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Topic: The grandfather and the golden coins (Read 2827 times) |
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antkor
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The grandfather and the golden coins
« on: Jan 9th, 2014, 1:39pm » |
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A grandfather decides to split an amount of coins equal to his age (in years) to his 3 grandsons. Since he is a mathematician and a backgammon champion, he gathers and tells them: We will play a game. i will give you two regular dices and you in order from the oldest to the youngest will take turns rolling them. the first one who gets the same number on both dices is the winner (for example two 5's, two 3's etc). If none of you make it, then the procedure will start over in the same order until somebody wins. If all of you calculate the theoretical probability each of you has to win the game, then and only then, i will divide these golden coins amongst you, proportionally to your individual probability.
The three grandsons tried hard and finally they got their individual probabilities of winning the game. Then they realised that their grandfather wanted to give each of them an amount of golden coins eqaul to his own age (meaning each grandson's age, not the grandfather's)!!!
What were the ages of the three grandsons and the age of the grandfather?
Notes: the ages are in years and are integer numbers, do not calculate months. Every grandson has a different age, there are no twins. The grandfather has not celebrated his 100th birthday yet.
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| « Last Edit: Jan 9th, 2014, 1:40pm by antkor » |
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jollytall
Senior Riddler
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Re: The grandfather and the golden coins
« Reply #1 on: Jan 9th, 2014, 10:47pm » |
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36, 30, 25, 91, but why is it in hard? Normally I would not categorise something hard, what can be solved in head without any advanced math.
Two solutions:
Whoever's turns it is, has 1/6 chance to win in that step and P probability to win sometime during the game. These numbers do not change, since the "dices have no memory".
In the first round A wins 1/6 probability, 5/6 B gets a chance and wins 5/36. In 25/36 C gets a chance and wins 25/216 chance. 125/216 starts the next round.
First solution
P=1/6+125/216*P ==> P=36/91.
Pa=36/91
Pb=30/91
Pc=25/91.
25, 30, 36 are relative primes, so you cannot get smaller numbers, and as the grandfather cannot be more than 100, so you cannot get larger multiples either.
Second solution
Somebody will win, so
1=Pa+Pb+Pc
1=P+5/6*P+25/36*P
36=36*P+30*P+25*P
P=36/91.
The rest is the same.
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| « Last Edit: Jan 9th, 2014, 10:48pm by jollytall » |
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antkor
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Posts: 30
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Re: The grandfather and the golden coins
« Reply #2 on: Jan 10th, 2014, 12:07am » |
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both of your solutions are correct. I posted it in hard because the source that i got this riddle from had it in the hard section. My personal opinion is that it is not that hard to solve this riddle, however, i did not know how to evaluate its difficulty, so i posted it here following the (possibly wrong) evaluation of the source site. If the riddle is considered medium/easy it can be moved to the suitable section, that is fine by me. Personally, i think it would be better if i had posted it in the medium section in the first place, so my apologies. 
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| « Last Edit: Jan 10th, 2014, 12:08am by antkor » |
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towr
wu::riddles Moderator
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Some people are average, some are just mean.
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Re: The grandfather and the golden coins
« Reply #3 on: Jan 10th, 2014, 5:03am » |
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The proportion of the probabilities of winning each round is the same as the overall proportions. So we have
proportion of wins is 1/6 : 5/6*1/6 : 5/6*5/6*1/6 = 36 : 30 : 25
36+30+25 = 91; done.
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Wikipedia, Google, Mathworld, Integer sequence DB
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UgoLocal02
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Re: The grandfather and the golden coins
« Reply #4 on: Jun 12th, 2014, 1:10am » |
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theri are two solutions to this riddle: Whoever's turns it is, has 1/6 chance to win in that step and P probability to win sometime during the game. These numbers do not change, since the "dices have no memory".
In the first round A wins 1/6 probability, 5/6 B gets a chance and wins 5/36. In 25/36 C gets a chance and wins 25/216 chance. 125/216 starts the next round.
First solution
P=1/6+125/216*P ==> P=36/91.
Pa=36/91
Pb=30/91
Pc=25/91.
25, 30, 36 are relative primes, so you cannot get smaller numbers, and as the grandfather cannot be more than 100, so you cannot get larger multiples either.
Second solution
Somebody will win, so
1=Pa+Pb+Pc
1=P+5/6*P+25/36*P
36=36*P+30*P+25*P
P=36/91.
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