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.

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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:

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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: Patterns #3

Problem of Serkanus

All the participants sit down on a round table that has ascending numbers on it. Last person sitting is the winner and claimer of the magnificent banana cake.

Rules:

Game can have as many players as wanted.

When players sit, each and every one of them gets a number between 1 and the #of players.

Elimination always starts from player 1 and goes clockwise.

Every second player gets eliminated. In other words, if it is your turn, you’ll eliminate the person who is after you in clockwise direction.

Game 1: One player.

It means that there is only one player who is the winner naturally.

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Winner: 1.

Game 2: Two players.

There are two numbers at the table: 1 and 2. Player 1 starts eliminating the player that is on his/her clockwise direction. This means player 1 eliminates player 2.

Since there is no other player left, player 1 claims victory.

Winner: 1.

Game 3: Three players.

There are 1, 2 and 3 sitting at the table. Game starts with player 1 eliminating player 2. Now it is player 3’s turn.

Player 3 eliminates the player that sits on his/her clockwise direction. That means player 1 is eliminated and player 3 is the last one sitting at the table.

Winner: 3.

Game 4: Four players.

Numbers 1, 2, 3 and 4 are sitting at the table. Player 1 starts the game with eliminating player 2. Now it is player 3’s turn and he/she eliminates player 4 who is sitting on the clockwise direction.

It is clear to see that only player 1 and 3 are left at the table and it is player 1’s turn to make a move. Player 1 eliminates player 3 and claims him/herself as the owner of the cake.

Winner: 1.

Game 5: Five players.

Player 1 eliminates player 2 and player 3 eliminates player 4. After these moves, it is player 5’s turn and he/she eliminates player 1 who is sitting at the clockwise direction.

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Finally player 3 and 5 are left alone. Since it is player 3’s turn, player 5 gets eliminated. Hence player 3 wins the game.

Winner: 3.

Game 6: Six players.

Player 1 eliminates 2, 3 eliminates 4 and 5 eliminates 6.

In the second tour there are only 1, 3 and 5 left at the table and it is player 1’s turn.

Player 1 eliminates player 3.

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Now it is player 5’s turn and he/she will eliminate player 1 to claim the righteous owner of the cake.

Winner: 5.

Game 7: Seven players.

Player 1 eliminates 2, 3 eliminates 4 and 5 eliminates 6 which gives player 7 right to eliminate.

Second tour at the table starts with Player 7 eliminating player 1 and player 3 eliminating player 5.

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It the third tour it is again player 7’s turn and he/she eliminates the only player left: Player 5. That means player 7 wins the game.

Winner: 7.

Game 8: Eight players.

In the first tour player 1 eliminates 2, 3 eliminates 4, 5 eliminates 6 and 7 eliminates 8.

Second tour begins with player 1 and players 3, 5 and 7 are the ones who survived the first tour. Player 1 eliminates 3 and 5 eliminates 7.

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Now only player 1 and 5 left at the table and it is player 1’s turn which makes him/her the victor.

Winner: 1.

Game 9: Nine players.

In the first tour player 1 eliminates 2, 3 eliminates 4, 5 eliminates 6 and 7 eliminates 8 which means player 9 will be starting the second tour.

In the second tour player 9 eliminates 1 and 3 eliminates 5. This means it is player 7’s turn.

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Player 7 eliminates player 9 which leaves player 3 and player 7 are the survivors. But since it is player 3’s turn, he/she will win the game with eliminating player 7.

Winner: 3.

Game 10: Ten players.

In the first tour player 1 eliminates 2, 3 eliminates 4, 5 eliminates 6, 7 eliminates 8 and 9 eliminates 10. Player 1, 3, 5, 7 and 9 will advance to the second tour.

Second tour starts with player 1 eliminating 3 and player 5 eliminating 7. Now it is player 9’s turn who will eliminate player 1.

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Finally, player 5 and 9 are left at the table and player 5 eliminates player 9.

Winner: 5.

Pattern

Let’s make a table and find out which player won in the first ten games.

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This table tells us really interesting facts. First of all, you probably realized that there is no chance for an even numbered player to win the game. In fact, they always get eliminated in the first tour. If you are a good observer, you might find two patterns in this table:

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First one is the whenever number of players and winner of the game has the same number, in case you add one more player to the game player 1 will win. Here lies the second pattern which is (if you exclude 1 player game) winners of the games will have odd ascending numbers until number of player will be equal to the number of winner. Check after game 3:

Winner of game 4 is player 1.

Winner of game 5 is player 3.

Winner of game 6 is player 5.

Winner of game 7 is player 7. They are equal so winner will be player 1 in the game 8.

If these two patterns are true, then we can write down on paper who will win in the next games even without playing the game. When it is applied it turns out the winner of game 15 is player 15. If our pattern is correct then winner of game 16 must be player 1.

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Let’s check it out:

In the first tour players 2, 4, 6, 8, 10, 12, 14 and 16 will get eliminated.

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Since it will be player 1’s turn, players 3, 7, 11 and 15 will get eliminated in the second tour.

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Now only players survived are 1, 5, 9 and 13. It is player 1’s turn in the third tour as well, so players 5 and 13 get eliminated.

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In the forth tour only survivors are 1 and 9 with player 1 has the right to start. Player 9 gets eliminated.

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Player 1 claims that he/she is the king of 16 players!

In other words: We have found a pattern that works!

Josephus Problem

This is a famous problem that is named after Flavius Josephus, an important figure in the Jewish history who lived in the 1st century AD.

Story is kind of a myth which was supposedly lived during the Roman-Jewish war. Josephus was trapped in a cave with his 40 soldiers and they were at the mercy of Roman army. Those 41 men had two choices: Surrender or commit suicide. They decided to commit suicide but Josephus his friend thought it was nonsense and therefore he quickly found a solution.

He convinced all men to make a circle. He suggested that when it is his turn; every man should kill the man who is third in the clockwise position. Josephus thought if they could be in the right position, he and his friend would be the last ones survived in the circle.

Question is: Where should Josephus and his friend locate inside the circle of 41 men?

M. Serkan Kalaycıoğlu

Real Mathematics: What are the chances?! #1

In 1987, Thomas M. Cover from Stanford University published an article which contained a bizarre result. Cover suggested the following question:

Player A and Player B will get into a competition.

Player A will be writing down distinct numbers on two pieces of papers, while Player B will be choosing one of the papers and will be reading the number written on it. Then Player B will have to decide if the bigger number is on his/her paper, or on the other one.

Naturally, Player B’s chance of finding the bigger number is 50%. In fact knowing one of the numbers has no use for his chance… or not?

Magic of Mathematics

Is there a strategy for Player B to use so that he/she will have more than 50% chance to win the game?

Although it sounds unlikely, there really is a strategy which Player B could use to win the game with a probability that is greater than 50%.

Strategy: Player B thinks of a number T on his/her mind. Every time Player B selects a paper, he/she uses T for comparison.

  • If T is greater than the selected number, Player B chooses the other paper.
  • If T is smaller than the selected number, Player B chooses the selected paper.

Theory

Let’s assume x and y are the numbers that are written on the papers. (x is greater than y) There are exactly three different scenarios for the game:

  1. T is smaller than x and y. In this case Player B has exactly 50% chance of winning the game.
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  2. T is greater than both x and y. In this case Player B again has 50% chance of winning the game.
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  3. T is greater than y, less than x. In this case if Player B picks y, since T is greater than y, he/she will be choosing the other paper. And if Player B picks x, since T is less than x, he/she will be choosing the paper he/she picked. In both situations Player B wins the game which gives Player B a 100% chance of winning the game.
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An Example

I will try to explain these three scenarios with an example.

Assume T is 80 and x and y are;

  1. 120 and 287,
  2. 1 and 2,
  3. 64 and 15000.

Solutions:

  1. Player B picks either 120 or 287. Since both numbers are greater than T (which is 80), Player B sticks with the paper he/she picks. In this case probability of picking 287 is exactly 50%.
  2. Player B picks either 1 or 2. Since both numbers are less than T, Player B claims the other paper has the greater number on it. In this case probability of picking 1 is again exactly 50%.
  3. Player B picks either 64 or 15000. T is between these numbers. If Player B picks 64, since it is less than T, Player B would pick the other paper. If Player B picks 15000, since it is greater than T, Player B would claim that the paper he/she picked is the greater number. In both cases Player B wins, which gives him/her a probability of 100% for winning the game.