Real Mathematics – What are the chances?! #8


This year’s cup final will be between two arc-rivals of the country; it is both a final and a derby game. Supporters of both sides are waiting anxiously for the game. Though, Steve has no interest in this game. His team was eliminated in the semi-finals.

Two of Steve’s close friends, Jack and Patrick, have been arguing over the final for the past two weeks. Jack thinks his team FC Ravens will be the winner as Patrick thinks his team AC Bolognese will be victorious. Meanwhile, Steve is thinking about what to eat at supper.


“We haven’t lost to them in years. The cup is 90% ours. But, this is football; anything is possible. Well, that makes 10% anyway. Yeah, I am pretty sure of my team. If you want, let’s bet on it!”


“We are the best team of the season. Plus, our striker scored 69 goals in 27 games. The cup is 75% ours. But, we have been unlucky against them. Which is why I am giving them 25%. Steve, I can bet on it right this second!”

Smart Steve

Steve is aware of the fact that either of these teams will be victorious as this is a cup final and there is no chance for a tie.

He would like to take advantage of his friends’ blindness. Steve would like to set such conditions for the bets such that he would profit whatever the result is.

What would you do if you were in Steve’s situation?

M. Serkan Kalaycıoğlu

Real MATHEMATICS – What are the chances?! #7

How Close?

Game: In a group, everyone is asked to pick a number between 0 and 100. Even though it is possible for more than one person to pick the same number, it is forbidden for participants to communicate with each other.

Winner: Winner is the person who is closest to the two-thirds of the average of the picked numbers.

Question: Is there any way for you to optimize your chance to win the game?

At first glance, one might think that it is not important which number you pick between 0 and 100. Because the winning number depends on the choices of others’. Although, if there is a player who has probability knowledge, he/she could maximize his/her chance for winning the game.

Step #1
Assume that we have a group of 12 people, and every individual selects 100. Then the average becomes:

12*100/12 = 100.

The winning number is the one that is closest to the two-thirds of the average. That means 100*2/3 = 66,666…

66,666… is the highest winning number for this game. If you are aware of this fact; then you would select a number that is between 0 and 66.


Obviously, there is a chance for you to win the game even though you select a number higher than 66. Then again; why would you select such a number if you know that the winning number is between 0 and 66?!

What if everyone realizes…?!

Let’s assume that you are aware of this fact. Then while others will pick a number between 0 and 100, you will be picking a number between 0 and 66. This is a huge advantage. But, suddenly you realized something else: What if everyone came to the same conclusion?

Step #2

If everyone knows that the winning number can’t exceed 66,666…, then no one will choose a number higher than 66. Hence, everyone will choose a number between 0 and 66.

In this situation, the highest average can be 66:

66*12/12 = 66 average.

66*2/3 = 44 is the highest winning number.


This means that if everyone selects between 0 and 66; the winning number can’t exceed 44. Then, why would you choose a number that is higher than 44?!

Step #3

If everyone comes to the same conclusion, then no one within the group will select a number that is higher than 44. This causes a new calculation. Since everyone knows that the winner will be between 0 and 44, the winning number can at most be:

44*12/12 = 44 (average)

44*2/3 = 29,333…

This means that the winning number is at most 29. Then no one will choose a number that is higher than 29.


If one follows the same logic, at the end of the 11th step he/she will find 0 (zero) as a result. This is why picking zero for everyone is the most logical move for the whole group. Using your probability knowledge, one will eventually conclude that zero is the most reasonable choice for each individual.


Mathematics can help a group find a solution that benefits everyone within the group, even though there is no communication inside that group of people.

One wonders…

You know a person inside the group who isn’t good at mathematics. In this situation, would you change your logic? Give your answers using probabilistic calculations.

Ps. You can click here and create yourself an example set.

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.


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)


  • 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:

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 – What are the chances?! #5

Youngest ones will approve this: It is usually youngest kid’s duty to run errands for home. This is why I was the minister of errands until I left home for college. Walking to our local market covered most of my daily duties. On an average day I was walking towards the market more than a few times. On top of that, my school was about 20 meters away from that market. I memorized every centimeter square of that street while growing up.

Actually I wasn’t bothered with this as I was able to create games in any kind of situation in my childhood. For example as I was walking down that street I dribbled with stones. I pretended that every little round stone is a foot ball and I was the famous French footballer Zinedine Zidane.

My favorite game was something I called “walking the line” which I still play by myself.

Walking the Line

In my old neighborhood, floor was tiled with these stones:

In the game of walking the line, goal is to step inside the boundaries of those stones. For every step that crossed the boundary I get -1 score as for each of the successful steps that landed inside the boundaries of the stone I got +1.

Q: Pick any step during a walk. What is the probability of that step being a successful one?


  • Assume that the shape of the foot is a rectangle.
  • Let the foot has sizes 30×6 cm.
  • Assume that the shape of the stone tiling is a square.
  • Let the square has size 60×60 cm.
  • Assume that foot always has the same direction when it lands on the square:


In order to take a successful step, foot can touch the boundary but never cross it. This actually means that the center of the foot (or the rectangle) must be inside a specific area.

Center of the rectangle.

Let’s assume that the top of the rectangle touches the top side of the square. In this case rectangle’s center would be exactly 15 cm away from the side of the square:


If this distance is less than 15 cm, it means that the rectangle crosses the boundary of the square:


Although, when this distance is between 15 and 45 cm, rectangle is considered to be inside the boundary of the square:


One can conclude the same for the lower side of the square.

It is also possible to observe that when the distance between the center of the rectangle and the boundary of the square is less than 3 cm, rectangle crosses the boundary of the square:


Although, when this distance is between 3 and 57 cm, one can know that rectangle is inside the square:


Thus, the probability of taking a random step that is successful (inside the boundary of the square) on a squared-tiling, foot’s/rectangle’s center must be inside the following area S:


Area of the square: 60*60 = 3600 cm2.

Area of S: 54*30 = 1620 cm2.

Probability of taking a successful step on a square:

1620/3600 = 0,45.

This means that for the given measurements it is possible to take a step that will not exceed the boundary of the squared-tiling 45% of the time.


One wonders…

  1. Take a coin and a chess board. Flip the coin on the board. What is the probability of this coin landing inside one of the squares on the chess board?
  2. What are the sizes of the coin and the square when there is more than 50% chance of winning?

M. Serkan Kalaycıoğlu

Real Mathematics – What are the chances?! #4

Coffee of Serkan

Certain days of the week (okay; at least six days a week) I visit a specific coffee shop. Almost all the baristas know what I drink because of my frequent visits… Or do they?

My preferences change every six months. In the period of October-March, I only drink either latte or filter coffee, while in the period of April-September I prefer iced latte or berry.

October-March: In case I drink latte today, there are 80% of chances that I will be drinking latte in the next day. If I drink filter coffee today, chances of me drinking filter coffee tomorrow are 60%.

April-September: If I drink iced latte today, tomorrow I will be drinking iced latte with %80 of probability. For berry that probability is 90%.

Diagram of my coffee selections in October-March period.

Question: If I drank filter coffee this morning, what are the chances that I will drink a latte 2 days later? (We are in February.)

This question resides one of the most crucial findings of mathematics with itself: Markov Chain.

A Markov Chain example. I will be explaining what it is in details inside the article.

It is clear to see that there are two different possibilities to drink latte two days from now. Sum of their probabilities will give us our answer:

Probability of drinking filter coffee (0,6) the next day, and latte (0,4) two days later: 0,6*0,4 = 0,24.

Probability of drinking latter (0,4) the next day, and again latte (0,8) two days later: 0,4*0,8 = 0,32.

Probability of drinking latte two days from now: 0,24 + 0,32 = 0,56.

It means the chance is 56%.

One wonders…

  1. Does it matter which day of February it is today?
  2. Would it change the answer if you learn that I drank filter coffee yesterday? Please elaborate your answer.
  3. If I drink iced latte on June 11th, what is the probability of me drinking berry on June 14th?

Driverless Cars

In case you make a simple web search you will see that there are thousands of pages of articles that question where flying cars are. A few generations including mine have been dreaming about flying cars whenever we were just kids. “Back to the Future” was one of the main reasons why we had such dreams. And it is not like we expect time travel. We just want flying cars!


It is 2019 and there are still no flying cars around. Technology developed as much as making driverless cars only. (Only?!)

Decision making systems are among the key technologies needed for building driverless cars. Because a self-driving car will make hundreds of decisions even if it travels short distances.

Gist of decision making systems is the Markov Chain I mentioned in the Serkan’s Coffee. A concept known as Markov Decision Process is the powerful tool that is being used for driverless cars.

Markov Decision Process (MDP): It is a mathematically formulation for decision and control problems with uncertain behavior.

Memory-less Probability

Markov Chain: If there is Markov Chain inside an event or system, future of that system depends only at the current state of the system; not to its past. And it is possible to predict the future of that system.

One of the examples of Markov Chain is Drunkard’s Walk. Reminder: A drunkard makes random decisions while he/she is trying to find his/her home. Assume that the drunkard had made these moves:


Drunkard’s next move will not depend on the previous moves he/she had made. This only depends on his/her current state and probabilities of the possible moves.

If drunkard will move from the point F, there are four possibilities and none of them depend on previous steps the drunkard took.

It goes the same for driverless cars: Decisions will not depend on the previous ones. For example if a driverless car is heading towards traffic lights its decision will depend on the color of the traffic light; not the left turn it made 200 meters behind.

Markov Chain was found more than 100 years ago and it is being used in economy, meteorology, biology, game theory and even modern technologies like driverless cars and voice recognition systems.

Mathematician Family

The person who gave Markov Chain its name is a Russian mathematician called Andrei Markov. His little brother, Vladimir Markov, was also internationally recognized mathematician. Vladimir died because of tuberculosis at the age of 25. Andrei’s son Andrei Markov Jr. was also a mathematician.

Politics and Andrei

Andrei Markov was involved with politics too. He was not in favor of Romanov dynasty which ruled Russia between 1613 and 1917. He showed his opposition with not participating in the 300th year celebration of Romanov dynasty in 1913. Instead he celebrated 200th year anniversary of the Law of Large Numbers! (I’ll get back to Law of Large Numbers later.)

M. Serkan Kalaycıoğlu

Real Mathematics – Puzzle #1

Crossing the River

A farmer is walking back home with a wolf, a goat and some lettuce. As he was on his way a river appears in front of him. Luckily for him, he sees an abandoned boat. Though it turned out the boat was not big enough for all of them. Farmer had to choose either one of the wolf, goat or lettuce in each time he crosses the river.

But there was a catch here: If they are left alone wolf would eat the goat and goat would eat the lettuce. What is farmer’s strategy in order to cross this river? (P.s: It is not possible to swim across the river.)


The Answer


Between these three only wolf and lettuce can be left alone. Then farmer should start the process with carrying the goat to the right side of the river.


Then he should return the left side and picks up either of the wolf or the lettuce. (Let’s assume that he picks the lettuce.) Farmer reaches the right side of the river with the lettuce. Now before he leaves he realizes that lettuce and goat can’t be left alone. So he takes the goat with him to the left side.


Now, lettuce is alone on the right side of the river as farmer, goat and the wolf are all on the left side. At this point farmer should carry the wolf to the right side and leaves it there with the lettuce as lettuce and wolf can be left alone without a problem.


In the end farmer can go back to the left side of the river and takes the goat to the right side. Hence the problem gets solved.

There wouldn’t be a solution for this puzzle if wolf was eating the lettuce. But, in order to understand that we had to check every possible variation. This would have been a waste of time and energy. Thus we haven’t found an actual method for the solution of this puzzle.

River Algorithm

Graph Theory is a relatively new subject of mathematics. In Graph Theory what matters is the placement and the connection; not the shape. In this part of mathematics objects are described with dots and connections with lines. Then wolf, goat and the lettuce can be represented as dots. Only connection between these three is whether they eat one another or not. In the puzzle goat eats lettuce and wolf eats goat. Hence connections (or lines) should be between them.

If dot disappears, then the connections within disappears as well. In the puzzle we can take only one dot at a time. In the first move we choose the goat. This can be shown in graph as follows:


Deleting the goat dot means to delete the lines connected to it. Graph has no line after the first move which is what we want to accomplish and we managed to do it with deleting only 1 dot. This means another thing: In order to solve the puzzle there must be as many spaces as the deleted dots in the boat.


Let me explain this with an example: Assume that we have to take at least n dots away from the graph in order to get rid of all the lines. This means if boat has n empty spaces the puzzle can be solved.

Let’s add a rabbit to this puzzle.

Second River Puzzle

Now farmer has a goat, a wolf, some lettuce and a rabbit. On top of the previous puzzle rabbit eats the lettuce and wolf eats the rabbit.


In the first puzzle we took goat’s dot and saw that there was no line left in the graph. But in the second puzzle none of the four dots is able to extinguish all the lines of the graph.

This is why in order to extinguish all the lines; we must try to remove at least two dots. For instance:


As seen above when we take goat and the rabbit, all the lines inside the graph disappears. Using graphs we found an algorithm and with that algorithm we know that the boat must have at least 2 spaces in order to solve the second puzzle.

One Wonders…

  1. Solve the second puzzle.
  2. Let’s add carrot, cat and mice to the second puzzle. Goat, rabbit and mice eats the carrot and cat eats the mice. Try to find how big the boat should be and solve the puzzle for that boat.

M. Serkan Kalaycıoğlu

Real Mathematics – Life vs. Maths #4

Where am I?

Mathematicians love to make generalizations. Personally I don’t enjoy that either, but generalizations are very useful in case you’d like to make some mathematical magic. Is there anything cooler than magic?!

Let’s assume (beware of the generalization that is coming towards you) that we had taken N random steps in one dimension. Even though it is very small there still is a possibility that those steps could well be taken into the right hand side (or left). That would have meant that after N random steps, we are standing at +N (or –N). In this case we understand that we are N steps away from the starting point.


If we take half of the N random steps to the right, and other half to the left hand side we would be standing right on the starting point. In that case we would be 0 steps away from the starting point.


These two scenarios are the furthest (N steps) and closest (0 steps) destinations to the starting point after N random steps are taken. Thus, we are finally aware of the fact that after N random steps in one dimension, we have to stop at 0 to N steps away from the starting point.

Q: Is there an algorithm to find out how far we would be to our starting point even before we take a certain number of random steps in one dimension?

For N random steps the answer is the square root of N. For instance if we take 100 random steps in one dimension, we would be √100 = +/- 10 steps away from the starting point.

Click here to learn why it is so.

Now you are wondering: Where and how can we use this information in life?

Place of The Basketball Team

There are 16 teams participating in the Euroleague, which is the most prestigious tournament of Europe. In the regular season of the Euroleague teams get to play with one another twice. In the end of the regular season top 8 teams advance into the playoffs where champion of the season is decided.

2018-19 regular season is still underway.

Let’s assume that you are supporting a team that is average which means your team would like to fight for the top 8 positions. Just before the season starts you look at the calendar and try to guess how many games your team could win in order to stay in this fight: “If we beat Barcelona at home, and Darüşşafaka on both games…”

You really don’t have to do that. Obviously I will show you how you can use mathematics in order to guess how many wins your team should get.

There are two possible outcomes for a basketball game: Win or lose. It doesn’t matter how strong your opponent is, a game will have two outcomes whatsoever.

Similarities with Random Walks

In one dimension we know that there are two outcomes for a random walk: Right or left. And this is why basketball games can be treated as a random one dimensional walk.

In the regular season each team will play 15×2=30 games. This is same as taking 30 random steps in one dimension.

Then the difference between win and loss column after 30 games can be calculated with taking square root of 30.

√30 = 5,47…

We will call it 6 games. Outcome of these 6 games depend on luck. Your team can win or lose each and every one of them. It proves that after 30 games you either won 6 games more than you lost, or you lost 6 games more than you won:



This information we found using one dimensional random walks tells us that if a team wins between 18 to 12 games in the regular season, that team will be fighting for the playoff positions.


These pictures show how two previous regular seasons ended in the Euroleague. As you can see teams that won between 18 to 12 games fought for a playoff spot.

One Wonders…

Try to apply what you learned for one dimensional random walks to a football team that is participating among 18 teams. If it is an average team, what kinds of predictions can you make for the team?

Don’t forget to include the third possibility: Win, lose or draw.

Ps. I didn’t forget about Steve the accountant. We are slowly heading towards the answer.

M. Serkan Kalaycıoğlu

Real Mathematics – Life vs. Maths #4 (Extra)

Q: Is there an algorithm for us to use in order to guess where a one dimensional random walk could end?

Let’s say we took N random steps in one dimension. I have to assign something for these steps so that we can distinguish each and every one of the steps. I’ll call the first step a1. In this case I can call;

second step: a2,

third step: a3,

forth step: a4,

Nth step: aN.

In the previous article I’ve mentioned that there are two possible values for every step: +1 or -1. I’ve also said that probabilities of these values are equal to each other and precisely ½. Now I can use all this information to calculate the average value each random step has.

In order to do that I can use arithmetic mean: Add both numbers together and divide the total into two. Let me show the arithmetic mean of a1 as <a1>. Then we get the following result:


Average value of a random step in one dimension is 0.

<a1> = 0

<a2> = 0

<a3> = 0

<aN> = 0

Summation of all the average values of N random steps would show us how far we are from the starting point. Let u be the distance from the starting point after N random steps. Then:

u = <a1> + <a2> + <a3> + … + <aN> = 0 + 0 + 0 + … + 0 = 0.

This result says that after taking N random steps in one dimension we should expect to stop at the starting point. This is a paradox because even if it is very small there is a possibility that all N steps could have the value (for instance) +1. In that case u would be +N. But our math shows us that if we increase the number of steps we would eventually end up at the starting point.

More Beautiful Mathematics Needed

This answer is not pleasing to neither to us nor to the mathematicians whom are dealing with this problem. Because 0 is only one of the many possibilities: Not the only possibility.

It would be better if we found an interval for the answer. We can use mathematical manipulations to achieve that.

Two possible outcomes for each step is either +1 or -1 and we just assigned a1 to the value of first step. Let me take the square of a1: It will be +1 in both cases. If I take the square of every one of them:

a12 = 1

a22 = 1

a32 = 1

aN2 = 1

Distance from the starting point was assigned to u. Now let’s square u. It will be a relatively long equation as follows:

u2 = (<a12> + <a22> + <a32> + … + <aN2>) + 2 (<a1a2> + <a1a3> + <a1a3> + … + <a1aN> + <a2a3> + … + <a2aN> + … )

First part of the equation makes N.

All the terms in the second part of the equation are equal to each other. That is why it will be enough if I calculate only one of the terms. Let me first calculate a1a2 so that I can take its average and find <a1a2>:

As you can see from the picture second part of the equation makes 0. Then square of the distance from the starting point is u2 = N. Take square root of both sides and we get our answer: +√N and -√N.

This result means that after taking N random steps in one dimension, one would stand at any point between +√N and -√N. For instance if one takes 100 random steps in one dimension, that person would be √100 = +/- 10 steps away from the starting point.

M. Serkan Kalaycıoğlu

Real Mathematics – Life vs. Maths #3

Drunkard’s Walk Back Home

Steve the accountant finished another working week. He usually spends his weekends in peace. But this specific weekend was different: He was supposed to meet his old college mates whom he hasn’t seen for ages. That night they talked about old days, laughed and drank until morning. Steve the accountant has never been a heavy drinker. At the end of the meeting even though he was barely standing he insisted that he can walk back to his home by himself.

indir (11)

Steve the accountant started walking in random directions: “This street looks familiar… Oh that building looks just like mine…”

Q: Can a drunkard make his way home using a random walk?

Random Walk in One Dimension

I have to talk about random walk in one dimension before I answer the fate of Steve the accountant. In one dimension walking path is something we are familiar since we are kids: The number line.

There are two directions on the number line: Right and left.


  • In one dimension one step to right means +1, one step to left means -1.
  • Let’s assume that the probability of choosing right and left is same.
  • Because of the previous assumption taking a step towards right or left has the probability of ½.


Now let’s draw a number line and choose zero as the starting point. First step can be taken towards +1 or -1. Their probabilities are equal: ½.

Taking two steps at once will be a little bit more complicated than taking one step. After taking only one step we concluded that there can be only two possibilities: +1 and -1. But when we try to take two steps at once, there will be four possibilities:

0 -> +1 -> +2

0 -> +1 -> 0

0 -> -1 -> -2

0 -> -1 -> 0.


In every possibility, probability will be equal: ¼.

After the second step, we may be standing on either one of +2, 0 or -2 with probabilities ¼, 2/4 and ¼ in order.

How about three steps?

We already know what the probabilities are after two steps. According to our findings third step can start either at +2, 0 or -2.

  • If we take our step from +2, we can go to +3 or +1. Their probabilities will be half of the probability of +2. Hence it will be 1/8 each.
  • If we take our step from -2, we can go to -3 or -1. Their probabilities will be half of the probability of -2. Hence it will be 1/8 each.
  • If we take our step from 0, we can go to +1 or -1. Their probabilities will be half of the probability of 0. Hence it will be 2/8 each.


Our calculations show that after the third step we could stand on:

+3 with the probability of 1/8.

-3 with the probability of 1/8.

+1 with the probability of 1/8 + 2/8 = 3/8.

-1 with the probability of 1/8 + 2/8 = 3/8.

In case we continue using the same logic, fourth and fifth steps would look like the following:

After 100 steps, final position and its probability is shown as follows:

Coming back to your starting point (which is zero) has the highest probability.

So far we can make these conclusions about a random walk in one dimension:

  1. When we take even number of steps, we stop on an even number. When we take odd number of steps, we stop on an odd number.
  2. As we increase the number of steps probability of stopping around the starting point gets higher.
  3. Previous argument indicates that if we take more steps, probability of returning to the starting point will increase as well.

Game of Random Walk in One Dimension

So far we understood that a random walk in one dimension has two possible outcomes. In order to simulate a one dimensional random walk we can use coin toss since there are only two possible outcomes for a coin toss: Heads or tails.


If we have a fair coin, probability of getting heads or tails will be equal to each other: 1/2. Then let’s assign tails to -1 and heads to +1.

  1. Toss a coin 10 times. Where did you finish your random walk?
  2. Do the same thing for 30 times and compare your result with the previous one.

I could hear you saying: “What about Steve the accountant?”

A little bit of a patience. We’ll get there in the upcoming articles.

M. Serkan Kalaycıoğlu

Real Mathematics – What are the chances?! #3

Thinking of a number

I am thinking of a number from 1 to 10. (It is 6.)

You have to guess this number.

You have only one shot which means that your success rate is 1 in 10. In other words it is 10%.

Let’s reverse this train of thought and only consider rate of failure which is 90%. (100-10=90)

Q: Assume that we have a group of five people. We ask everyone to write down a number from 1 to 10 simultaneously. What is the probability of having at least 2 people writing down the same number?

First answer comes to your mind is 50%, isn’t it? Because 5 out of 10 makes 50%. Although things may not be as they seem in probability theory.

We’ve already talked about this case for two people. Probability was 9/10 or 90%. How about for three people?

For the first person, there are 10 available numbers from 1 to 10: 10/10.

For the second one, there are 9 available numbers from 1 to 10: 9/10.

For the third person, there are 8 available numbers from 1 to 10: 8/10.

Probability of these three cases to happen can be calculated with multiplication of them: (10/10)*(9/10)*(8/10) = 0,72. This means that %72 of the time three people will write down different numbers.

For four people probability can be calculated like this: (10/10)*(9/10)*(8/10)*(7/10) = 0,504. This means that 50,4% of the time four people will write down different numbers.

For five people probability will be: (10/10)*(9/10)*(8/10)*(7/10)*(6/10) = 0,3024. This means that 30% of the time five people will write down different numbers from 1 to 10.

Probability is way above 50%. Are you surprised?

Birthday Paradox

Whenever you meet someone new, the day you were born will be one of the subjects of this first conversation. If there is at least one female in that conversation your horoscope signs will definitely be mentioned. Now think for a second: Have you ever met someone who shared the same birthday with you?

You probably haven’t thought about it since you know that it is a highly unlikely thing. Because if you consider one year as 365 days (sorry 29th of February)this probability is 1 over 365 which makes 0,27%.

Q: How many people do you need in a group to have 50% chance that two of them will have the same birthday?

At first you might think that you would need 365/2 people for that. But this is not the true answer.

This question can be answered with the same logic we used for the thinking of a number game. And when you apply the same method, you will end up with a bizarre result: Only 23 people are needed for such probability!

Let’s start analyzing this result slowly as we did in the thinking of a number game.

For the first person there will be 365 available days out of 365 total days.

For the second person there will be 364 available days out of 365 total days.

For the third person there will be 363 available days out of 365 total days.


For three people, not having same birthday is 99,2%.


For four people it is 98,4%.


For ten people it is 88,3%. As we continue calculating we can see for 23 people it is 49,3%.


This means that 50,7% of the time two people will have the same birthday among a group of 23 people.

This is an astonishing result!

One wonders…

Determine how many people do you need in a group such that 99% of the time you will be able to find two people who will have the same birthdays?

M. Serkan Kalaycıoğlu