bonanova said:For the purposes of this puzzle, consider our old friend Albert to have the shape of a rectangular paralellepiped (when he was born the doctor remarked to his mother, I don't explain them, ma'am I just deliver them) just meaning a solid having (six) rectangular sides. At a recent physical exam, Albert was found to be 2 meters tall, 1 meter wide and .20 meters thick (front to back.) He maintains his geometric rectitude by never leaning forward when he walks or runs.
So anyway, Albert, alas, has found himself caught in a rainstorm that has 1000 raindrops / cubic meter that are falling at a constant speed of 10 meters / second, and he is 100 meters from his house.
Just how fast should Albert run to his house so as to encounter as few raindrops as possible?
This is a thought exercise I've considered in the past. The answer that I've always come up with is that it doesn't matter when disregarding thickness. If the raindrops are constant and fill any cubic meter equally, he'll encounter the same number of raindrops on his "face" regardless of speed.
The .20m thickness, though, needs to be considered. We know that when he moves at 0m/s, he will encounter some amount of rain along that thickness and he will have made no progress. So to minimize the number of raindrops that lands on top of him, we just increase his speed infinitely. We also know that this can cause severe issues, however.
If you'd like some math to support this argument:
...you should probably ask someone else.


Named for the first wherever you go
Hundreds of cycles in column and row.
Grouped into teams oft coloured the same.
Hardly one in each will ever earn fame.
The race will begin when ball touches ground
Millions of cheers can be heard all around.
You expect it to start when northern half cools,
But location may often vary the rules.
Too fast or too slow, not so much a race,
As all are required to maintain one pace. 
6 hours ago, BMAD said:There is a machine with 20 pieces of candy. Five of those candies are butterscotch. If you put in a 25 cents, one candy is provided at random. If you put in 75 cents, two candies are dropped at random but you may give the machine back one candy in exchange for a 25 cents. And if you put in $1.50 you receive 5 pieces of candy at random but are guaranteed at least one butterscotch.
How much should I expect to spend to get all of the butterscotch?
Clarification question:
If I put in 75p, can I return a candy I received from a previous purchase for the 25p rebate?

You could represent the bazillion dollar coin as a binary number using the first 10 coins. That covers the 2
^{10} possibilities. You can make it more efficient by using the 11th bit to define whether heads is 0 or 1. This saves flips in the event that you need to flip more than half of the coins to make the correct binary number. Depending on the number of coins, you could extend or reduce the amount of bits you use by using the fewest bits that can represent the highest possible number. 
14 minutes ago, harey said:Yeah, I was assuming no false paths. Allowing for false paths makes this difficult problem yet more difficult.

I'm assuming there are no false paths, which would likely make this challenge impossible for n>4
Case: n=1
Solution: We did it!Case: n=2
Solution: 3 moves (RDR or DRD), We did it!Case: n=3
Solution: RRDDRRDDRD (alternately: DDRRDDRRDR), We did it!At n>3, we get to the tricky part where we have to start incorporating ups and lefts, so the instruction set will definitely look different. We have to ask ourselves how long this list of instructions is going to be for n>3. That's a question I'm equipped to answer.
We can phrase it this way: For n=5, the instruction set will be ______ long.
Answer:
Hella

1 hour ago, Quantum.Mechanic said:I must be thick today, I can't follow the given solution. Can someone paste a little bit more hand holding? Like the first 5 values in the sequence?
EDIT: For clarity, I use "number" to reference an individual digit. I use "sequence" to reference the string of numbers.
EDIT2: Added spoiler tag.Sorry, I'm pretty bad at explaining things. You're basically building two sequences: n and 2n. I arbitrarily chose 1 as the first number of the first sequence and we don't know the length of the sequence, so it looks like this:
First: 1????...???
Second: ?????...???Since 1 is the first number of the first sequence and the second sequence is double the first sequence, we can update our second sequence.
First: 1????...???
Second: 2????...???From the OP, we are promoting the least significant number of the first sequence to the most prominent position. So we now know that the first sequence ends with a 2. Let's update that.
First: 1????...??2
Second: 2????...???Again, since our second sequence is double the first, we know that the second sequence ends in a 4.
First: 1????...??2
Second: 2????...??4Remember that the second sequence is the first sequence except that the last number is moved to the front. So the second to last number of the first sequence is a 4.
First: 1????...?42
Second: 2????...??4Four doubled is 8.
First: 1????...?42
Second: 2????...?84The 8 moves up to the first sequence:
First: 1????...842
Second: 2????...?84Eight doubled is 16.
First: 1????...842
Second: 2????...684You can keep doing this until your second sequence can begin with a 2 without having to carry over a number. It's a neat puzzle where you build two sequences simultaneously using new information from one sequence to build the next part of the other sequence.
I hope that clears it up.
 1

Still looking for better, but I think I've done it with 7.
3 4 5 6 7 8 9
4 5 6 7 8 almost works, but I can't counterbalance 5+4.

Initial observation: it definitely needs more than 4. I'll see if I can find a solution for 5.

Hmmm...
Ketamine?
EDIT: Amphetamine?

10 minutes ago, bonanova said:@Molly Mae gets the coveted bonanova gold star for finding an algorithm and getting the answer in less than one day.
Signature updated!

Ouch
I wasn't expecting a number this long and I almost gave up thinking it wouldn't work. After hunting for a few random 2 and 3digit numbers, I decided to try it mathematically. I started with a number beginning with the number 1, so I know the last number must be 2, which means the last number of the rearranged number must be 4. Then, since the number would have to be doubled, I multiplied the 4 by 2, so the rearranged number must end in ...84. I kept this up until I arrived at 105,263,157,894,736,842. Shifting the 2 to the front returns 210,526,315,789,473,684, which is double the original number.
I can't guarantee it's the smallest number with this property, but it's the first one that begins with a 1.
EDIT: Another note, I use Google's calculator function to verify that the second number is indeed double the first and discovered that these are called parasitic numbers. I think that's a fun name.

interestingly:
The fact that Al pairs off with the winner is a red herring. They each have a 1/3 chance of winning, since the distribution of votes is decidedly equal.As an example, take 3 voters (the first odd number divisible by 3). Their preference will be expressed as XY, where X is their first choice and Y is their second. There are two possibilities:
1. AB or AC
2. BC or BA
3. CA or CBInterestingly, I originally ran with 9 voters and figured the odds would carry over to a smaller (easier) number. That isn't the case. I'd like to try with 15 voters. The odds definitely change based on the number of voters, though. With 3 voters, Al always wins. With 9 voters, I have more than one case where Al doesn't win.

Another factor to possibly consider...
If enemies have, for example, 7 HP. The pistol (and the rail gun and even the rifle) would clearly be better than the shotgun. If they had 6 HP, the shotgun is clearly better than all of them. If their HP is high enough, these considerations aren't weighted as heavily.
 1

8 hours ago, bonanova said:With a strategy of always betting when the count is high, it doesn't matter. Foiled again by practical application? I can see that for any variation of that strategy, I'll gain cases that win but I'll lose an equal number that used to win.
N=2 is 50%
RRBB  Lose
RBBR  Win
RBRB  Lose
BBRR  Lose
BRRB  Win
BRBR  WinN=3 is 50%
RRRBBB  Lose
RRBRBB  Lose
RRBBBR  Win
RRBBRB  Lose
RBRRBB  Lose
RBRBRB  Lose
RBRBBR  Win
BRRRBB  Win
BRRBBR  Win
BRRBRB  Win
BBRRRB  Lose
BBRRBR  Lose
BBRBRR  Lose
BRBRRB  Win
BRBRBR  Win
BRBBRR  Win
RBBRRB  Win
RBBRBR  Win
RBBBRR  Lose
BBBRRR  Lose 
Further:
Counting cards in blackjack will get you kicked out of a casino. The same principle applies here. Using the count to your advantage in blackjack can actually give you an edge over the house, with optimal play. That's one reason they use multiple decks in a casino.

38 minutes ago, ThunderCloud said:I remain convinced by my initial argument, but to expound a little:
At any point in time, the probability that the next card drawn will be red is equal to the probability that the final card will be red. This is because both cards are drawn from the same wellshuffled pile. Therefore, at any given juncture, it is never more (nor less) advantageous to call "Stop!" than it is to allow all of the cards to be dealt out.
I agree with this statement.
I imagine it is implying that the last card is always at probability 50%, and that I'll disagree with. Since we can evaluate the probability after a new card is revealed, we can update the probability with that new information. If, for example, after 26 cards, we've only drawn black, we can say with certainty that the next card is red. And we can say the same about the last card. The next card and the last card will always have the same probability, but it won't always be 50%.

On 2/22/2018 at 6:49 PM, bonanova said:Isn't this fun?
The { real numbers } between 0 and 1 comprise two infinite groups: { rationals } and { irrationals }. Rationals (expressible as p/q where p and q are integers) are countably infinite. We can order them, by placing them in an infinite square of pq space and drawing a serpentine diagonal path. But the irrationals are not countable. The cardinality (notion of size used for infinite sets) of the rationals is called Aleph_{0} and for the irrationals (and reals) it's called Aleph_{1} or C (for continuum.) So first off, what we can do with the { numbers } between 0 and 1 depends on { which numbers } we want to deal with.
Next, there's a problem that points are not objects that can be moved from place to place, as coins can. Points are more descriptions of space than of autonomous objects. If 0 and 1 are on a number line, the location midway between them is the point denoted by 0.5. It can't be removed. (We could erase the line, I guess, but that would not produce an interval [0 1] devoid of numbers, either.) But we can finesse this matter by "painting" a point. Kind of like what we did when we inscribed a number on each coin. Painting does a little less  it puts a point into a particular group or class but it does not distinguish among them. We can tell coin 2 from coin 3 and also distinguish both from a coin with no inscription. Two blue points, however, are distinguishable from unpainted points, but not from each other. But for our purpose here, painting is actually enough.
Since the { rationals } are countable, we can paint them, in sequence, and if we do that at times of 1, 1/2, 1/4 .... minutes before midnight, all the rationals between 0 and 1 will be painted blue by midnight. (I don't know of a similar scheme for completing uncountable task sets, so the irrationals alas must remain unpainted.) Next, instead of asking whether all the points in [0 1] were removed, we can ask, with pretty much the same meaning, whether at this point the entire interval [0 1] has been painted blue. It should be, right? After all, between any two rationals lie countably infinite other rationals. So the paint must have covered the entire interval, right?. Well ... actually ... the answer is counterintuitively No.
And this is because even though the { rationals } are "dense" meaning there are no "gaps" between them, as noted above, the { irrationals } are even "more dense." That is, between any two irrational numbers there are uncountably many other irrationals. That means, for every blue point we created in the interval [0 1] there are uncountably many unpainted points. We can't magnify the line greatly enough to "see" individual points, but "if we could," if we were lucky enough to come across a single blue point we'd have to pan our camera over uncountably many unpainted points before we saw another blue one. In measure theory, the measure of the rationals over any interval is zero. The measure of the irrationals over [0 1] is unity.
So what of our quest of emptying [0 1] of points? It's the same as our quest to paint [0 1] blue by painting all the rational numbers in that interval. (Reminding ourselves that there were too many irrationals to paint one at a time.) Turning again to measure theory, the measure of the "blueness" of the interval would be zero. That means we would not see even the hint of a faint blue haze. Back to our "empty the interval by removing the rationals" quest, Nope. [0 1] would not be empty: uncountably many points (numbers) would remain.
NEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEERD!
This is actually great. I learned something! Explanations like this are the reasons I keep coming back to BD. If I had the power, I'd give you a gold star.

Interestingly, betting on the last card is always the same probability as betting on the secondtolast card. I'll say that the
bestbetter odds are likely to wait until the count of red cards is higher than black.This doesn't fall victim to the inverse probability problem.
Example:
1. BRR... should bet after the first card.
2. RBB... should bet after the third card.Any time you don't bet a red will come up, you're betting that the odds of a black coming up are equal or better, which is only the case if more (or an equal number of) reds have already appeared.

It takes 8 moves. All of the second player's moves are forced, but he usually has two ways to respond. In either case, though, the second player can't prolong the game any longer than move 8 (that I can find).
1. f6, f3
2. d6, g3
3. d4, g7
4. e4, g4
5. g6, d3
6. e6, c6
7. e3, resignsOne of 8. e2 and 8. e7 is unstoppable.

Every cut increases the number of separate pieces by 1. Since you can't cut more than one piece at a time, you will always need n1 cuts to separate every square. EDIT: You'll also never be able to do it in more than n1 cuts.
Random passengers
in New Logic/Math Puzzles
Posted · Report reply
Ah.
The odds of Al moving seem to be 50%. If we test it against a smaller number of passengers, we can see how often these closed loops form and how often Al is in his own assigned seat. It seems as the likelihood of the latter decreases, the likelihood of the former increases appropriately.
Here I have the layout for 3 and 4 passengers. 1 is Al and the last number is Bert (3 for 3 passengers, 4 for 4 passengers). Any time 3 can take the third spot, Al stays put. Any time Al is already in his assigned seat, Al can stay put. Any time the displaced person finds their seat, Al can stay put. Al staying put is denoted with a 'y'. Al moving gets an 'n'.
Al stays put 3/6 times with 3 Passengers:
123  y
132  y
213  y
231  n
312  n
321  n
Al stays put 12/24 times with 4 passengers:
1234  y
1243  y
1324  y
1342  y
1423  y
1432  y
2134  y
2143  y
2314  y
2341  n
2413  n
2431  n
3124  y
3142  n
3214  y
3241  n
3412  y
3421  n
4123  n
4132  n
4213  n
4231  n
4312  n
4321  n