die 2 wins over die 1: 2 wins 1/3 of the time, 4 wins 1/2 of the time: The total probability of die 2 winning over die 1 is 1/6 x 1/3 + 2/3 x 1/2 =1/2.
Let $A_i$ be the number of times $i$ occurs in die $A$ for $i = 1,2,\ldots,N$, and similarly for $B_i$ and $C_i$. Let $$Q_N(A,B) = \sum_{j=1}^N \sum_{i=j+1}^N A_i B_j$$
therefore for die A to beat die B with >50% probability, $Q_N(A,B)$ must be >18
hey, could someone check if there are only minor mistakes or is it totally wrong? mathbin.net/93009 (Poisson process) I have a feeling that i made some illegal operations..
Suppose there is a polygon, and there is a point inside it, how to prove that the sum of distance of that point from the sides of that polygon is actually equal to the area of polygon !
Suppose there is a polygon, and there is a point inside it, how to prove that the sum of distance of that point from the sides of that polygon is actually equal to half of the area of polygon ! (ignore them dimensions)
Suppose there is a regular polygon, and there is a point inside it, let the sum of that point from the sides of the polygon be X, and the area of the polygon be Y, how to prove that X=2Y/a (where a is the side of the regular polygon)
@5T0M The polygon is (by assumption) regular and therefore convex. So whatever the point is, draw lines from it to each corner of the polygon. That gives you a set of triangles that work.
Speaking of white noise: I always need to have the fan on in my room, cold or not; if it isn't on, I hear my clock, and the clock's ticking is more annoying than the hum of the fan...
do you guys know of other math chatrooms? i don't mean to insult or anything but I really need help setting this up and there seems to be a lot of nonresponse on this website
@JM You could look for a digital clock with an analog face (actual hands) which self-syncronizes via NTP when it is powered up. Then wire it so it feeds from the room light so it turns off when you shut off the light to go to sleep and resynchronizes in the morning!
Well, back in olden times the computations one could contemplate before needing to print the result in human-readable notation were so simple that the extra complexity of implementing native BCD arithmetic was small compared to converting between binary and decimal at every input and output. Today the ratio between arithmetic operations and decimal I/O events is different.
This may also be related to why BCD was especially popular in administrative/financial computing (very little arithmetic), whereas it never caught on in scientific computing.
@JM I actually wondered when writing the answer: do (4-function) calculators still even exist? I assume graphing/programmable calculators use general purpose hardware and IEEE754 arithmetic internally. But even those are rapidly being displaced by smartphone apps, I think.
@HenningMakholm If memory serves, boh TI and HP are still using BCD internally. I've wondered about why none of these high-end calcs know IEEE myself...
...and yeah, there are still those simple calculators in supermarkets and low-end electronic shops.
(I don't know about the other calc makers, but those two are usually the ones who do a lot of changes.)
Ah, I knew it. The current high-end ones still use Motorola 68k processors, and those do BCD arithmetic....
A_[N] * (sum from i=1 to N-1) B_[i] is like saying "Number of faces on A that have the value N times the count of all faces on B that have numbers lower than N"
which makes sense intuitively
my current attempt at making F() although for only the scenarios where A>B>C>A: pastebin.com/QcEV1r9V
@AOAOne: Sorry to change the thread here, but did you get the following: $ \begin{align} Q(A,B) &= 19\\ Q(B,A) &= 14\\\\ Q(A,C) &= 15\\ Q(C,A) &= 19\\\\ Q(B,C) &= 20\\ Q(C,B) &= 14 \end{align} $
when i say a die wins more than 50% of the time i mean that there are 36 possibilities for two dice (6 faces vs. 6 faces). The die that rolls a higher number wins. ties don't count in either person's favor -- all we care about is how often we can expect to beat the other die
conditional probability: if A is 1, how many can it beat? 0. if A is 2, how many can it beat? 1. If A is 3, how many can it beat? 2. If A is 4, how many can it beat? 3. If A is 5, how many can it beat? 4. If A is 6, how many can it beat? 5. That's 15/36, or <50%
i am operating purely within the confines of the problem's definition here as to how it's counting wins
it's only counting conditional wins based on how often rolls of one die will be greater than the other. 111111 vs. 111112 just means die 1 is never going to win but die 2 wins 6/36 of the time. That means it's only rolling a higher value 6/36 of the time, which is far less than the problem asks for
as it is defining a nontransitive die as rolling greater values >50% of the time
perhaps the question is just worded poorly but the output matches their statement for N=7 under this assumption
It depends on how you interpret "larger than 50% chance of winning". If we consider only the wins, then we would use conditional probability. If we consider the ties, then there are lots of dice that are incomparable.
anyhow, my problem is about finding the number of sets of dice triplets with maximum potential face value N that fits all the criteria
i think Q(A,B)>18 is what they are looking for if only because the program i wrote yields 9780 -- if you did it purely by Q(A,B)>Q(B,A) I think you'd have a ton more solutions
@MattN I think you need to look up what a metric space is and what a topological notion is. Both questions you suggested for metric spaces did involve (possibly) non-metric spaces and I still think the question you wanted to re-tag did not even involve a topology. You can't speak of a topology when you don't know this and if you speak of a topology you don't have to ask it. (My last word on this matter, no time for silly arguments).
@robjohn Hi, robjohn. Small browser hickup, didn't intend to leave immediately again...
@robjohn Anyhow I have the counting function f() and I am modeling it based off of the recursive definition of Q. However, my function seems to yield a lot more values than it should
@tb The metric-spaces retag was (obviously) (an attempt at) humour. As for topological "notion": no, indeed I've not heard that one before but I think "notion" is synonym for thingamajig. And as for the other retag: it involved a metric space (if I remember correctly) hence a topology. (I can't just sit by while being accused of having to look up what a metric space is. (Even if you star robjohn's comment.))
Definitely in favor of bird calls. Telephone calls depend on whether I'm the caller. Margin calls ... depend on whether we're going to have a derivatives market at all.
@robjohn But that won't help the 5/10 phase (since you cannot delete an accepted answer). Instead you should keep a vaguely-interesting question with an odd number of votes around for these occurrences.
