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The problem of the unfinished game
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on 8 May 2016Transcript of The problem of the unfinished game
The problem of the unfinished game
Pacioli, 1494
Divided the stakes in proportion to the number of rounds won by each player, and the number of rounds did not enter his calculations at all.
Proposed that the stakes should be split in the ratio of 5 to 3.
Since the score is 53, the player who has won 5 games takes the 5/8 of the money and the player who has won 3 games takes the 3/8 of the bet.
What would the division be if the score was 50? Would that be fair?
Cardano, 1539
Conclusion
Introducing the topic
Two friends start a six winning round board game.
They place a 20 euro bet (they give 10 euros each).
The game gets interrupted when the score is 53.
QUESTION:
How do they divide the money?
History
This problem is in fact a version of the problem of points.
It is also known as “the unfinished game” or the "division of the stakes".
It is a famous problem that served as the backdrop for the birth of Probability Theory.
It was solved by Blaise Pascal and Pierre de Fermat.
In 1654, Chevalier de Méré asked Pascal to solve a number of problems related to games of chance and the one that drew his attention was the problem of points.
HOW COULD THE GAME PROGRESS?
Tartaglia, 1556
Believed that the solution to this problem is to count how many games each player needs in order to win the whole match.
5 (
Xenia)
 3
(Lyda) (6 wins needed)
Xenia needs 1 game to win
Lyda needs 3 games to win
Total games: 4
Xenia
Lyda
5
3

Xenia:
I should take all 20 euros.
Xenia
Lyda
5  3
Lyda:
Let's take our money back.
Lyda: 0 euros
Xenia: 20 euros
Lyda: 10 euros
Xenia: 10 euros
Xenia
Lyda
5  3
Xenia:
I should get the 5/8 of the bet and Lyda the 3/8.
Xenia
Lyda
5
3

Lyda:
I should take the 1/4 of the money and Xenia the 3/4.
Lyda
Xenia
5
3

Both
: Xenia should get the 7/8 of the money and Lyda the 1/8.
Lyda: 7,5 euros
Xenia: 12,5 euros
Lyda: 5 euros
Xenia: 15 euros
Lyda:
2,5 euros
Xenia:
17,5 euros
Xenia
Lyda
5  3
Lyda: 6,67 euros
Xenia: 13,33 euros
Xenia:
I should take the 80/6 of the bet and Lyda the other 40/6.
Tartaglia constructed a division that takes into account the size of the lead and the length of the game.
Lyda: x
Xenia: x+2/6*20
(Xenia) x+2/6*20 = 40/3
This division would award the two players in the exact same way regardless the distance to victory, that means 53, 42, 31 and 20 all have the same leadsize.
x+x+2/6*20=20
(Lyda) x=20/3
Xenia: (1  1/4) * 20 = 3/4 * 20 = 15 euros
Lyda: (1  3/4) * 20 = 1/4 * 20 = 5 euros
Frances Gorou
Xenia Livieratou
Lyda Milidoni
The Moraitis School
Euromath 2016
Using Pascal's triangle or,even better, the formula we are able to find the fair division for any possible score.
Probability Theory helped Jan de Witt to carry out an accurate analysis of annuities pricing.
Pascal and Fermat's correspondence created our modern view of the future.
The ability to calculate probabilities of future events transformed in practice the field of Statistics. Without the ability to quantify risk, there would be no liquid capital markets and global companies.
The pundits and pollsters who today tell us who is likely to win the next election, make direct use of mathematical techniques developed by Pascal and Fermat.
After 1654 our lives have become much easier due to Probability Theory developed by Pascal and Fermat.
Each player has 1/2 chances of winning so the score could become 63 or 54.
It is reasonable to assume that the probability for each outcome is 1/2.
How many ways (paths) exist that lead Player 1 to victory?
Player 1: Xenia(red)
Player 2: Lyda(green)
53 63
53 54 65
53 54 55 65
Paths for Player 1:
Paths for Player 2:
53
54
55
56
However each path has a different weight for its realization. That weight is the probability of its occurrence.
Path 53 63
Probability
1/2
Path 53 54 64
Probability
1/2*1/2=1/4
Path 53 54 55 65
Probability
1/2*1/2*1/2=1/8
Total probability that Player 1 eventually wins:
1/2+1/4+1/8=7/8
Whereas for Player 2:
Path 53 54 55 56
Probability
1/2*1/2*1/2=1/8
Fair Division: 7:1 that is 7/8 (Player 1) and 1/8 (Player 2).
For the score 53.
Fermat repeated the same procedure for every part score. So for the score 43, where the graph is:
He would compute the probability for each path and add them for every player.
Paths for Player 1:
Path 43 53 63
Path 43 53 54 64
Path 43 53 54 55 65
Path 43 44 54 64
Path 43 44 54 55 65
Path 43 44 45 55 65
Total: 1/4+2*1/8+3*1/16=11/16
Paths for Player 2:
Path 43 44 45 46
Path 43 44 54 55 56
Path 43 44 45 55 56
Path 43 53 54 55 56
Total:1/8+3*1/16=5/16
Now what about 42 or even 10?
And if the score was 113 in a game of 100?
It seems impractical to compute the probability for each player every time from the beginning
Pascal's solution:
Paths for Player 1:
Path 42 52 62
Path 42 52 53 63
Path 42 52 53 54 64
Path 42 52 53 54 55 65
Path 42 43 53 63
Path 42 43 53 54 64
Path 42 43 53 54 55 65
Path 42 43 44 54 64
Path 42 43 44 54 55 65
Path 42 43 44 45 55 56
Total:1/4+2/8+3/16+4/32=26/32
Paths for Player 2:
Path 42 43 44 45 46
Path 42 52 53 54 55 56
Path 42 43 53 54 55 56
Path 42 43 44 54 55 56
Path 42 43 44 45 55 56
Total:1/16+4*1/32=6/32
The depth of the tree equals the maximum games that can be played.
Here: 1 more for Player 1 to reach 5 and 2 more for Player 2 to reach 5.
And another 1 for some player to win the game
Therefore 1+2+1=4 max rounds.
Pascal's triangle computes the total number of paths that lead to a certain point.
Hence, 1+4+6 paths would lead Player 1 to victory, and 4+1 paths for Player 2 from a total of 16.
Probability for Player 1 to win:
Probability for Player 2 to win:
4
(4+1)/2
4
(1+4+6)/2
Generalising this result
r represents the number of rounds Player 1 needs to win.
Here r=64=2
s represents the number of rounds for Player 2 to win.
Here s=63=3
Maximum number of rounds to be played in order to determine a winner is r1+s1+1=r+s1.
Here r+s1=4
Notation:
n!= 1*2*3...(n1)*n (n factorial)
(combinations)
stands for the usual summation for all values of the index from 1 to n.
Then Pascal proposed that the division of the stakes should follow this ratio:
to
(and the probability for each player would have the values above as nominators and 2 as denominator.
r+s1
Let's see how this would work for the score 42.
r = 64 = 2
s = 62 = 4
Max depth: r+s1 = 5
Therefore the probability for each path would be 1/2 =1/32
Number of paths that lead Player 1 to victory: 1+5+10+10=26
Number of paths that lead Player 2 to victory: 5+1=6
5
Using the formula:
and
Nowadays, in order to quantify risk we are able to plan our future events, with far greater precision than formerly.
Probabilities have been helpful in many fields such as Statistics.
Probability Theory was first applied to mortality tables and improved their accuracy using the same data.
In 1709 Nikolaus Bernoulli applied probability theory to legal and court matters. Bernouilli proved the law of "large numbers", which says that large samples are representatives of the entire population.
Full transcriptPacioli, 1494
Divided the stakes in proportion to the number of rounds won by each player, and the number of rounds did not enter his calculations at all.
Proposed that the stakes should be split in the ratio of 5 to 3.
Since the score is 53, the player who has won 5 games takes the 5/8 of the money and the player who has won 3 games takes the 3/8 of the bet.
What would the division be if the score was 50? Would that be fair?
Cardano, 1539
Conclusion
Introducing the topic
Two friends start a six winning round board game.
They place a 20 euro bet (they give 10 euros each).
The game gets interrupted when the score is 53.
QUESTION:
How do they divide the money?
History
This problem is in fact a version of the problem of points.
It is also known as “the unfinished game” or the "division of the stakes".
It is a famous problem that served as the backdrop for the birth of Probability Theory.
It was solved by Blaise Pascal and Pierre de Fermat.
In 1654, Chevalier de Méré asked Pascal to solve a number of problems related to games of chance and the one that drew his attention was the problem of points.
HOW COULD THE GAME PROGRESS?
Tartaglia, 1556
Believed that the solution to this problem is to count how many games each player needs in order to win the whole match.
5 (
Xenia)
 3
(Lyda) (6 wins needed)
Xenia needs 1 game to win
Lyda needs 3 games to win
Total games: 4
Xenia
Lyda
5
3

Xenia:
I should take all 20 euros.
Xenia
Lyda
5  3
Lyda:
Let's take our money back.
Lyda: 0 euros
Xenia: 20 euros
Lyda: 10 euros
Xenia: 10 euros
Xenia
Lyda
5  3
Xenia:
I should get the 5/8 of the bet and Lyda the 3/8.
Xenia
Lyda
5
3

Lyda:
I should take the 1/4 of the money and Xenia the 3/4.
Lyda
Xenia
5
3

Both
: Xenia should get the 7/8 of the money and Lyda the 1/8.
Lyda: 7,5 euros
Xenia: 12,5 euros
Lyda: 5 euros
Xenia: 15 euros
Lyda:
2,5 euros
Xenia:
17,5 euros
Xenia
Lyda
5  3
Lyda: 6,67 euros
Xenia: 13,33 euros
Xenia:
I should take the 80/6 of the bet and Lyda the other 40/6.
Tartaglia constructed a division that takes into account the size of the lead and the length of the game.
Lyda: x
Xenia: x+2/6*20
(Xenia) x+2/6*20 = 40/3
This division would award the two players in the exact same way regardless the distance to victory, that means 53, 42, 31 and 20 all have the same leadsize.
x+x+2/6*20=20
(Lyda) x=20/3
Xenia: (1  1/4) * 20 = 3/4 * 20 = 15 euros
Lyda: (1  3/4) * 20 = 1/4 * 20 = 5 euros
Frances Gorou
Xenia Livieratou
Lyda Milidoni
The Moraitis School
Euromath 2016
Using Pascal's triangle or,even better, the formula we are able to find the fair division for any possible score.
Probability Theory helped Jan de Witt to carry out an accurate analysis of annuities pricing.
Pascal and Fermat's correspondence created our modern view of the future.
The ability to calculate probabilities of future events transformed in practice the field of Statistics. Without the ability to quantify risk, there would be no liquid capital markets and global companies.
The pundits and pollsters who today tell us who is likely to win the next election, make direct use of mathematical techniques developed by Pascal and Fermat.
After 1654 our lives have become much easier due to Probability Theory developed by Pascal and Fermat.
Each player has 1/2 chances of winning so the score could become 63 or 54.
It is reasonable to assume that the probability for each outcome is 1/2.
How many ways (paths) exist that lead Player 1 to victory?
Player 1: Xenia(red)
Player 2: Lyda(green)
53 63
53 54 65
53 54 55 65
Paths for Player 1:
Paths for Player 2:
53
54
55
56
However each path has a different weight for its realization. That weight is the probability of its occurrence.
Path 53 63
Probability
1/2
Path 53 54 64
Probability
1/2*1/2=1/4
Path 53 54 55 65
Probability
1/2*1/2*1/2=1/8
Total probability that Player 1 eventually wins:
1/2+1/4+1/8=7/8
Whereas for Player 2:
Path 53 54 55 56
Probability
1/2*1/2*1/2=1/8
Fair Division: 7:1 that is 7/8 (Player 1) and 1/8 (Player 2).
For the score 53.
Fermat repeated the same procedure for every part score. So for the score 43, where the graph is:
He would compute the probability for each path and add them for every player.
Paths for Player 1:
Path 43 53 63
Path 43 53 54 64
Path 43 53 54 55 65
Path 43 44 54 64
Path 43 44 54 55 65
Path 43 44 45 55 65
Total: 1/4+2*1/8+3*1/16=11/16
Paths for Player 2:
Path 43 44 45 46
Path 43 44 54 55 56
Path 43 44 45 55 56
Path 43 53 54 55 56
Total:1/8+3*1/16=5/16
Now what about 42 or even 10?
And if the score was 113 in a game of 100?
It seems impractical to compute the probability for each player every time from the beginning
Pascal's solution:
Paths for Player 1:
Path 42 52 62
Path 42 52 53 63
Path 42 52 53 54 64
Path 42 52 53 54 55 65
Path 42 43 53 63
Path 42 43 53 54 64
Path 42 43 53 54 55 65
Path 42 43 44 54 64
Path 42 43 44 54 55 65
Path 42 43 44 45 55 56
Total:1/4+2/8+3/16+4/32=26/32
Paths for Player 2:
Path 42 43 44 45 46
Path 42 52 53 54 55 56
Path 42 43 53 54 55 56
Path 42 43 44 54 55 56
Path 42 43 44 45 55 56
Total:1/16+4*1/32=6/32
The depth of the tree equals the maximum games that can be played.
Here: 1 more for Player 1 to reach 5 and 2 more for Player 2 to reach 5.
And another 1 for some player to win the game
Therefore 1+2+1=4 max rounds.
Pascal's triangle computes the total number of paths that lead to a certain point.
Hence, 1+4+6 paths would lead Player 1 to victory, and 4+1 paths for Player 2 from a total of 16.
Probability for Player 1 to win:
Probability for Player 2 to win:
4
(4+1)/2
4
(1+4+6)/2
Generalising this result
r represents the number of rounds Player 1 needs to win.
Here r=64=2
s represents the number of rounds for Player 2 to win.
Here s=63=3
Maximum number of rounds to be played in order to determine a winner is r1+s1+1=r+s1.
Here r+s1=4
Notation:
n!= 1*2*3...(n1)*n (n factorial)
(combinations)
stands for the usual summation for all values of the index from 1 to n.
Then Pascal proposed that the division of the stakes should follow this ratio:
to
(and the probability for each player would have the values above as nominators and 2 as denominator.
r+s1
Let's see how this would work for the score 42.
r = 64 = 2
s = 62 = 4
Max depth: r+s1 = 5
Therefore the probability for each path would be 1/2 =1/32
Number of paths that lead Player 1 to victory: 1+5+10+10=26
Number of paths that lead Player 2 to victory: 5+1=6
5
Using the formula:
and
Nowadays, in order to quantify risk we are able to plan our future events, with far greater precision than formerly.
Probabilities have been helpful in many fields such as Statistics.
Probability Theory was first applied to mortality tables and improved their accuracy using the same data.
In 1709 Nikolaus Bernoulli applied probability theory to legal and court matters. Bernouilli proved the law of "large numbers", which says that large samples are representatives of the entire population.