Real MATHEMATICS – Game #11

Sprouts

Sprouts is a multiplayer game which was created by M. S. Paterson and brilliant J. H. Conway back in 1967. All you need for playing sprouts are just a piece of paper and a pen/pencil.

  • Game starts with 3 dots on a paper:
    20190429_140440.jpg
  • Players take turns and draw lines from one dot to another (a line can be drawn to the same dot as well). Lines don’t have to be straight and a new dot must be placed on each line:
  • Lines can’t cross one another:
    20190429_140912.jpg
  • A dot is called “dead” if it has 3 lines coming out of it. In other words any dot can be connected to at most 3 lines:

    On right: A, B and E have 3 dots. This means A, B and E are all “dead”.

  • Player who draws the last possible line is the winner.

Brussels’ Sprouts

Brussels’ sprouts is a different kind of sprouts game. It is a multiplayer game just like regular sprouts and all it needs are a paper and a pen/pencil as well. But this time game starts with dots that have thorns. Assume that we will start a Brussels’ with two dots with 3 and 4 thorns on them:

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Players take turn to draw lines between thorns. When a player draws a line, he/she should mark a new dot that has two thorns on it:

Just like regular sprouts, lines can’t cross in Brussels’ sprouts. And the player who draws the last line wins the game:

Euler and Sprouts

You might wonder how on Earth I get to mention Euler in a game that was created about 200 years after he passed away. I recommend you to check Euler characteristics article.

Euler says:

Let’s imagine that V dots and E lines (which don’t cross one another) are sitting on a plane. If the number of faces on this shape is F (don’t ever forget to count the whole plane as one face), then the equation

V – E + F = 2

will always be satisfied.

Take a finished Brussels’ sprouts game on hand:

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Find the numbers of the dots, lines and faces:

Apply Euler’s formula:

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Euler will always be right!

Four Colors

Now take any Brussels’ sprouts sheet and color the faces on it. (Neighboring faces have different colors.)

You will see that four different colors will be enough to color any Brussels’ sheet:

One wonders…

  1. At most how many turns can there be in a regular sprouts game that starts with 3 dots?
  2. Is there a winning strategy for sprouts?
  3. Start a Brussels’ sprouts with 3 dots. If each of them has 3 thorns, at most how many turns can there be?

M. Serkan Kalaycıoğlu

Real MATHEMATICS – What are the chances?! #6

How much luck?

Some people talk about how useless mathematics is. Maybe not all but most of these people lack the ability to understand probability of events, and they don’t even have any clue on how much damage they get because of that. Oddly enough whomever you ask, people will tell you how confident they are about probability.

One of the greatest examples for this is a casino game called roulette. If you have been to a casino before, you already know what I am going to talk about. For those who have no idea what roulette is, it is a gambling game in which a ball is dropped on to a revolving wheel with numbered compartments, the players betting on the number at which the ball comes to rest. Numbers on the wheel are between 0 and 36. Out of 37 numbers; 18 of them are colored in red, another 18 are black and only 1 of them (zero) is neither.

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In roulette one of the things you can gamble is color of the number. You can choose either red or black. It is not that hard to realize that the probability of red or black number winning the round is equal to each other and it is 18/37.

Q: Let’s say you are sitting on a roulette table and you observe that the last eight winners were all red. What would you bet on the next round: Red or black?

I asked ten of my friends and got these answers: 2 reds, 6 blacks and 2 “whatever”. Strange thing is nine of them said (without me asking them) that they know the probability is same in each round. It means that their choices were made instinctly, not mathematically.

The answer should be “whatever” as the probabilities stay the same in each round. Previous rounds have no effect on the next round. Except your psychology…

Warning: Gambling is the mother of all evil. Don’t do it!

“Are you a gambler?”

It is a multiplayer game. For “Are you a gambler?”, all you need is a standard dice.

indir (12)

Rules:

  • Players roll the dice in turns.
  • Outcomes of the rolls are added together. Goal is to be the first player who passes 50.
  • Whenever a player rolls 6, round ends for that player. Player also loses all the points he/she got in that round.
  • Players can stop their round whenever you want; as long as they don’t roll 6.

For instance if player A rolls 4 in the first round, he/she will get to choose one of the following two options: Roll the dice for the second time or end the round and add earned points (4 in this case) to his/her total. Let’s say player A decided to roll the dice second time and got 5. Again there will be two options for player A: Roll the dice for the third time or end the round and add (4+5=9 in this case) earned points to his/her total.

What to learn from all this?

There is only one number you avoid to roll and that is 6. This means whenever you roll the dice there is 5/6 = 0,833… (in other words 83%) chance that you will survive your round. Surviving two rolls consecutively has (5/6)*(5/6) = 0,694… probability. It means your chances have dropped to below 70% just after two rolls.

Imagine that you rolled a dice for 10 times. The probability of getting a 6 gets higher. (Almost 60%)

Although if one takes each roll individually probability of avoiding 6 never changes: 5/6. When you consider cases individually, you miss the bigger picture.

Unlikely events turn into most likely events within time. For instance getting a 6 after 100 rolls is almost certain:

Screenshot_20190326-155011_Calculator.jpg
99,99999879% you will get a 6.

Conclusion: If you roll a dice 100 times, it is almost certain that you will get a 6. But if you take a look at each roll, probability of getting a 6 is always 1/6. This is why unlikely event becomes the most likely event within time.

Then I ask you: When do you stop rolling while playing “Are you a gambler?”?

One wonders…

Let’s change “Are you a gambler?” a bit. Assume that:

  • You don’t have to avoid 6 anymore.
  • Whenever a player hits the total of 5, 10, 15, 20, 25, 30, 35, 40 or 45, player loses all of his/her points.

What kind of strategy would you use? What do you think is different from the original game?

M. Serkan Kalaycıoğlu

Real MATHEMATICS – Game #10

Free the Oreos

Free the Oreos is a multiplayer game for everyone aged over 6. Here is what you need for the game:

  • 9 identical squares drawn inside a bigger square:
    20190317_200302
  • 24 matches in order to construct those 9 little squares:
    20190317_200651
  • Placing an Oreo inside each little square:
    20190317_200631

Rules:

  1. Players will remove matches in turns. A player will remove exactly one match whenever it is his/her turn.
  2. In order to free an Oreo inside a square, all the matches of that square must be removed:

    To free the Oreo sitting on top-left, all four sides of the square must be removed.
  3. The player who removed the last side of a square is known as the one who freed the Oreo.

Goal is to free more Oreos than your opponent.

Defense vs. Attack

Attacking Strategy: You should remove the last match surrounding the Oreo in case you want to free it.

Defending Strategy: In case there are two matches around a specific Oreo, you should avoid removing either of those matches. Otherwise your opponent can and will free that Oreo in his/her turn.

Two experienced players will use the defending strategy as long as possible. But in the end the game will be in a position where defending won’t be possible:

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In such situation, game can continue as follows:

At the end of this sample game first player freed 6 Oreos as the second freed only 3.

One wonders…

  1. Can you find a winning strategy?
  2. At most how many Oreos can a player free in a single turn? Show your answer with a sample game.

M. Serkan Kalaycıoğlu

Real MATHEMATICS – Game #9

Worst Kind of Games

Football is the most popular game in the world right now. It has fairly simple rules which are easy to understand. In my opinion it is the greatest game ever. Although there is something which I despise in football: Draw. I really don’t feel right when a football match ends in draw. Nevertheless there is always only one champion which makes football great again.

zidane-juve
Do they look like they will be happy to get a draw?!

A game ending with a draw is something I have issue with. I just can’t understand how a person can enjoy the possibility of a tie. I am sure some (or possibly most) of you disagree with me. I’ve mentioned about this in the previous posts: I am a gamer. I try to turn everything in my daily life into games, and I like to see a winner at all times. For example when I was a child I played tic-tac-toe literally thousands of times. Perhaps you played it too. What is strange is that I would make deliberate mistakes whenever I realized the possibility of the game ending in draw.

Solution is Hex

  • Make a diamond-shaped board out of nine regular hexagons as follows:
    20190308_152834
  • Let one of the opposite sides be red and the other blue:
    rb.jpg
  • One of the players becomes red team as other becomes blue team.
  • Throw a coin in order to choose who goes first.
  • Players can select any unoccupied hexagon.

Goal: Red player tries to build a red road between red sides as the blue one tries the same with blue color.

Hex has a significant difference from tic-tac-toe: There can never be a draw in Hex.

One of the players will always win the game as other one loses.

Example: Red goes first.

Let’s try to draw in Hex. This can only happen if there aren’t any connections between red-red and blue-blue.

Assume that red made his first two moves as follows:

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In order to avoid a red bridge blue must play his/her second move like this:

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Now red must prevent blue to have a connection. But there is a problem here as blue has two options for a win:

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This is why no matter what red does, blue will win the game:

Our goal is to prevent a win. So let’s assume that blue is an amateur at Hex and he/she can’t see this possibility. In this case red will not only prevent the loss, he/she will win the game:

There is no escape from a win.

Randomness

Let’s use randomness to show that there can’t be a draw in Hex. I numbered the hexagon boxes from 1 to 9 as follows:

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Then I placed 5 red and 5 blue papers in a bag and started selecting papers one by one. I placed the first paper in the hexagon box number 1 and so on…

In the end I ended up with this:

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As you can see the blue team won.

I tried the randomness one more time and ended up with red team winning:

Even when the play is random, there will be no draw.

Some History
Hex was first invented by Danish architect and mathematician Piet Hein in 1942. Although it is known that great mathematician John Nash found the same game independently from Hein in 1948. The name Hex was given by the Parker Brothers, the company that popularized Hex as a board game.

One wonders…

Original Hex board is in 11×11 sizes. You can click here for an original Hex board and play the game. Try to think of winning strategies for Hex.

M. Serkan Kalaycıoğlu

Real Mathematics – Game #6

Oreo Placement Game – OPG

Every week Steve and Tanya meet for coffee and play a game in the coffee shop with the Oreos they brought. They take turns placing those Oreos on a napkin that lies on the table before them. While they do that Oreos must not overlap. Also Oreos must stay within the napkin’s boundary: Overflow is forbidden. The player who places the last Oreo wins the game.

Rules:

  • Two players take turns and place one Oreo each time on a napkin.
  • Napkin has a circle shape.
  • Oreos can’t overlap.
  • Oreos can’t overflow the boundary of the napkin.
  • Winner is the player who puts down the final Oreo.

Is there a winning algorithm/strategy for either of the players?

Yes, there is an algorithm Steve and Tanya can use in order to win at OPG. But this algorithm works only for the player who puts down the first Oreo.

Winning Algorithm:

  • Be the player who starts the game.
  • Place your Oreo at the center of the napkin in the first move.
  • For the rest of the game wherever your opponent places his/her Oreo, play your next move with taking its symmetry by the center Oreo.
  • Eventually second player will run out of space on the napkin.

Place it to the center.

Wherever your opponent places his/her Oreo, rotate that location 180 degrees and place your Oreo there.

Continue the same strategy. Eventually your opponent will run out of space and you will win.

Other Shapes

In OPG, you can use the same algorithm with a napkin that has a regular polygonal shape. But first player might need adjustments in certain situations. Assume that you play OPG in a triangular-shaped napkin.

If you can fit even number of Oreos on the napkin, placing the first Oreo on the center of the triangle will be a losing strategy:

When the first Oreo placed on the center and exactly four Oreos fit inside the triangle, you will not win no matter what.

In order to avoid that you must place the first oreo slightly above the center:

Placing the first Oreo slightly above the center will divide the rest of the triangle into two isolated spaces where you can fit one Oreo each. First player will win no matter what happens with this strategy.

For pentagonal napkins you will have to use the same strategy which you used for triangular napkins:

It is possible to fit 6 Oreos on this pentagon. If you start first and place your Oreo on center, you will lose.

However, when you place the first Oreo slightly closer to one of the corners, algorithm will guarantee a win in every single time.

Q: Is there a winning algorithm in OPG when you don’t get to start the game?

Yes, there is. But in order to accomplish that second player must select a shape that is not convex. Let’s say the napkin is shaped as follows:

Here, first player tries to place his/her Oreo such that the triangle will be divided into two isolated places. No matter what the second player does, first player will win.

Second player must decide the shape of the napkin like following:

nankon4

In these situations winner should leave the napkin in two isolated regions such that these regions must have space for only one Oreo. Thus, second player must play his/her second to last move in such way:

No matter where the first player places his/her Oreo, second player will win if he/she leaves two isolated areas on the napkin.

One Wonders…

Would the same algorithm work if Oreos were in squared shapes? Test this with napkins that are in triangular and pentagonal shapes.

M. Serkan Kalaycıoğlu