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bobbym

  • 2 years ago

Hi; There are 4 bins. I throw balls at them until there are 3 balls in one of the bins. Then I stop. The probability of hitting a bin are all the same, 1 / 4. What is the expected number of throws?

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  1. Edutopia
    • 2 years ago
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    prob of hitting one bin is 1/4, times how many bins = 1/16 times three = 1/48...

  2. bobbym
    • 2 years ago
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    Hi rajathsbhat; Thanks for replying but 64 throws is too high. You can not distribute more than 9 balls in the 4 bins without there being 3 in one bin. This implies that that 9 throws is the maximum. Of course you can do it in a minimum of 3 throws. So the expected number of throws is somewhere in between 3 and 9. I should have been clearer about the fact that the balls stay in the bins after each throw.

  3. rajathsbhat
    • 2 years ago
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    Oh my goodness, you're right!

  4. rajathsbhat
    • 2 years ago
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    Sorry.

  5. bobbym
    • 2 years ago
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    Hi; That is okay, this problem has killing me for a week now. I should have been clearer.

  6. rajathsbhat
    • 2 years ago
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    No, you couldn't have have been any clearer. I mean, if the balls were removed as soon as they fell into a bin, there is no way that you can end up with three balls in one bin...i wonder what i was thinking when i was typing that reply....

  7. bobbym
    • 2 years ago
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    Hi; I have a computer program that gives the exact answer but I can not figure out how to do it.

  8. rajathsbhat
    • 2 years ago
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    maybe if the question was something like "what is the probability of getting three balls into one bin in consecutive throws, regardless of how many balls are in the bin already", my answer would be right.

  9. rajathsbhat
    • 2 years ago
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    computer program huh? that's so cool..

  10. bobbym
    • 2 years ago
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    Hi; It is good for getting answers when you need them but sometimes you need the math.

  11. rajathsbhat
    • 2 years ago
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    yeah lol

  12. bobbym
    • 2 years ago
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    Hi; Thanks for looking at the problem. If you get anything... See ya.

  13. mathmate
    • 2 years ago
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    What did you get using your computer program, did you get approx. 4.3 throws?

  14. Edutopia
    • 2 years ago
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    wost case ide say its 36, suposing there would have to be 2 balls in all 4 bins before one would have 3 in it

  15. mathmate
    • 2 years ago
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    With 9 balls, it is a certainty.But they need to fill only any one bin, so we don't need to multiply by 4 to get 36.

  16. Edutopia
    • 2 years ago
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    |dw:1356470323583:dw| it takes 4 thoughts to gaurentee that there will be a ball in a given bin, that times 4 means that you can expect, worst case to have a ball in each bin by the time you throw 16 balls, times 2 for 2 balls in each bin and you have 32, then + four to make sure that one has at least 3

  17. Edutopia
    • 2 years ago
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    *throws, lol

  18. mathmate
    • 2 years ago
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    The worst case is 9 throws, as you have illustrated, 2 in each bin, and the 9th can land in any bin to get 3 balls. The best case is 3 all in the same bin. The answer is somewhere in-between.

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