Some people here are incredibly smart mathematicians and I've never had any luck obfuscating an otherwise straightforward problem, so I imagine this challenge will not prove as difficult as I would hope.
Prove that
where n is a positive integer. (I'm actually pretty sure n can be any positive real number, if that helps.)
I hope I transcribed that correctly...
(4 choose 3)*8/52*1/51*6/50*1/49*4/48*1/47 = 1 in 19 million
thommy3 wrote:
Four possible ways to chosse three players from the four who get dealt KQ suited.
That number overcounts the cases though (since it counts the case when all four players have KQs four times).
For (1), the number I have is:
(I ignore the order of the two cards in each hand, but I specify the four players. I choose the three players that get KQs, then the three suits, then order them among the players, then give the fourth player any two cards. I subtract the cases where the fourth player also has KQs.)
[C(4,3)C(4,3)3!C(46,2)-4*4!] / [C(52,2)C(50,2)C(48,2)C(46,2)], or about 5.23435×10^-8 (1 in 1.91046×10^7).
If you allow the fourth player to have KQs (and count it only once) then:
[C(4,3)C(4,3)3!C(46,2)-3*4!] / [C(52,2)C(50,2)C(48,2)C(46,2)], or about 5.23561×10^-8 (1 in 1.91000×10^7).
For (2), I assume that the fourth hand has been folded and plays no further part. I assume that the fourth player could have had the last KQs, or anything else other than the other kings or queens that the other players hold, and so is no different from being randomly dealt two cards. If all four players have KQs, it is counted four times (once for each player that folds).
(I also ignore the order of the five cards on the board.)
At a minimum, AJT is on the board. Then it is impossible to have a full house or four of a kind. The only higher hands possible are flushes and straight/royal flushes. Since three players hold KQs, there can't be three cards of those suits on the board. The board also can't have all cards being the fourth suit.
So the cases for the ranks of the cards on the board are:
(x and y denote ranks below ten (T), z denotes K or Q, u and v denote A or J or T)
@ AJTxy, C(8,2) * [4^5 - C(3,1)[C(5,3)3^2+C(5,4)3+C(5,5)] - 1] = 19740
@ AJTxx, C(8,1) * [4^3C(4,2) - C(3,1)[6+3*9]] = 2280
@ AJTzx, C(2,1)C(8,1) * [4^4 - C(3,1)[C(4,3)3+C(4,4)] - 1] = 3456
@ AJTKQ, 1 * [4^3 - C(3,1) - 1] = 60
@ AJTux, C(3,1)C(8,1) * [4^3C(4,3) - C(3,1)[6+3*9]] = 3768
@ AJTuz, C(3,1)C(2,1) * [4^2C(4,2) - C(3,1)3] = 522
@ AJTuv, C(3,2) * [4C(4,2)^2 - C(3,1)3^2] = 351
@ AJTuu, C(3,1) * [4^2C(4,3) - C(3,1)3] = 165
Total: 30342
So the probability is thommy3's answer times 30342 / C(46,5), which is about 1.15976×10^-9 (1 in 8.62249×10^8).
Now, you might argue that, for example, having AJTKQ on the board violates the spirit of the problem, because that means everyone's KQs is useless for making an A-high straight. So I'll look at the extreme: suppose that we require that not only do all three players have nut hands (hands which can never lose from their standpoint), but also that each player is guaranteed to win but for another player having KQ (this is the most exciting scenario, and the one which Warp saw as indicated by the screenshot, which I will post here again).
Warp wrote:
That means the only permissible board layout is AJTxy with no three cards of the same suit. In that case there are C(8,2) * [4^5 - C(4,1)[C(5,3)3^2+C(5,4)3+C(5,5)]] = 16800 possibilities, and so the probability is thommy3's answer times 16800 / C(46,5), which is about 6.42144×10^-10 (1 in 1.55728×10^9).
Edit: Fixed the denominator for (2). Obviously it's supposed to be C(46,5), not C(44,5), since we were discarding consideration of the fourth player's hand.
OmnipotentEntity wrote:
This is known as the "curve of pursuit." Also known as the "mice problem."
I don't know the solution to the curve, but it should be rather easy to set up the parametric differential equations. I'll leave that to someone else though.
Seeing this as a pursuit problem is very useful. For example, this blog discusses square pursuit, and an image of square pursuit from that blog is shown below:
Suppose that four animals are placed at the corners of the unit square and each animal walks at unit speed directly toward the one on its left. Since the motion vectors always form 45-degree angles to the radial displacement (from the center of the square), radial velocity is always 1/sqrt(2) inward and tangential velocity is always 1/sqrt(2) clockwise (and we also know that tangential velocity is equal to radial distance times angular velocity). In terms of differential equations:
r=distance from center, t=time, θ=angle
dr = -1/sqrt(2) dt
r*dθ = -1/sqrt(2) dt = dr
Solving the last equation:
dθ = 1/r dr
θ+C = ln(r)
r = eθ+C = Aeθ.
Therefore each of the four animals traces a logarithmic spiral with polar equation r = Aeθ for some A depending on which one of the four is considered.
When I was playing poker last week I ran into a situation where I lost with quad tens against a royal flush, but that situation presented above is far more unlikely.
I don't know the solution to the curve, but it should be rather easy to set up the parametric differential equations. I'll leave that to someone else though.
Many tasks are easy from first view, but later turns out - they not.
marzojr wrote:
I will leave the rest as an exercise :-)
Leaved most hard part :)
marzojr wrote:
n(theta) = (theta-theta_0) / arctan(r)
This move is nice! (switching theta and n)
I had troubles exactly with finding how many steps I need to reach point parametrized by t. (not theta)
Even though, I've solve it.
Moving all into exponent nice too.
marzojr wrote:
Yeah, only sqrt(2) is diagonal of unit square, sqrt(2)/2 is distance from corner to center, and 1/sqrt(2) - idk.
FractalFusion wrote:
Since the motion vectors always form 45-degree angles to the radial displacement (from the center of the square), radial velocity is always 1/sqrt(2) inward and tangential velocity is always 1/sqrt(2) clockwise (and we also know that tangential velocity is equal to radial distance times angular velocity).
Can you explain in details?
FractalFusion wrote:
Suppose that four animals are placed at the corners of the unit square and each animal walks at unit speed directly toward the one on its left.
In the end: can you prove that this task is same/related?
Some people here are incredibly smart mathematicians and I've never had any luck obfuscating an otherwise straightforward problem, so I imagine this challenge will not prove as difficult as I would hope.
Prove that
where n is a positive integer. (I'm actually pretty sure n can be any positive real number, if that helps.)
I hope I transcribed that correctly...
The binomial series tells us that, if |x| < 1, then we have
By multiplying each factor of the product by n and then dividing by n^k, and then substituting y = -x we get
As y → 1, the terms of the infinite sum monotonically decrease, and they all have the same sign, so we can interchange the sum and the limit and so the left hand side converges to the sum in Bobo's post, plus 1. Meanwhile if α > 0 then the RHS converges to 0. Now subtract 1 from both sides and set α to 1/n.
Some people here are incredibly smart mathematicians and I've never had any luck obfuscating an otherwise straightforward problem, so I imagine this challenge will not prove as difficult as I would hope.
Prove that
where n is a positive integer. (I'm actually pretty sure n can be any positive real number, if that helps.)
I hope I transcribed that correctly...
The binomial series tells us that, if |x| < 1, then we have
By multiplying each factor of the product by n and then dividing by n^k, and then substituting y = -x we get
As y → 1, the terms of the infinite sum monotonically decrease, and they all have the same sign, so we can interchange the sum and the limit and so the left hand side converges to the sum in Bobo's post, plus 1. Meanwhile if α > 0 then the RHS converges to 0. Now subtract 1 from both sides and set α to 1/n.
Absolutely correct!
And wow! I drew out a lurker! Welcome to the boards!
The binomial series tells us that, if |x| < 1, then we have
Most likely I'm missing something here, but if you require |x| < 1 to use this series, then later let y → 1, doesn't that then imply that you've used x=-1 and hence |x|=1?
Does this all work out because you're taking limits? (Is this similar to finding f(0) when f(x) = sin(x)/x ?)
The binomial series tells us that, if |x| < 1, then we have
Most likely I'm missing something here, but if you require |x| < 1 to use this series, then later let y → 1, doesn't that then imply that you've used x=-1 and hence |x|=1?
Does this all work out because you're taking limits? (Is this similar to finding f(0) when f(x) = sin(x)/x ?)
Yes, the formula with y in it only needs to hold for y < 1, and then I take a limit (which can give you a proof that the formula holds for y = 1, i.e. x = -1, but by then I'm already done). This means that you do need a brief justification for why taking the limit is valid here, which I do give but don't fully spell out.
(It's a tiny bit different from the sin(x)/x thing, there you're saying that if f is a continuous function around 0 that is equal to sin(x)/x for x =/= 0, then f(0) = 1)
Since the motion vectors always form 45-degree angles to the radial displacement (from the center of the square), radial velocity is always 1/sqrt(2) inward and tangential velocity is always 1/sqrt(2) clockwise (and we also know that tangential velocity is equal to radial distance times angular velocity).
Can you explain in details?
Because of 90-degree rotational symmetry around the center of the square, the four animals always form the corners of a square at any given time. The velocity vector of the top-left animal is shown in the diagram above but this applies similarly to all of them. Because the animals form the corners of a square at all times, θ is always 45 degrees, and so |vr|=(sqrt(2)/2)*|v| and |vt|=(sqrt(2)/2)*|v|.
For simplicity, I assumed that |v| was always 1 (unit speed). However, it doesn't actually matter what |v| is, as long as all four animals have the same speed at the same time (so that it always remains a square). For example if instead we let speed be a function of t, so that we use |v(t)| instead of unit speed, then:
dr = (-1/sqrt(2))*|v(t)| dt
r*dθ = (-1/sqrt(2))*|v(t)| dt = dr
and the rest is the same.
r57shell wrote:
FractalFusion wrote:
Suppose that four animals are placed at the corners of the unit square and each animal walks at unit speed directly toward the one on its left.
In the end: can you prove that this task is same/related?
The side length of the first square inside the outer (unit) square is sqrt((1-a)2+a2)=sqrt(2a2-2a+1), which we call c. Also, the diagram is self-similar, so a rotation and scaling maps the first square to the second, the second to the third, and so on. The corners of the squares can thus be modelled by:
@ Four animals are placed at the corners of the unit square.
@ 1st step: Each of them faces the one on the left and moves distance a.
@ 2nd step: Each of them faces the one on the left and moves distance ac.
@ 3rd step: Each of them faces the one on the left and moves distance ac2.
...
@ nth step: Each of them faces the one on the left and moves distance acn-1.
...
Each step generates all the corners of the squares. Letting a→0 causes all the distances above to go to 0, so the corners converge to a curve, and this curve is the one formed where each animal walks directly toward the one on its left (continuously).
Say, I don't know if this has been answered in the thread already, but I have a question about this pursuit problem.
It would appear to me that there is a certain scale+rotation symmetry in this problem. The condition after a certain time t matches the initial condition, just scaled down and rotated slightly. As such, it seems logical to me that the curvature of the path should not vary with where the animals are on the path. This is an obvious contradiction with the approximating figure and the known result, since constant curvature would imply circular paths. Does anyone have a relatively simple resolution to this contradiction?
Staring at the figure, I can almost see how this symmetry would imply the differential equation r' = r, characterizing the curve. Can someone put the last pieces together?
Edit: Aaaand, immediately after posting, I see that the scaling increases the curvature, just like scaling a circle down produces a circle of greater curvature. Derp.
I'll let my post stand and hang my head in shame.
Yeah, only sqrt(2) is diagonal of unit square, sqrt(2)/2 is distance from corner to center, and 1/sqrt(2) - idk.
sqrt(2)/2 = 1/sqrt(2)
Shame on me :(
FractalFusion wrote:
Because the animals form the corners of a square at all times, θ is always 45 degrees, and so |vr|=(sqrt(2)/2)*|v| and |vt|=(sqrt(2)/2)*|v|.
Ok now with v and image - clear enough.
FractalFusion wrote:
@ nth step: Each of them faces the one on the left and moves distance acn-1.
Up to here is fine.
FractalFusion wrote:
Each step generates all the corners of the squares. Letting a→0 causes all the distances above to go to 0, so the corners converge to a curve, and this curve is the one formed where each animal walks directly toward the one on its left (continuously).
Bad reasoning.
Each of corner tends to starting corners. It's easy proovable:
Fix n, thus corner at distance acn. Now limit a->0 distance is zero.
So, nor distance, nor sequnce of sequences {{p0n,p1n,p2n...}} in any lp tends to curve. :)
In lp they tends to 0 = {0,0,0,0,0...}
This is what I had troubles with. Even though I had in my mind possible way to get whole distance passed by.
r57shell,
You've agreed that there is a velocity along the radius. If there is a velocity along the radius then the radius must change. Because the velocity is monotonic constant and in the negative direction it must constantly decrease. But in order to arrive at the next corner, it must attain the original radius. So therefore it cannot.
Build a man a fire, warm him for a day,
Set a man on fire, warm him for the rest of his life.
Yes, the formula with y in it only needs to hold for y < 1, and then I take a limit (which can give you a proof that the formula holds for y = 1, i.e. x = -1, but by then I'm already done). This means that you do need a brief justification for why taking the limit is valid here, which I do give but don't fully spell out.
Thanks, that makes a lot of sense.
I didn't manage to solve Bobo the King's original problem, but here are some things I discovered along the way while trying to attack it:
Here's the original again:
We can transform the product index to "simplify" the product:
And we can add an extra sum term:
This second step works because the k=0 case always equals 1 (if we accept that an empty product equals 1 vacuously). I like this form because summing to zero looks more elegant, but seems to take you further away from the original binomial series.
Each of corner tends to starting corners. It's easy proovable:
Fix n, thus corner at distance acn. Now limit a->0 distance is zero.
Distance traveled: acn.
But total distance: ac0 + ac1 + ac2 + ... + acn = a(1 - c^(n+1))/(1 - c)
When a->0, this is zero.
OmnipotentEntity wrote:
r57shell,
You've agreed that there is a velocity along the radius.
No, I haven't.
It was his interpretation of task, and I just ask him prove relation if he want.
Results are same, but it doesn't mean that you can replace one task with another:
00 = 1
12 = 1
Even though result is same (1), this doesn't mean you can say: 00 is one because 00 = 12, which is one.
Each of corner tends to starting corners. It's easy proovable:
Fix n
The intention was not to fix n, but to consider the set of all (infinite) corners as a whole.
Then:
r57shell wrote:
But total distance: ac0 + ac1 + ac2 + ... + acn = a(1 - c^(n+1))/(1 - c)
When a->0, this is zero.
● The total distance is an infinite sum: ac0 + ac1 + ac2 + ... + acn+ ..., which equals a/(1-c).
● c depends on a; c=sqrt(a2+b2)=sqrt(a2+(1-a)2)=sqrt(2a2-2a+1).
● a/(1-c)=a/(1-sqrt(2a2-2a+1))=(1+sqrt(2a2-2a+1))/(-2a+2); when a→0, the limit is 2/2=1.
Suppose I have two values of the form A=a*10b and B=c*10d and I would want to interpolate between the two (using an interpolation factor between 0.0 and 1.0) in such a manner that the mantissas get interpolated linearly and the exponents get interpolated linearly. In other words, I would end up with this:
x = a*(1-factor) + c*factor
y = b*(1-factor) + d*factor
result = x*10y
I have the strong feeling that there's a simple function that allows doing that without having to deal with the mantissas and exponents separately, ie. calculating with the values A and B directly. In other words:
factor = someFunction(factor)
result = A*(1-factor) + B*factor
However, I'm not at all sure what the function is. Probably something dealing with exponentiation or logarithms, but what exactly?
Suppose I have two values of the form A=a*10b and B=c*10d
Given your use of the term "mantissa" in this context, I'm guessing you're talking about a base-10 floating point number? In which case we have the extra constraints that 1 ≤ a,c < 10 and that b and d are integers (I'm assuming A and B are strictly positive for this, for simplicity).
Restating the problem, we're looking for a function f(A,B,t) such that
Such a function has striking discontinuities in it. For example:
but:
These discontinuities occur every time one of the exponent values b or d goes up by 1. Intuitively it feels to me like the simplest way to achieve this behaviour is by splitting A and B into mantissas and exponents, like you originally had, because the functions that split A and B into mantissas and exponents provide precisely the discontinuities that f(A,B,t) has. It's hard to imagine how you would replicate this discontinuous behaviour without the splitting functions.
Incidentally, this is an odd form of interpolation, in that it can easily have a result that is outside the range from A to B. Consider that f(9,10,0.5) = 5√10 ≅ 15.81
In which case we have the extra constraints that 1 ≤ a,c < 10 and that b and d are integers
Obviously they can't be integers because else you wouldn't be able to interpolate between them. (Well, I suppose you could, but you would be rounding it to the nearest whole number, which just wouldn't work very well as an interpolation.)
Thinking about a bit more, I realized that my approach to the problem was a bit off. Well, I was kind of there, but not quite. I realized that what I was really attempting was interpolating in logarithmic coordinates rather than linear ones.
If I have two numbers A and B, and I want to interpolate between them in the manner I described (ie. logarithmically), I should calculate the x and y in A=10x and B=10y, then interpolate between x and y linearly, ie. exp=x*(1-factor)+y*factor, and then number I'm looking for is 10exp.
By definition x=log10(A), and y=log10(B).
So, I suppose, the answer to my question is:
result = 10log10(A)*(1-factor)+log10(B)*factor
This ought to give me the answer to my original question.
1) Well that's a classic conundrum. I don't know how much mathematical training you've had (you need to know some calculus, specifically the chain rule and product rule), but if you rewrite the equation as y = e^lnx/x then it's a relatively trivial problem:
y = elnx/x
dy/dx = d/dx(lnx/x)*elnx/x (chain rule)
d/dx(lnx/x) = 1/x^2 - lnx/x^2 (product rule)
dy/dx = (1-lnx)elnx/x/x^2
Set dy/dx = 0: 1-lnx = 0, x = e
Therefore y = e1/e
Suppose that four animals are placed at the corners of the unit square and each animal walks at unit speed directly toward the one on its left. Since the motion vectors always form 45-degree angles to the radial displacement (from the center of the square), radial velocity is always 1/sqrt(2) inward and tangential velocity is always 1/sqrt(2) clockwise (and we also know that tangential velocity is equal to radial distance times angular velocity). In terms of differential equations:
r=distance from center, t=time, θ=angle
dr = -1/sqrt(2) dt
r*dθ = -1/sqrt(2) dt = dr
Solving the last equation:
dθ = 1/r dr
θ+C = ln(r)
r = eθ+C = Aeθ.
Therefore each of the four animals traces a logarithmic spiral with polar equation r = Aeθ for some A depending on which one of the four is considered.where n is a positive integer. (I'm actually pretty sure n can be any positive real number, if that helps.) I hope I transcribed that correctly...
By multiplying each factor of the product by n and then dividing by n^k, and then substituting y = -x we get
As y → 1, the terms of the infinite sum monotonically decrease, and they all have the same sign, so we can interchange the sum and the limit and so the left hand side converges to the sum in Bobo's post, plus 1. Meanwhile if α > 0 then the RHS converges to 0. Now subtract 1 from both sides and set α to 1/n.
Because of 90-degree rotational symmetry around the center of the square, the four animals always form the corners of a square at any given time. The velocity vector of the top-left animal is shown in the diagram above but this applies similarly to all of them. Because the animals form the corners of a square at all times, θ is always 45 degrees, and so |vr|=(sqrt(2)/2)*|v| and |vt|=(sqrt(2)/2)*|v|.
For simplicity, I assumed that |v| was always 1 (unit speed). However, it doesn't actually matter what |v| is, as long as all four animals have the same speed at the same time (so that it always remains a square). For example if instead we let speed be a function of t, so that we use |v(t)| instead of unit speed, then:
dr = (-1/sqrt(2))*|v(t)| dt
r*dθ = (-1/sqrt(2))*|v(t)| dt = dr
and the rest is the same.
The side length of the first square inside the outer (unit) square is sqrt((1-a)2+a2)=sqrt(2a2-2a+1), which we call c. Also, the diagram is self-similar, so a rotation and scaling maps the first square to the second, the second to the third, and so on. The corners of the squares can thus be modelled by:
@ Four animals are placed at the corners of the unit square.
@ 1st step: Each of them faces the one on the left and moves distance a.
@ 2nd step: Each of them faces the one on the left and moves distance ac.
@ 3rd step: Each of them faces the one on the left and moves distance ac2.
...
@ nth step: Each of them faces the one on the left and moves distance acn-1.
...
Each step generates all the corners of the squares. Letting a→0 causes all the distances above to go to 0, so the corners converge to a curve, and this curve is the one formed where each animal walks directly toward the one on its left (continuously).
