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Math Puzzles (from FiveThirtyEight - new puzzle every Friday) (1 Viewer)
- Thread starter Ignoratio Elenchi
- Start date
heckmanm
Footballguy
Initial thoughts (some probably blindingly obvious but it helps me to state them)
The game ends when there are X more tails than heads, or Y more heads than tails. E.g., if X = -4, then you need 4 straight tails to start, or 5 tails and a head in some combination, or 6 and 2, etc.. The same applies for Y, except it needs more heads than tails to come up.
The game ends when either of them win, so I think the expected duration would be until a net difference of (Y-X/2) between heads and tails is achieved. Could also be expressed as (|X|+|Y|)/2, I suppose.
So: what is the expected number of random coin flips until the difference between heads and tails reaches any given number?
I ran a 4000-flip trial in Excel. It bounced around between 0 and about 12 for the first 1400 flips or so, then made a quick climb close to 60 around #1700, dropped back near 0 by 2400, and fluctuated between 0 and 30 the rest of the way, finishing exactly at 0 (i.e., 2000 heads, 2000 tails). Not that this particular trial would be the least bit predictive; I was just curious to see if I could simulate the game.
The game ends when either of them win, so I think the expected duration would be until a net difference of (Y-X/2) between heads and tails is achieved. Could also be expressed as (|X|+|Y|)/2, I suppose.
So: what is the expected number of random coin flips until the difference between heads and tails reaches any given number?
I ran a 4000-flip trial in Excel. It bounced around between 0 and about 12 for the first 1400 flips or so, then made a quick climb close to 60 around #1700, dropped back near 0 by 2400, and fluctuated between 0 and 30 the rest of the way, finishing exactly at 0 (i.e., 2000 heads, 2000 tails). Not that this particular trial would be the least bit predictive; I was just curious to see if I could simulate the game.
oldmanhawkins
Footballguy
I did what I could working out basic examples. I had some fun figuring out ways to sum a few of the series, but unfortunately had to start brute forcing the expectation at X=2 and Y= 3. Thankfully, by the time I did X=3, Y=3 I saw the answer:
it's just XY
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oldmanhawkins
Footballguy
This write up: http://web.mit.edu/neboat/Public/6.042/randomwalks.pdf
does a pretty good job of deriving the answer but I couldn't follow a lot of it. If anyone knows where I can learn more about this I'm all ears.
does a pretty good job of deriving the answer but I couldn't follow a lot of it. If anyone knows where I can learn more about this I'm all ears.
heckmanm
Footballguy
My analysis:
Assume planet has radius r miles. If the guardian moves r mph, she can reach any spot on the planet (180 degrees) in π hours
This means the aliens move r/20 mph, and if they started directly opposite each other (180 degrees apart; worst case), they would each need to go 1/4 around the circumference, which is 1/2*π*r miles. This would take them 10π hours, which is at least 10x as long as it would take the guardian to get to their meeting point.
In order to meet in less than π hours, then, they must be less than 18 degrees apart. Put another way, they must be < 9 degrees from their meeting point.
The surface area of the portion of a sphere with an angle of 9 degrees from the pole is:
A = 2*π r2* (1 - cos Θ)
For r = 1 and Θ = 9, A = 0.078
The area of the entire planet is 12.56. So the aliens must land in an area of (0.078 / 12.56) of the surface, which is 0.56%. The guardian can cover the other 99.44% of the surface.
And as long as the aliens do not BOTH land in the same 0.56% area, she will intercept them. The probability of that is 0.562, or 0.3136, so the probability that the guardian can save the planet is 99.6864%
This means the aliens move r/20 mph, and if they started directly opposite each other (180 degrees apart; worst case), they would each need to go 1/4 around the circumference, which is 1/2*π*r miles. This would take them 10π hours, which is at least 10x as long as it would take the guardian to get to their meeting point.
In order to meet in less than π hours, then, they must be less than 18 degrees apart. Put another way, they must be < 9 degrees from their meeting point.
The surface area of the portion of a sphere with an angle of 9 degrees from the pole is:
A = 2*π r2* (1 - cos Θ)
For r = 1 and Θ = 9, A = 0.078
The area of the entire planet is 12.56. So the aliens must land in an area of (0.078 / 12.56) of the surface, which is 0.56%. The guardian can cover the other 99.44% of the surface.
And as long as the aliens do not BOTH land in the same 0.56% area, she will intercept them. The probability of that is 0.562, or 0.3136, so the probability that the guardian can save the planet is 99.6864%
Didn't know where else to put this and I didn't think it was worthy of starting a new thread. Today's Google doodle celebrates Gottfried Leibniz, an all-time brilliant human and the co-developer of one of the greatest achievements of the mind, Calculus. The image on the Google portal shows a hand scribing a series of six binary numbers (8-bits) but the very last digit is not completed. Is it going to become a zero or a one, and why? Happy birthday, Leibniz!
Interesting sudoku puzzle - only 4 numbers given
Normal sudoku rules apply.
In addition, both marked diagonals must also contain the digits 1 to 9.
Cells that are a knight's move apart (in chess) cannot contain the same digit.
The central 3x3 box must form a magic square.
Normal sudoku rules apply.
In addition, both marked diagonals must also contain the digits 1 to 9.
Cells that are a knight's move apart (in chess) cannot contain the same digit.
The central 3x3 box must form a magic square.
Ignoratio Elenchi
Footballguy
Forgot about these. This week's Riddler Express looks like a quick one:
Question 1: 25%
Question 2: 25%
Question 2: 25%
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I understand the 1st answer - or at least I came up with the same one.
2nd answer has me confused - given that there are 7 tiles with a 6 on at least one side. I assume that the reveal changes the odds - probably similar to the Monte Hall problem. But I don't know why...
Ignoratio Elenchi
Footballguy
The way I thought of it is
There are 8 sixes in a set of dominoes (including the 2 on the double-six domino). 2 of them have a six on the other side. So 2/8 chance the six you're looking at has a six on the other side.
I solved the first one on my own - but it took about 45 minutes, so I was a little disappointed to see the video was only 20 some minutes, and he was using some of that time to explain the easy stuff...Walking Boot said:
ETA - I'll give this one a go in the morning.
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