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Mathematical Theory of Magic Fruits

   

Interesting patterns of fractions

japan deco Munetoshi Sakaguchi
Satoshi Hashiba
Daisuke Minematsu
Kohsuke Morimoto
Naoya Urakawa

Ryohei Miyadera
japan deco 2
 

In the year 1560, many warlords were trying to take control of Japan. At that time, a small region called Owari was ruled by the Oda clan. Powerful warlords in surrounding areas were invading Owari, and people thought that the Oda clan would soon be lost. Nobunaga, who was the young leader of the Oda clan, secretly summoned the bravest samurais in his army.

He told them, "I am going to reveal the utmost secret of our clan. There is a deep and dangerous cave near the shrine of our guardian god. Inside the cave there grows a special fruit tree. Some of its fruits will give you the power to rule Japan. Other fruits are just bitter. There may be more than one magic fruit, but once a person eats a magic fruit and obtains power, the other magic fruits will turn into bitter fruits".

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Problem 1
Only 2 samurais survived to reach the tree. Their name were Hideyoshi and Mitsuhide. The tree had 6 fruits. The two samurai wanted the magic fruit very much, but they were wise enough not to fight for the magic fruit. They took turns and began to eat the fruits. Since Mitsuhide was older, he ate first. If there is only one magic fruit, what is the probability of Mitsuhide's eating it?

Answer
We denote Mitsuhide by 'M' and Hideyoshi by 'H'. Let's say that M is going to eat first. In the first round, the probability of M's eating a magic fruit is 1/6. Let's calculate the probability of M's eating the magic fruit in the third round. The probability of M's not eating the magic fruit in the first round is 5/6 and in the second round H should not eat the fruit. Its probability is 4/5, since there are only 5 fruits, one of which is the magic fruit. There are 4 fruits remaining, including the magic fruit. The probability of M's eating the magic fruit is 1/4. Therefore, the probability of M's eating the magic fruit in the third round is $ \frac{5}{6} \times \frac{4}{5} \times \frac{1}{4}$. As to the probability of M's eating the magic fruit in the fifth round we can do almost the same calculation and we get $ \frac{5}{6} \times \frac{4}{5} \times \frac{3}{4} \times \frac{2}{3} \times \frac{1}{2}$

Therefore the probability of M's getting the magic fruit is $ \frac{1}{6} + \frac{5}{6} \times \frac{4}{5} \times \frac{1}{4} + \frac{5}{6} ...
...s \frac{3}{4} \times \frac{2}{3} \times \frac{1}{2} = \frac{3}{6} = \frac{1}{2}$

We see then that both M and H have the same chance of eating the magic fruit.

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Problem 2
Suppose that we have the same situation as in Problem 1 except that there are two magic fruits among the 6 fruits.

Answer
Clearly, the probability of M's eating the magic fruit in the first round is 2/6, and the probability of his not eating it is 4/6. We can use almost the same method we used in Problem 1, and we come out with an answer of $ \frac{2}{6} + \frac{4}{6} \times \frac{3}{5} \times \frac{2}{4} \times + \frac...
... \frac{2}{4} \times \frac{1}{3} \times \frac{2}{2} = \frac{9}{15} = \frac{3}{5}$

So far we have studied the case of 6 fruits, but we can change the numbers. For example, we can study the case of 100 fruits, where 53 of them are magic fruits. We denote by F[n, m] the probability of the first man's eating the magic fruit when there are n-fruits and m-magic ones among them. For example, from Problem 1 we have F[6,1] = 1/2, and from Problem 2 F[6,2] = 3/5. Similarly we have F[n, m] = $ \frac{m}{n} + \frac{n-m}{n} \times \frac{n-m-1}{n-1} \times \frac{m}{n-2} + \f...
...times \frac{n-m-2}{n-2} \times \frac{n-m-3}{n-3} \times \frac{m}{n-4} + \dotsb $

By this formula you can calculate F[n, m] for any natural numbers n and m.

If you find this formula too difficult to understand, do not worry about it. If you can understand Problem 1 and 2, you can understand the rest of our article. You just have to understand that there is a way to calculate F[n, m] for any n and m. Now we are going to show you a new discovery in mathematics! With F[n, m] for any n and m we make a triangle (see Figure 1).

Figure 1
\begin{displaymath}\begin{array}{c} \text{Figure 1}\\ \\ F[1,1,] \\ F[2,1],F[2,2...
...\\ F[7,1],F[7,2],F[7,3],F[7,4],F[7,5],F[7,6],F[7,6] \end{array}\end{displaymath}

By calculating F[n, m] we obtain a second triangle (Figure 2) from the above triangle. Let's compare these two triangles. F[6,3] is the third entry in the 6th row of the above triangle. In the same position in the triangle below we have 13/20. Therefore F[6,3]= 13/20.

Figure 2
\begin{displaymath}\begin{array}{c} \text{Figure 2}\\ \\ \frac{1}{1} \\ \frac{1}...
...ac{24}{35}, \frac{16}{21}, \frac{6}{7}, \frac{1}{1} \end{array}\end{displaymath}

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Problem 3
Can you find any pattern in Figure 2 above?

Answer
Let's compare Figure 2 to Figure 3 below. The pattern is quite obvious in Figure 3. For example look at 6th row. F[6,2] = 9/15 and F[6,3] = 13/20 the second and third entries in the row. F[7,3] = 22/35 = $ \frac{9 + 13}{15 + 20}$, which is the third entry in the 7th row, reminds us of Pascal's triangle. In general, there exists the same kind of relationship among F[n, m], F[n, m+1], F[n+1, m+1] for any natural number n and m with m < n. We omit the proof.

Figure 3
\begin{displaymath}\begin{array}{c} \text{Figure 3}\\ \\ \frac{1}{1} \\ \frac{1}...
...ac{24}{35}, \frac{16}{21}, \frac{6}{7}, \frac{1}{1} \end{array}\end{displaymath}

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Problem 4
Can you find any other pattern in Figure 2?

Hint
In fact there are several patterns. Please look at the following Figure 4, 5 and 6 below. The patterns are obvious, aren't they?

Figure 4
\begin{displaymath}\begin{array}{c} \text{Figure 4}\\ \\ \frac{1}{1} \\ \frac{1}...
...ac{24}{35}, \frac{16}{21}, \frac{6}{7}, \frac{1}{1} \end{array}\end{displaymath}
Figure 5
\begin{displaymath}\begin{array}{c} \text{Figure 5}\\ \\ \frac{1}{1} \\ \frac{1}...
...ac{24}{35}, \frac{16}{21}, \frac{6}{7}, \frac{1}{1} \end{array}\end{displaymath}
 
Figure 6
\begin{displaymath}\begin{array}{c} \text{Figure 6}\\ \\ \frac{1}{1} \\ \frac{1}...
...ac{24}{35}, \frac{16}{21}, \frac{6}{7}, \frac{1}{1} \end{array}\end{displaymath}

Remark
We can also study the case with more than two people. For example, if 4 people play the game, then the probability of the third person's eating the magic fruit forms the following triangle. Can you find any pattern in this triangle?
Figure 7
\begin{displaymath}\begin{array}{c} \text{Figure 7}\\ \\ \frac{0}{1} \\ \frac{0}...
...frac{4}{35}, \frac{1}{21}, \frac{0}{7}, \frac{0}{1} \end{array}\end{displaymath}

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Thanks
We would like to thank Ms. Carolyn Ashizawa for her help in correcting our English.

We omit the mathematical proof of our result. If you want to know the proof, please download "Elementary Russian Roulette.pdf" that is a PDF file from: http://library.wolfram.com/infocenter/MathSource/5710/

 

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You're encouraged to expand and improve the article featured on this page. Send your comments, feedbacks or suggestions to Dr. Ryohei Miyadera or to contact@archimedes-lab.org.

 
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