A Square of Grids

E4
According to Sam Loyd, the 19th century puzzle maker, the following dice game is very popular at fairs and carnivals, but since two persons seldom agree on the chances of a player winning, he offered it as a problem in the theory of probability instead. On the counter are six squares marked 1, 2, 3, 4, 5, 6. Players are invited to place as much money as they wish on any one square. Three dice are then thrown. If your number appears on one die only, you get your money back plus the same amount. Two dice showing your number gets you your money and twice the amount you placed on the square. If your number appears on all three dice, you get your money back plus three times the amount. Of course, if the number is not on any of the dice, the operator gets your money.
Now, a player might reason: the chance of my number showing on one die is 1:6, but since there are three dice, the chances must be 3:6 or 1:2, therefore the game is a fair one. Of course, this is the way the operator of the game wants everyone to reason, because it's quite fallacious. But is the game favourable to the operator or the player, and how favourable is it?
 
DEAR MS
(The problem concerned balancing four types of solids - spheres, cubes, cylinders and cones - against each other, along with the least number of solids that could be used. - MS)
Solid-Solution Dept:
Working out the three given simultaneous relations, we get the ratio of the masses/weights of the cones:cubes:cylinders:spheres as 3:8:13:4. When we put 4 cones in one pan of the balance, we have to put either 3 spheres or 1 cube and 1 sphere in the other pan to restore equilibrium. Weight of 4 cones = 4 x 3 units = 12 units. Weight of 3 spheres = 3 x 4 units = 12 units. Also weight of a cube = 8 units, and weight of a sphere = 4 units. Sum = 12 units. Hence, the least number of solids needed to bring equilibrium when we put 4 cones in one pan is 2 - that is, 1 cube and one sphere.
- Balagopalan Nair,
balagopalannair@gmail.com

(The other problem was about a 3ft wide carbon nanotube ribbon, anchored in Ecuador and stretching 72,000km into space with the centrifugal force of Earth's rotation keeping it taut. The problem was - if I snip off the ribbon at its base, what will happen? - MS)
 
Ribbon-Cutting Dept:
(1) If the carbon nanotube ribbon is not stiff, then it will remain tangential to Earth's surface, and will wrap around the surface along the equator (because Ecuador lies on Earth's equator). (2) If, however, the tube is stiff, and is able to withstand the weight of the house/station at the other end away from the pivotal point at Ecuador, then it will always remain tangential to the equator pivoted at Ecuador. (3) The winds due to Earth's rotational effect will wear out the tube in no time. (4) The tube would remain taut due to Earth's centrifugal force.
- Altaf Ahmed, ctrlaltaf@yahoo.in

(The third problem was a logic one concerning mothers and daughters, but is too long to be repeated here... - MS)
 
A-Mother-Figure Dept:
The required mother/daughter combination is as follows: (1) Mothers: Beth Arnold, Amy Connor, Candice Botti and Dot Densgrove; (2) Daughters: Carol Arnold, Delia Connor, Beverly Botti and Ann Densgrove.
- Saifuddin S F Khomosi,
saif_sfk@hotmail.com
 
ENDGAME
A 3x3 grid has only the number '1' already inserted in the third row, middle square. Complete it by putting 8 different prime numbers in the remaining 8 squares so that the rows, columns and diagonals add up to the same total and it must be the smallest possible total under the conditions. To help you, the number in the middle square is the average of the 2 numbers directly above and below it and the third largest number is not in the right-hand column, and every square contains 1 or 2 digits.

(Mukul can be reached at
mukul.mindsport@gmail.com)

• Mindsport