z[0,25],z[1,23],z[2,21],z[3,19],z[4,17],z[5,16] > 25

<t>For the general case, For simplicity,we can write z[-1,N]=0 and z[M,-1]=z[M,-2]=z[M,-3]=-1And for M>=0 and N>=0, z[M,N]=1+1/3*Max(z[M-1,N],1+z[M,N-1])+1/3*Max(z[M-1,N],1+z[M,N-2])+1/3*Max(z[M-1,N],1+z[M,N-3]).1 corresponds to dice=4,5,6, second term to dice=3, third dice=2 and last dice=1.We can ...

<t>Using symetry works for the circle problem.For example, if we have 4 points A0 B0 C0 D0 and their symetrics A1 B1 C1 D1,We can always find a combination Aa Bb Cc Dd that they are on the same semi-circle.Suppose it's A0 B0 C0 D0 in this order.The following quadruplets are in a same semi-circle:A0 ...

<t>For the circle and 4 points problem, I've found 3/8.The idea is the extension of outrun's one.Fix first point p1, define second:p2 by angle alpha between o and pi and divide the 2pi circle in 4 area for p3 with angle betaArea [0;alpha] -> the 4th point must be in [pi;pi+alpha]Area [alpha;pi] -> t...

I think the answer is 50% because the problem is over if someone seats at the place of the first or the last passenger.

<t>I was waiting wileysw's answer and i' m not disappointing...For people who prefers matrix (to enable an easier understanding in Excel for example) rather than vertex, you can translate the answer.If you write the mathematicians answers in a N*6 matrix with 0 and 1: A,the 6*6 matrix transpose(A)*A...

<t>Agree that my example is wrong if it's strictly more than 2/5.Agree that for 5 mathematicians, you must have 4 mathematicians at 5 good answers (it's also true if you have 5*i mathematicians) You can show it directlty by the number of couple of problems:-> you have 15 couples with 3 (2*i+1) good ...

Of course, this can be lower but if you find a counter example with a guy at 3 answers, give him another good answer and you have another counter example.

<t>For N=7, you don't have counter example because for each couple of problems, you need 3 correct answers.As the total number of wrong answers is 13 (2*6+1), for one problem, you have 3 wrong answers.And in the 4 mathematicians who have the right answer, you can't have 2 who fail at the same other ...

<t>If not, you can find an indication on the number of mathematicians to try to find a counter example.If N is this number, you will have one mathematician with 5 corrects answers and N-1 with 4.So for each couple of problems, the N-1 will have a 6/15 probability and the better one 10/15.0.4*N+4/15>...

The more than 2/5 is more or equal or strictly more?If it's a more or equal, you have a solution with 5 mathematicians and only one mathematician with 5 correct answers. A 1 1 1 1 1 0B 1 1 1 0 0 1C 1 0 0 1 1 1D 0 0 1 1 1 1E 0 1 0 1 1 1

No.P(A/HH)=1-P(B/H)=1-P(A/H)Because the proba he wins with H followed by B/H is 1-the proba that B wins with an H : 1-P(B/H)The notation used is P(A/...) is the proba that A wins and P(B/...) is the proba that B wins.

The board is infinite only on two directions, you have a corner.

<t>I think the answer is yes.First step, put the rook outside the opponent's king (outside means both x and y are superior)Second step, put your king outside the opponent's king. It's feasable because your king can move in diagonal (increase both x and y) and you can force his king not to increase b...

<t>If N->infinity, it's quite easy to calculate P(inf)(i): P(inf)(i)=(i-1)/2^iSo, if N is finite, we can obtain P(i)=sum k P(inf)(i+kN)=sum k (i+kN-1)/2^(i+kN)After some calculation:P(i)-P(i-1)=1/2^(i-1)*2^N/(2^N-1)-P(i)Let's note X=2^N/(2^N-1)P(i)=P(i-1)/2+X/2^iP(i)=P(1)/2^(i-1)+X*(i-1)/2^iP(1)=P(N...