BobbyGo
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Posts posted by BobbyGo
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I get 17/60.
If you win the first match (2/5 chance), you have 5/12 chance of winning the second match, equaling 1/6 chance of winning the first two (WWx).
If you tie the first match (1/5 chance), you have 1/3 chance of winning the second match. Of those 1/3 possibilities, you then have 1/2 chance to win the third match. The final probability of TWW is 1/30.
If you lose the first match (2/5 chance), you have 5/12 chance of winning the second match. Of those 5/12 possibilities, you then have 1/2 chance to win the third match. The final probability of LWW is 1/12.
1/6 + 1/30 +1/12 = 17/60
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Non-Returning: 1, 5, 2, 6, 3, 7, 4 ,8 for 42.957
Returning: 1, 4, 7, 5, 2, 8, 3, 6, 1 for 47.278
You have the longest non-returning path.
The returning path can be slightly longer.
I'm not sure which order of 7 and 8 you are using, but I see one path that is too short to use.
There are no (chess) Knight moves, like 7-5 and 2-8 (using the correct 7 8 labeling).
If you are using the original 8 7 labeling you are not using knight moves, but too many 1-4 type moves.
Clue for those still working on this the longest total length is 47.97, and from that you can get the distribution of city-city distances used, and work out the path.
Try my next, harder puzzle just posted.
Doh... mixed up 7 and 8 on the returning...
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Non Return: 1, 5, 2, 7, 3, 8, 4, 6 for 49.698
Return: 1, 5, 2, 7, 3, 8, 4, 6, 1 for 56.023
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Non-Returning: 1, 5, 2, 6, 3, 7, 4 ,8 for 42.957Returning: 1, 4, 7, 5, 2, 8, 3, 6, 1 for 47.278
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Hmm... just noticed that the coordinates listed for points 7 (4,0) and 8 (2,0) are switched on the graph. It labels 7 at (2,0) and 8 at (4,0). Which set is correct?
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My first instinct is to have the salesman criss-cross as much as possible. The route I would take (just a guess, haven't verified) is 1, 5, 2, 6, 3, 8, 4, 7 for a total of 41.989 units traveled. -
Just realized the section:
a1=sqrt ((a2-b2)*(a2-b2));
b1=sqrt ((c2-b2)*(c2-b2));
c1=sqrt ((c2-d2)*(c2-d2));
d1=sqrt ((a2-d2)*(a2-d2));
should actually be:
a1=sqrt ((a2-d2)*(a2-d2));
b1=sqrt ((a2-b2)*(a2-b2));
c1=sqrt ((c2-b2)*(c2-b2));
d1=sqrt ((c2-d2)*(c2-d2));
I don't think it really it effects the outcome, since it stayed in the same order. But still.
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Can you share the code?
I'm sure there are more elegant ways of coding this, and it only gives back the first longest set it finds (it won't return 8,21,45,1 after it finds 1,8,21,45) so I'm sure there is more than one answer.
I think 5,29,42,49 might be another set, but haven't tested them yet.
int a,b,c,d,a1,b1,c1,d1,a2,b2,c2,d2,a3,b3,c3,d3; int z3=0,z,x; int main() { for(a=1; a<50; a++) { for(b=1; b<50; b++) { for(c=1; c<50; c++) { for(d=1; d<50; d++) { a1=a;b1=b;c1=c;d1=d; x=1;z=0; while(x==1) { if(a1==b1 and c1==d1 and a1==c1) { x=0; }else { a2=sqrt ((a1-b1)*(a1-b1)); b2=sqrt ((c1-b1)*(c1-b1)); c2=sqrt ((c1-d1)*(c1-d1)); d2=sqrt ((a1-d1)*(a1-d1)); z+=1; } if(a2==b2 and c2==d2 and a2==c2) { x=0; }else { a1=sqrt ((a2-b2)*(a2-b2)); b1=sqrt ((c2-b2)*(c2-b2)); c1=sqrt ((c2-d2)*(c2-d2)); d1=sqrt ((a2-d2)*(a2-d2)); z+=1; } } if(z>z3) { z3=z; a3=a; b3=b; c3=c; d3=d; } } } } } cout<<z3<<endl<<endl<<" "<<a3<<endl<<b3<<" "<<d3<<endl<<" "<<c3<<endl; return 0; } -
I cheated; with less than 6 million different starting combinations, I wrote some code to brute force the answer.1
8 45
21
Lasted for 12 iterations (possibly 11 or 13, counting isn't my strong suit
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If the big and little balls are touching the same walls and the floor counts as one of the walls (if the big ball has the little ball "trapped" inside the corner of a rectangle), then the first objective is to find length of the vacant diagonal left from the big ball. The midpoint of the big ball acts as the corner of a cube, each length being 3.5 feet. The diagonal of that cube is sqrt(3)*3.5, and subtracting the radius of the big ball leaves a line of approximately 2.562... feet in length.The ratio of the diagonal from the corner to the outer edge of the big ball compared to the vacant length in the corner should be equal to the ratio of the diagonal from the corner to the outer edge of the small ball compared to the vacant length in the corner it does not take up. Once we know the length of the vacant diagonal left from the little ball, we can subtract that from the length of the vacant diagonal left from the big ball which will leave us with the diameter of the little ball.
= 1.875644......I think...
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If I'm reading this right, the longest possible ladder to fit through the passageway would touch both outer walls of each hallway at a 45 degree angle and will also touch the corner where the hallways meet. This would form two right triangles, and the ladder would form the hypotenuse of both of them. Since the other two angles in each right triangle are both 45 degrees, then the non-hypotenuse sides (forgot what they were called) in each triangle would be the same as well. So the total length of the longest ladder would be
= 14.14213562373...So yes, the 14 foot ladder should pass through from one hallway to the other.
Of course, that doesn't factor in the thickness of the ladder.
The ladder as a whole would act as the hypotenuse of right triangle with each outer hallway wall. For a 0 width ladder, the length of each wall segment would be
= 10. For a 14-foot ladder, they would each be 
= 9.89949...If I'm mathing correctly, the difference between these two side lengths should be the hypotenuse of another right triangle, where the other two sides are equal to the maximum thickness of the 14-foot ladder.
= 0.0710678...So for a 14-foot ladder, if it is thinner than 0.0710678... feet it would pass, but if it was thicker it would not pass.
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If I'm reading this right, the longest possible ladder to fit through the passageway would touch both outer walls of each hallway at a 45 degree angle and will also touch the corner where the hallways meet. This would form two right triangles, and the ladder would form the hypotenuse of both of them. Since the other two angles in each right triangle are both 45 degrees, then the non-hypotenuse sides (forgot what they were called) in each triangle would be the same as well. So the total length of the longest ladder would be
= 14.14213562373...
So yes, the 14 foot ladder should pass through from one hallway to the other.
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Ahh... that pesky unexposed side. Gets me every time.Maybe the slice starts by intersecting the midpoints of two connecting edges and progresses through the opposing side's inverse edges midpoints (crude drawing attached to act as translation).
I'm not sure about the math, but the small face of each piece should be stable enough to act as the base and not tip over?
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When you say, "so that it will melt the quickest", should we assume it refers to the whole cube and not just the slice? Otherwise you could just slice off the smallest possible chunk from a corner.If the slice must be a straight one, then, starting from two adjacent corners, cut diagonally through the cube to the two opposite corners. This should give us the most surface area on each of the two halves and allow for the quickest natural melting of the cube.
Or cut it with fire. That might be quicker.
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Yes. Answer the question. At worst, Alex asks an unanswerable question and you get no money -same as if you had refused. At best, Alex gets to embarrass you with whatever truth you told and you get $1,000,000. (Just make sure it's not in Somalia Shillings, or Monopoly money.)
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It isn't indeed. And English is my second language, so all this time I've thought that it meant the same thing when applied to division.
So, BobbyGo has it then...
Unless superprismatic is right. I didn't even consider that the third number might be less than the primes instead of greater.
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*Tried to edit original answer to fix mistake, but was too late.
Funny Number 1: 42857
142857 * 3 = 428571
Funny Number 2: 15625
15625 = 1252 = 253
(729 = 272= 93 is the first)
Funny Number 3: 4619046190 = (11 * 13 * 17 * 19) + 1
int main() { int x = 20, y = 0; for(x; y<1; x++) { if(x%11==1 and x%13==1 and x%17==1 and x%19==1) { y=1; } } x=x-1; cout<<x<<endl; return 0; }Product of Funny Numbers: 30930700468750
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Funny Number 1: 42857
428571 * 3 = 142857
Funny Number 2: 15625
15625 = 1252 = 253
(729 = 272 = 93 is the first)
Funny Number 3: 46190
Product of Funny Numbers: 30930700468750
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Diameter of intersecting circle = 48 mmRadius of intersecting circle =
= 24-
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Calling Dr. Bayes ... Where is Bushindo when you need him?
being heads biases the probability above 1/2 that she is using the double-headed coin. And more so in the second case of two heads. So the asked probability is greater, in both cases, and more in the second case, than the unbiased probability of 3/4.
I won't, on principle, plug numbers into B's formula, so I ask myself this question: if she flips it once and says it's heads, what is the likelihood she flipped the double-headed coin? Seems it would be 2/3. Similarly if it was heads twice, it would be 5/6.
So, if I'm thinking right,
the first answer would be 1/3 x 1/2 + 2/3 x 1 = 5/6
and the second answer would be 1/6 x 1/2 + 5/6 x 1 = 11/12.
I agree with your first answer, but not with your second...
If the biased coin was flipped once, it is 100% likely that the result was heads. If the balanced coin was flipped once, it is only 50% likely that the result is heads. This gives us 2:1 odds (2/3 chance, just like you said) that the biased coin was flipped in the event that heads was reported.
1/3 x 1/2 + 2/3 x 1 = 5/6 probability of next toss being heads
If the biased coin was flipped twice, it is 100% likely that both resulted in heads. If the balanced coin was flipped twice, it is only 25% likely that the results were both heads. This gives us 4:1 odds (4/5 chance, instead of 5/6) that the biased coin was flipped in the event that both reported flips landed as heads.
1/5 x 1/2 + 5/6 x 1 = 9/10 probability of next toss being heads
Of course, I have a knack for being terribly wrong about these types of things.
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Alice can total the value of all the odd numbered coins and for all of the even numbered coins. If the odd numbered coins are greater, her first pick will be the first coin. If the even numbered coins are greater, her first pick will be the last coin. After that, she will always pick a coin from the side that Bob last picked from.
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My guess is 8.
I think the main thing is that the triangles in each of the two objects should be a different size. If they were the same size, the square pyramid would have a slightly shorter vertex than the tetrahedron. Joining the two triangles of dissimilar size would only cancel the smaller, not both.
Too late to edit:
The last sentence should have read:
Joining two similar triangles with different sizes would only cancel the smaller, not both.
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Play dead. When an eagle comes down to eat you, lasso that bird and fly across the quicksand.
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My guess is 8.
I think the main thing is that the triangles in each of the two objects should be a different size. If they were the same size, the square pyramid would have a slightly shorter vertex than the tetrahedron. Joining the two triangles of dissimilar size would only cancel the smaller, not both.
Paper Scissors Rock
in New Logic/Math Puzzles
Posted
Does the OP want to know what the chances are of you as a player winning two in a row are? Or what the chances are of either player winning two in a row?