Coin Conundrum
The chances of a tossed coin falling head or tails is almost 50-50 with an extremely small chance of it falling on its edge.
How will you make a coin so that it has a probability of 1/3 to fall heads, 1/3 to fall tails, and 1/3 on its edge?
How will you make a coin so that it has a probability of 1/3 to fall heads, 1/3 to fall tails, and 1/3 on its edge?
Labels: logic, mathemagic, outside-the-box
posted by Rajesh Lal at
10:31 PM
21 Comments:
1) If you consider a coin to be a cylinder, then make the two faces further apart (making the edge bigger. I don’t know how large you have to make it in comparison to the surface area of the faces (heads/ tails). But at some point, if the surface area of the edge is large enough, then one third of the time it will land there.
2) Make a three sided coin… Make a triangular prism, but instead of having the two triangular faces, condense those faces into a single point (attach two triangular pyramids on top of the faces). So that in the end you have a “coin” with 3 sides each with the probability of 1/3. I know I am not explaining this shape very well…
~Cybersurf~
I think it's impossible, because for it to have exactly 1/3 probability, the shape would need to have 3 equal sides, which is impossilbe.
Flip it over soft mud
ok so picture a cylinder.
the are of the defining face [pi x r^2] must equal the area of the curved surface of the cylinder [2 x pi x r to give perimeter, then times the height of the cylinder which we shall call h]
pi x r^2 = 2 x pi x r x h
pi x r x r = 2 x pi x r x h
cancel the pi and one r from each side and you get
r = 2 x h
so when the radius is equal to twice the height of the cylinder, we have an even playing field so to speak. now to prove it. let the radius or r be 10 and the height or h be 5.
pi x 10 x 10 = 2 x pi x 5 x 10
100pi = 100pi
approx 314.15926 = approx 314.15926
hooray
ps. can someone check this for me?
Assume the coin must be a cylinder. It must have an equal chance of falling on either face or standing straight. Assuming it is likely to touch at any angle when dropped, it must have its center of gravity exactly over the edge when dropped at 30 degrees off the vertical, as this would give two 60 degree arcs for landing on the edge and two 120 degree arcs for landing on either side. If you build a right triangle with the legs from the center of the con on the surface to the edge and to the middle of the coin, the angle between the surface leg and the hypotenuse must be 30 degrees. If you take the hypotenuse to be one, then the surface leg is sqrt(3)/2 and the other leg (from the center of a face to the center of gravity) is 1/2. So the coin's thickness must be 1/sqrt(3) of its diameter. If the coin is 1 inch across, it will be about .577 in. thick.
Make its width = radius of coin.
In this case, it will become cylinder. and will have equal probability for all three possibilities.
width=half the radius
x=width=r/2
area of 1 side=(pi)r^2
area of edge=2(pi)rx, or 2(pi)r(r/2), which is (pi)r^2
idk, spin it!!!
Since we've the freedom to change the shape of the coin, make coins in cubic shapes (like dice). Call opposite faces heads, tails and edges.
My brain just exploded.
roll a dice, if its a 1 or 2, place the coin on heads, 3-4 tails and 5-6 Place it on the edge (by creating a passage that makes the coin land on the edge
~ghostrider655
put it on a table with cracks in it so i could land in on and be on its side =D
1) Rejesh is sneaky and cruel. He tortured everybody's brain.
2) Cybersurf is probably thinking of a pyramid with triangular base (instead of square base). But that wont work, since the "coin" will have 4 sides.
3) Those thinking of equal surface areas of cylinder pls be reminded: IF 3 sides don't have equal geometry surfaces, then the chance of landing on ea side IS NOT equal.
4) Joel has a big brain. soft sticky mud will do. superglue would do btter.
5) Conclusion: GhostRider has biggest brain. a regular dice will do, with 2 "Heads, 2 "Tails", and 2 "Edges". But, it doesnt have to be 12, 34, 56. Could be 14, 25, 36, or any other combo. (Make sure GhostRider's not cheating & dice is not weighted...)
With the cylinder why cant you make an edge with an equal probabilty of landing on its side, equal to that of landing on its head or tail? I do believe you can make an edge thick enough but not too thick. But how do you figure out the perfect thickness??
no..triangular prism...
so the 3 faces are rectantagles..and the two triangles are replaced by pyramids so that it cant be landed on.
~Cybersurf~
In layman's terms... Make it chunkier. :P
Make it not exist. That way, the chances it landing on each side are 0, which is 1/3 of a surface area of 0.
You people have too much time on your hands, especially Daniel lol. But I must agree with the idea of a 6 sided die. Heads would cover 2/6, tails would cover 2/6 and 'the edge' could cover 2/6.
2/6 = 1/3
1. To CyberSurf: Thx 4 ur explanation. Yes, I raise both hands 4 praise & applaud.
Yes, "triangular prism" will do.
2. Ur brain is bigger than GhostRider's (dice).
However, in order to avoid landing prism on tips, both ends have to be beveled, & ea end has 3 triangular tips.
Unfortunately, this gives the "coin" 9 sides.
3. No, I still vote for dice.
4. Daniel, I think Kylie is interested in you..
Shark
I am going to glue or melt some coins together until they will land on the edge 1/3 of the time,1/3 heads and 1/3 tails. I will through it up in the air 10 feet spinning and let it hit the ground and see what happens.
Wow, Daniel i need help with my triginometry home work think u can help me out :-)
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