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Sorry to post this really simple question about probability here but my hands are forced here.My little brother came home with a question which I needed to help him solve. The only problem here is that I started been confused myself.

here is the problem statement. there are 4 players of marbles, and 20 marbles in total. the rule of the game is that he who manage to get the absolute maximum of marbles is the winner(absolute in the sense that no other player should have the same number of marbles).

Question 1 what is the minimum number of marbles should one player have to be assured of the victory.

Question 2 what is the probability that the player with that number of marbles will win

so to me the question 1 is kind of logical and common sense.one player should have at least 11 marbles to be assured. So I got that because it's kind of obvious to me but I don't know how to explain to my brother how I end up with that number 11.

the second question is completely confusing for me.I know it's not permutation because there is not order of any arrangement here. it's not just a combination.

I really need help here. Thanks for reading this

EDIT i think i didn't understand the topic very well, the question 2 was really about the probability of any of them been assured of winning. That as well i feel like it's 1/4 but just in case i would like to cross check with you.Thanks for those who quickly responded

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Regarding the edit: do you mean the probability that there is a player with an absolute majority, if the marbles are distributed at random? –  Brian M. Scott Jan 7 '13 at 23:01
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Regarding the edited question for part two: are you looking for (1) the probability of the existence of a player with absolute majority (6 or more marbles) - or (2) if we call the players A, B, C, and D, are you looking for the probability that a particular player, say player A, will have an absolute majority (will win), or (3) are you looking for the probability that (any) player obtains 11 marbles (guaranteed of winning)? –  amWhy Jan 8 '13 at 0:38
    
You did not specify any rules of the game, just a (sufficient?) condition for somebody to be the winner. There is just no way to give a reasonable answer with no information. –  Marc van Leeuwen Jan 8 '13 at 15:31

2 Answers 2

If you have $10$ marbles or fewer, it’s possible that another player has all the rest, in which case you have not won: that player has at least as many as you have. If you have $11$, however, there are only $9$ other marbles, so even if one player has all of them, you win.

If you have $11$ marbles, you are certain to win, so your probability of winning is $1$.

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Hi thanks for your input. I just edited the question –  black sensei Jan 7 '13 at 22:57

(1) You are correct: The minimum number of marbles to be ASSURED of the victory = $11$ marbles.

  • Then the total number of marbles that the other three players can have is $20 - 11 = 9$. No matter how to split up $9$ among three players, even if one player has all the other 9 marbles, $11 > 9$, ($11 > 20/10 = 10$). So a player with 11 marbles is guaranteed to win...

(2) So the probability that the player with $11$ marbles who is ASSURED of victory will win = $1$

  • After all, the rules of the game are "he who manage to get the absolute maximum of marbles is the winner" (that means WIll certainly win, not might win.)

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Hi thanks for your comments , i just edited the question.thanks –  black sensei Jan 7 '13 at 22:57

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