Why do people insist that opposite sides of a die add up to the same number? : dice (4 children) More important than pairing numbers on opposite sides, you need a good spread of numbers so that all the high values aren't all on one side. If a die has more high values on one half than on the other then a player can attempt to do a weak roll of only half a turn, a full turn, 1-1/2 turns, etc to get it to land on the higher value half of the die. An ideal die will have an equal total number of points on both halves no matter where you choose to divide it (not always possible geometrically for dice with fewer numbers of faces, like a D6). A natural byproduct of this goal is opposing faces with a consistent sum of N+1. Some D20s have all the low values adjacent to each other. This is common on dice intended as life counters, such as a Magic: The Gathering counter. It's easier for a player to find the 19 from 20 if similar values are near each other instead of spread out. You generally shouldn't use such dice for fair rolls.

Counters also tend to have opposing faces that equal the same number because ...reasons. Somewhat related: pay attention the next time you see a dart board. The opposite wedges don't add up to a consistent total, but they do exhibit a spread of numbers such that aiming for a large area full of high numbers isn't possible. The 1 is between the 20 and 18, for example. I don't know the precise rationale for their placement, but it's clear they don't want similar values as neighbors. If you're aiming for 7 you shouldn't get 6 or 8 with a near miss. Source: I have no credentials. I just think about the geometry of dice a LOT. Why must opposite faces of dice always add up to 7 (1/6,2/5,3/4)? Would it make any difference if they didn't? | Notes and Queries | guardian.co.uk Why must opposite faces of dice always add up to 7 (1/6,2/5,3/4)? Would it make any difference if they didn't?

Bill Thomson, Watford, Herts The design as a whole is to provide a randomly determined integer from one to six, each of those values being equally likely to militate against concerns that the faces of dice cause a small bias. For a single roll of a fair s-sided die, the probability of rolling each value is exactly 1/s-an of a discrete uniform distribution. For n multiple rolls, with a s-sided die the possibility space is equal to sn. So, for n rolls of an s-sided die the probability of any result is 1/sn. As the number of dice increases, the distribution of the sum of all numbers tends to normal distribution by the central limit theorem. GOTETI MVSR KRISHNA, TADEPALLIGUDEM,ANDHRAPRADESH INDIA If you take the numbers 1,2....6 then add opp faces! You could have 1+2=3. 4+5=9. remember you could have other combination,,this is random, but for the example chosen.......It only leaves,the two numbers we have not used 6,3,,6+3=9 Zen Silvestri, melrose Scotland I want an answer to this Question!

The laws of physics come from an empirical point of view or from a theoretical point of view but they must satisfy the empirical reality if they come from theoretical consideration. So it looks like to me that the empirical came first and from that we get the laws/rules that fit that empirical relationship.If that is the case we can never ""INVENT"" anything all we do is find out what is there and adapt it to fit our needs//NOO?? I look forward to an answer to this DILEMA Zen Zen Silvestri, Melrose Scotland dice - Why do the opposing sides of a D20 generally add up to 21 - Board & Card Games Stack Exchange Why do the opposing sides of a D20 generally add up to 21 up vote 16 down vote favorite 1 Most D20s I own (with the exception of poorly made foam ones or MTG spin-down dice) seem to have all opposing sides add up to 21 (20 + 1, 15 + 6, etc). Why is this? Who started this?

5 You'll notice this on most other dice, too: a d6, a d8 and many others follow the practice of putting the highest number opposite the lowest, the second highest opposite the second lowest, and so on, so that any given side added to its opposite adds up to the same number as any other choice. – doppelgreener Nov 4 '13 at 11:29 D6 always add up to 7 too. You have to arrange the face some way, and having opposite faces add up to N+1 seems to be the traditional way. – Nick Nov 4 '13 at 13:14 up vote 30 down vote accepted Opposite sides add up to 21 on a d20 for the same reason that opposite sides add up to 7 on a d6: if there's a manufacturing or design defect, and the die ends up slightly flatter than intended, then the average result of the die will not change. Consider a d6 that ends up a bit flat: two sides opposite each other will both have an increased chance of being landed on.

But since these two sides' numbers are "opposite" to each other, they will not affect the average of the die. For example, a d6 might have a 1/3 chance of landing on a 1, a 1/3 chance of landing on a 6, and, and a 1/12 chance of landing on each of the other faces. The average of this seriously-deformed die is: 1 * 1/3 + 2 * 1/12 + 3 * 1/12 + 4 * 1/12 + 5 * 1/12 + 6 * 1/3 = 3.5, the same as a normal die. Of course, various other deformities (malformed corner, air bubble in the plastic, some enterprising fellow put a little lead in the die, etc.) will still play merry hob with the probabilities in an unfair way, but a flattened die will be somewhat fair in that its biases favor both high and low numbers equally. The layout for d8, d10, d12, and d20 dice were probably created from this same idea to make the dice somewhat fair if they're flattened.

up vote 7 down vote I may be misinformed, however my understanding of the situation is that this is a result of trying to normalize the value of each region on the d20. The best comparison is between an MTG spindown (where the numbers count down from 20-1 in sequence spiraling around the d20) and a regular d20. A spindown has all the high values clustered at one end, and the low values clustered at the other. with practice, you could easily learn to roll a spindown to get 10 or more the vast majority of the time, for a regular d20 this is impossible as the numbers are spread evenly, so high numbers are surrounded by lower numbers. It is very hard (if not impossible) to roll a d20 for an exact individual side, so spacing them out in this way is the best way to "balance" the values, such that a practiced roll aiming for 20 has a high chance to land on much lower numbers that are adjacent, where a spindown would have other high numbers adjacent.

by laying out the numbers such that all sides add up to 21, you give a memorable way for users to navigate a d20, whilst also giving a reasonable (if not optimal) proportionality in value accross the areas of the d20. as for who started it, I couldnt answer that, d20 themselves have been around since the Romans ( ), but I can find no information on who decided the numbering scheme. up vote 3 down vote Another reason might be that originally 6-sided dice used little holes to mark the numbers. The material removed to create a 6 is more than that to create a 1. As such, the different sides of a die had different weights. To balance the weight across all three directions of the die, opposing numbers were chosen to balance out the different directions. The material taken out of 6 and 1 would be the same weight as the material taken out of 5 and 2, and 4 and 3. All three directions would lose material equal to 7 holes. This, combined with distributing the numbers as far as possible from one another would lead to the current distribution of the numbers on a 6-sided die.

However, on dice with more than 6 sides, the sum rule would not be enough to guarantee that all numbers are as far as possible from each other. Also, on 20-sided dice, numbers are not usually created by holes but using digits. As such, the material argument doesn't hold anymore. up vote 3 down vote The tradition of numbering a d6 with 1 opposite 6, etc. goes back to the Ancient Greeks, if not before. As far as I know, the original reasoning behind that numbering isn't really known. It is, however, a way to improve fairness for a slightly flattened die, as noted above. Other defects such as voids could cause a particular region of a die to preferentially land facing up. In order to minimize the effect of such defects on fairness, dice would ideally have the faces surrounding a vertex (a point where three or more faces come together) summing to the same number. Bob Bosch, Henry Segerman and I have just produced such a numerically-balanced d20 for the first time. For more details, please visit .

Number Gossip: 7 Tanya Khovanova Number Gossip (Enter a number and I'll tell you everything you wanted to know about it but were afraid to ask.) Unique Properties of 7 7 is the smallest number that cannot be represented as a sum of 3 squares 7 is the smallest integer that is not the difference of two primes There are 7 frieze groups, infinite discrete symmetry groups for patterns on a strip (infinitely wide rectangle) The numbers on opposite sides of a regular die always add up to 7 7 is the minimum number of colors that is always sufficient to color any map on a torus 7 is the only prime followed by a cube 7 is the smallest n for which a regular polygon with n sides cannot be constructed with ruler and compass The Fano plane is the projective plane with the least number of points and lines: 7 each (submitted by Qiaochu Yuan) Rare Properties of 7 Dice Dice ... ... The dice have 21 points. The numbers 1 to 6 appear on the six sides of a cube. - You throw the dice by hand or by a leather dice cup. You turn the cup upside down and you let the dice roll to a standstill.

It is by chance which number appears on top. This is the attraction of throwing the dice. The dice are used in many board games. They often determine the playing. They give the game unexpected turns and add excitement to the games. There are 30 Dice top ... ... If you give the numbers 1, 2, 3, 4, 5, and 6 and form all permutations of the six numbers, you get 1*2*3*4*5*6=6!=720 cubes. Now there are also those dice among them, which are the same, because you can get them by turning around 13 axes. There are 24 turnings. Hence you only have 720:24 =30 different cubes. The following drawing illustrates this. (The numbers under the cubes give you the number of turnings.) Appearance of the Standard Die top The die has six sides for six numbers. If you try all possible ways to spread them out, you get 30 dice. - Every die has its reflection. You can find the possibilities if you reverse the numbers beside the 1. ... ...

You can mark every die by a number with six digits: Start with 1, go upwards, go around 1 counter clockwise and add the number that appears opposite the number 1 on the die. - The red die has 123546 for instance. The dice on the left are ordered by these numbers. You find more about the number 30 at my web page Macmahons Coloured Cubes . There is only the red die in the Western World. ... This die has two characteristics: >The sides opposite each other always add to 7. >If you look at the corner with the numbers 1, 2, and 3 they are arranged anticounter clockwise. Most of the dice makers follow these rules. ... ...... There is a way the dice can be different. There are two ways of placing the points of number 2, 3, or 6, which change into themselves by turning through 180°. This leads to eight pictures. The red die is the most frequent. But I also found the green ones. Four puzzles with dice will follow now. Guessing the Sum top ... Build a tower of several dice.

Is it possible to find the sum of the points of the hidden horizontal cube sides? In this case the tower is formed by three dice. Five surfaces would be counted: The under side of the top die, and the top and bottom sides of the two lower dice. Solution: You take 1 at the top and count the number of the lower cubes. Number 6 is opposite the 1 side. You add 7 for each die. The sum is 6+2x7=20. ... You can make a nice cube puzzle, if you cut the die in nine bars. The vertical bars are in the middle layer, the horizontal ones on the outside. Then the puzzle is more interesting. ... ... You can also form a 3x3x3-cube with 9 V-Triominos (drawing) or with Soma cubes. Draw points on them. Rubik's Dice top Rubik's dice is a black plastic dice 7cm x 7cm x 7cm with circular holes instead of points. There are seven white squared tiles with red circles, which you can lay inside the cube. The tiles directly lying at the holes stick to walls inside. You can remove them with two connected sticks.

The tiles close the holes in red or in white. You have found a solution, when all points have become white. You will soon notice, that you can`t solve the puzzle by trying. There are too many possibilities with seven tiles for closing the holes. Moving the tiles is difficult, too. Act like this: You can look inside the cube and can recognize the patterns of both sides of the tiles. Draw them. The tiles show a number with two digits on the top right (meaning?), which can be used as names. There is the following image: Now the solution is simple. If you like, don't read further and solve the puzzle yourself... Solution: Tile 78 (left side) only fits on the six, then tile 65 (left side) only fits on the five, then tile 34 (left side) only fits on the four, then tile 58 (right side) only fits on the three, then tile 12 (right side) only fits on the two, and then tile 14 (left side) only fits on the one. Tile 47 is useless. There is only one solution.

In the meantime I opened the cube with a saw :-( to scan one side of a tile. Besides I was curious to see how the tiles stick inside. The tiles are made of iron. They are held by six ring magnets. They are inside the cube in the centres. You can see their contours from outside. ... ... Give eight dice. Build a 2x2x2 cube, so that the sum of the points on each side is the same. Here is one of 20 736 solutions with the sum 14. You find more at the German magazine <bild der Wissenschaft> 3-1980. More Dice Shapes top The cube belongs to the five Platonic solids. These are solids, which are only formed by regular polygons. They are called tetrahedron, hexahedron (cube), octahedron, dodecahedron, and icosahedron. You can build dice after their model. You can find 4,6, 8,12 or 20 numbers by throwing the dice. The number at the top counts. The tetrahedron is an exception.

The hidden number counts or the one, that is on top as seen in the scan. I found more interesting solids, which are possible to play the dice from the mathematical view. They are convex, are formed by congruent polygons and have the same position related to the centre of gravity. ("Properties of Dice", URL below). There are more dice forms. I chose two of them. ... ... The die d10 is interesting. It has the shape of a double pyramid with a pentagon as the mutual base. It shows the numbers 0 to 9, the digits of the decimal system. So you can find random numbers with several digits by several dice of this kind. ... .... The die in ball form is funny: There is an iron ball inside the die, which only rest in one of the six hollows. So the points lie at the top. ... ... There are many possibilities to find random numbers. You can lay six balls marked with number 1 to 6 in a box similar to the national lottery. You can mix them and then take one without looking at. The chosen number is the random number. If you play the "dice" a second time, you must put back the ball in the box first.

This is the normal model for a die in mathematics to classify the die in a theory. In spite of the inflation of dice forms: The normal d6 is and will stay the standard die. It has several advantages: >The number 6 is not too small and not too large. You can play well with six random numbers. >The normal die rolls best, especially if the corners are rounded. It is common. >You can make them easily, because the cube sides are perpendicular to each other. Escape from the Castle : nrich.maths.org Can you rank these quantities in order? You may need to find out extra information or perform some experiments to justify your rankings. Escape from the Castle Stage: 2 Challenge Level: We were overwhelmed with solutions for this problem - thank you everyone. Many of you sent good explanations for each part. Daniel from Kings School, New Zealand sent a very clear solution: The First Room: To solve this room, you must know that opposite sides of a dice always add up to seven. For example, if someone puts a dice on the floor and asks what number is on the other side, you can immediately tell them that the number is six if the number on the top of the die is one.

Since all opposite sides add up to seven, you must multiply seven by the number of dice there were (three). Because the sum of the numbers on the top of the dice is eight, you must subtract $8$ from $21$ to reveal the sum of the numbers that you can't see. The key to the first room is 13. The Second Room: The first card is $8$. The second card is $5$. The third card is $4$. The fourth card is $7$. The fifth card is $9$. First card and second card add up to $13$ (key to the first room). Second and third card add up to $9$. Third card and fourth card add up to $11$. Fourth card and fifth card add up to $16$. The key to the second room is $9$. The Third Room: The diameter of the circle is $5$cm. Halve this to get the radius ($2.5$cm). The centre point of both circles is $2.5$cm away from the edge of the diagram. Because there are two circles, you must double the distance and subtract that from $9$ (the key to the second room).

$4 \times 4 = 16$ $15$ is the key to room three. The Fourth Room: $6 + 7 + 3 + 8 = 24$ $24 - 10 = 14$ The key to the fourth room is $14$. The Last/Fifth Room: {There are also two special trays used to cook them in. These hold the key number of tarts each. There are enough tarts to fill one tray but not enough to fill the other one as well. } These two sentences suggest that the key number is between $14$ and $28$ (not including $14$ and $28$). $15$ cannot be the number of tarts as $15$ is divided by $4$ to leave a remainder of $3$. It also is divided by $3$ to leave a remainder of nothing. We need the remainder to be three when divided by four and a remainder of one when divided by three. $16$ will leave a remainder of zero when divided by four, so it cannot be the correct number.

$17$ will leave a remainder of one when divided by four, so it cannot be the correct number. $18$ will leave a remainder of two when divided by four, so it cannot be the correct number. $19$ will leave a remainder of three when divided by four and a remainder of one when divided by three, so it must by the number of the key. The key to the last/fifth room is $19$. The Last Question: The keys to all the rooms were $13$ The 13th number of the alphabet is m. The 9th number of the alphabet is i. The 15th number of the alphabet is o. The 14th number of the alphabet is n. The 19th number of the alphabet is s. Mions Unscramble the word above to get Simon. Skippy's real name is Simon. Sam from Shrewsbury House used some slightly different reasoning.

Here is Sam's solution: Room 1: The numbers on the top of the dice add up to $8$ below are possible numbers $1, 2, 5$ - $1, 3, 4$ - $1, 1, 6$ - $2, 2, 4$ - $ 2, 3, 3$ Opposite sides of a die add up to $7$ below are the bottom numbers taking into consideration this fact $6, 5, 2$ - $6, 4, 3$ - $6, 6, 1$ - $5, 5, 3$ - $5, 4, 4$ They all equal $13$ so it doesn't matter which set of numbers is on the dice Therefore the first key is $13$ Room 2: Cards 1 and 2 = $13$ Cards 2 and 3 = $9$ Cards 3 and 4 = $11$ Cards 4 and 5 = $16$ To work out what the number on the last card is we need to find out what the number on every card is.

To do this we need to use trial and improvement The number nine is the key to the second room Room 3: The radius of the circle is half the height of the rectangle $5 \div2$ is $2.5$ Therefore the radius of the circle is $2.5$cm. The distance between the centres is $9$cm - twice the radius so $9 - 5 = 4$. The distance between the circles therefore is $4$cm $4 \times 4 = 16$ The third key is $15$ Room 4: The numbers all have to add up to $15$. So this means that there either has to be the number $8$ and two numbers that add up to $7$ or vice versa. The $8$ must go with the $6$ and $1$ because the one has to go with the two highest even numbers. There cannot be three even numbers because an even + an even + an even = an even.

Therefore, The fourth key number is $8 + 6 + 7 + 3 - 10=14$ Room 5: $4 + 4 + 4 + 4 + 3 = 19$ $3 \times 6 +1 = 19$ The key number for room 5 is $19$ The key numbers are $13, 9, 15, 14, 19$ M, I, O, N, S Simon! Hannah from Leicester High School for Girls gave lots of detail about how she tackled each room's challenge. Here is the Word document which Hannah sent us. I really like the way Hannah checks her solutions. Lilith, Bradley, Emily and Jake from Great Torrington Junior clearly communicated their thinking too. The methods they have used to tackle the problems in the second and fifth rooms are particularly interesting. Here is Lilith, Bradley and Emily's work: Key 1: The top and bottom of a die always add up to $7$, so if you had two dice the top and bottom faces would always add up to $14$, three dice would add up to $21$ and so on.

All these numbers are multiples of $7$. So for the three dice with the top faces adding up to $8$ then the bottom faces must add up to $13$ as $21 - 8 = 13$. Key 2: We made lists of the different combinations for each of the totals. As there are only two possible cards for the last card it is easier to work backwards. If you have 5th = $7$, then the 4th = $9$, 3rd = $2$, 2nd = $7$ as you can only use each card once this does not work so it must be a $9$ for the last card. So the order is: $8, 5, 4, 7, 9$, so the answer is $9$. Key 3: We knew the length of the rectangle was $9$cm and its width was $5$cm. From the bottom of the circles to its centre is half way, which is $2.5$cm, so from the sides to the centre of each circle is $2.5$cm. We took these away from the $9$.

So, $9 - 2.5 - 2.5 = 4$. $4 \times 4 = 16$. $16 - 1 = 15$. Key 4: We worked out all the possible combinations of three numbers from $1$ to $8$, which totalled $15$, then we arranged them in the tiles, we soon found out that you could not have the four highest or lowest numbers for the corners, it had to be a combination. $8, 1, 6$; $6, 2, 7$; $7, 3, 5$ and $3, 4, 8$. We found out that the four corners were $8, 6, 3$, and $7$ this totals $24$, we subtracted $10$ and got $14$. Key 5: We made 2 grids.

We knew the number had to be more than $14$ and less than $28$ ($2 \times 14$) and must appear in both tables so the answer is $19$ Final question: $13$ = M, $9$ = I, $15$ = o, $14$ = n and $19$ = s when rearranged they spell out SIMON which is Skippy's real name.

A special mention should to the following people who also sent in well-explained solutions: Aaminah, Sujan, Ahmaan, Bilal, Hasan and Hassan from Arboretum Primary School; Alexi from St Stephen's Junior School; Olga, Amy, Rachel, Bethany, Adam, Abigail, Vicky, and Jade from St Gile's School Killamarsh; Harley and Edward from Great Barford Lower School; Martin from Sutton Grammar School; Year 6 at Highfield School Liphook; Antony from Gaer Junior School; Luke and Alisha at Deneholm Primary, Tom, Thomas, Joshua, Lauren, Cameron, and Jack from Balksbury Junior School; Joe and Kyle at Rudyard Kipling Primary School; Austin and Rushton from Little Heath School; Year 5 at Aboyne Lodge School; Lily, Jess, Andrew and Josiah from Myton; Year 5 at St. Mary's R.C.

Primary School, Whickham; Leah at Cavendish Close Junior School; Yasmina, Abidah, Tahmina, Azmina, Samuel, Omar, Miraj and Thanbir from Burnley Brow Community School, Oldham; Makhan and James H from Newtown Maths Group Droxford; Grace, Caitlin, Katie, Brad, Jack and Abbie from Sir Thomas Boughey (Staffordshire); Tom and Ruhan at Dobcroft Junior School, Sheffield; Claire and Simon from Lyneham Primary School. How the dice got their spots How the dice got their spots How the dice got their spots Most people know that the opposite sides of a six-sided die always add up to seven. But what is the relationship between all six sides of a die? How do they wrap around each other? Set a die down so the six face is on top and the two is facing you. You automatically know the one is on the bottom, and the five is facing away from you. What you can't automatically know is if the four is on the left or the right. Some dice manufacturers place the four on the left of the dice. Others put the four on the right side.

The same dice company may even change faces for different sizes and styles of its dice. Of course, the three is always on the opposite side of the four. Yellow die has four on the viewer's right side. Middle white die also has a four on viewer's right side. Right white die has a three on viewer's right side. Another way to think about how the spots are arranged on the dice is to view the three sides of the one, two, and three spots. Some people refer to the different layouts as left or right-handed. I prefer to think of the two layouts as clockwise or counterclockwise. The one, two, and three faces run in a counterclockwise direction on the yellow die and a clockwise direction on the white die. There are also some interesting differences in how the spots of individual faces are laid out. For example, some Asian dice place the three spots for the three face in a triangle, rather than the traditional diagonal line. You may also see the pair of spots for the two face in a vertical or horizontal line, rather than a diagonal line. For most six-sided dice the only two faces with spots that are not symmetrical are the two and the three.

Both of these numbers usually have the spots run in a diagonal line on that face of the dice. But in which direction? Do the diagonal lines have positive or negative slopes? It turns out that that there are four different ways to lay out the spots for these two faces. Two of the possibilities are when the line formed by connecting the three spots meets the line formed by connecting the two spots. These two lines can either meet at the corner of the six face or the one face. The green die has the two and three sloping up to meet the one. The red die has the two and three sloping up to meet the six. The lines made by the spots of the two and the three faces don't have to meet at one corner. They can run parallel to each other. Again, there are two different ways in which the two lines can run parallel. Assume the six face is on top and the die is placed so you can see the six, three, and two faces. The lines created by the two and three faces can either be positive or negative slopes. The blue die shows the two and three with positive slopes to meet the six. The white die shows the two and three with negative slopes.

Finally, while the six face is symmetrical those six spots can have two different orientations on that face of the die. The two rows of three spots can run in the same two tracks after bending around to align with the four spots of the four face. The two rows of three spots for the six face can also be placed so they run at a right angle to the four spots of the four face. The six spots on the blue die run across to the four. The six spots on the green die run up to the five. One year the International Jugglers' Association (IJA) held its annual juggling festival at a hotel/casino in Primm, Nevada. You had to walk through the casino to get from one's hotel room to the exhibition hall where the juggling took place. One early morning, about three a.m., I was walking back to my room, through the almost-empty casino. Except for the pit crew, one of the craps tables was empty. Thinking they might know about the history, I asked the three employees about these subtle differences in the possible orientations of spots on dice.

They hadn't given any thought to the idea that the two and three faces could have positive or negative slopes, and they confessed to me that they probably wouldn’t notice the change if someone slipped in a pair of dice that only had that difference. As we were talking their boss walked up. He had overheard just enough of our conversation to explain that he would notice the difference and wanted to know why his employees wouldn't. With the realization that the training manual was soon going to be updated, I thought it an appropriate time to leave. I later contacted a dice manufacturer and was told that there is no reason to explain the difference between various sets of dice. It’s up to the individual designer. Outline Mathematics: A Math trick with two dice A "Math trick" with two dice In introductory algebra or advanced arithmetic classes pupils often face the following problem: Take 2 dice. Dice are made up so that opposite faces add up to 7. Make sure that 6 is opposite 1, the 5 is opposite 2, and the 4 is opposite 3.

Roll the dice and then multiply the top two numbers multiply the bottom two numbers multiply the top of one die by the bottom of the other and now multiply the other top and bottom Now add up all of your (four) answers, and it always adds up to 49. Why? Copyright © 1996-2016 Alexander Bogomolny Solution Take 2 dice. Dice are made up so that opposite faces add up to 7. Make sure that 6 is opposite 1, the 5 is opposite 2, and the 4 is opposite 3. Roll the dice and then multiply the top two numbers multiply the bottom two numbers multiply the top of one die by the bottom of the other and now multiply the other top and bottom Now add up all of your (four) answers, and it always adds up to 49. Why? (In the text below, some words are omitted. These have been underlined. Click just above the line.

See what happens.) First note that each die, a cube,cube,square,triangle,circle,pyramid, has six,four,five,six,seven,eight faces which are paired - a face and its opposite - into three,three,four,six pairs. Let's call a pair of faces relevant to a roll of a die, if this happens to be the top/bottom pair of faces after the roll. Thus, for a roll, one of the three possible pairs is relevant. When two dice are thrown, there are two relevant pairs: one for each die. An important question is, How many different combinations of two relevant pairs are there? There are three,three,four,six possibilities for one die and also three,three,four,six possibilities for the other. In all, there are 3×3 = 9,6,8,9 ways to combine the three pairs from one die with the three pairs from the other. Let's take an example.

Assume one die rolled 6 on the top, the other 2, which means that the first die's relevant pair is 6/1, that of the second 2/5, 1,1,2,3 and 5,4,5,6 being the numbers on the bottom,side,adjacent,bottom faces. Following up on the instructions leads to the following sum: 6×2 + 1×5,3,4,5 + 6×5,3,4,5 + 1×2,1,2,3 which is the sum 12 + 5 + 30 + 2 = 49. Which we should have expected, right? Now, if the first die showed 1 at the top, the relevant pair would be the same and the calculations would be just a little different: 1×2 + 6×5 + 1×5 + 6×2 with the sum 2 + 30 + 5 + 12 = 49. A simple rearrangement of the sum before: the four,four,five,six products are the same, their order in the sum is slightly different.

What is the point of all this pair counting? Well, be patient, you'll see the light shortly. Denote the pairs (of faces) A, B, C. The possible combinations are: A-A, A-B, A-C, B-A, B-B, B-C, C-A, C-B, C-C. The pairs A-B and B-A lead to the same calculations (please verify this), as are the pairs A-C and C-A,A-B,C-B,C-A and B-C and C-B,C-A,A-B,C-B. The conclusion is that, in order to solve the problem, it is only necessary to carry out the calculations for six pairs: A-A, A-B, A-C, B-B, B-C, C-C. Not a big task actually and a good exercise, too. To sum up: in order to solve the problem, all one has to do is verify that in six,two,four,six,eight cases indicated above the result is always 49. This is one solution. A more advanced (one can say algebraic) solution goes like that.

Assume T and B are the top and the bottom numbers on one die, and their lower case counterparts t and b,t and b,t and 2,t and 3,t and t are the top and the bottom numbers on the second die. The instructions lead to the following sum: T×t + B×b + T×b + t×B.

This sum can be modified using the commutative law (both for addition and multiplication): T×t + T×b + B×b + t×B,B×b + t×B,T×b + t×B,B×b + t×T = T×t + T×b + t×B + B×b = T×t + T×b + B×t + B×b,B×b + t×T,B×t + B×b,B×t + T×b Now, the distributive law enters the fray twice: T×t + T×b + B×t + B×b = T×(t + b) + B×(t + b) = (T + B)×(t + b),(T + b)×(t + B),(t + B)×(t + b),(T + B)×(t + b) And finally comes the punch line: recollect that the dice are created so that in any pair of the opposite faces the numbers add up to 7,5,6,7! Therefore, T + B = 7

Question: What number do the numbers on opposite sides of a die always add up to?
Answer: 7