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Looking for the logic behind this

  1. Aug 15, 2013 #1
    We got a bunch of balls numbered 1 to n. And we got a bunch of bowls numbered also 1 to n.

    What is the chance of ball x hitting bowl x after dropping each ball randomly in a bowl one by one?

    The answer involves saying 1/n after going (n-1)!/n! and I wonder, what is the full logic behind it?

    I think I can get the denominator as a collection of all the shuffling results, but what is the exact logic that leads to the numerator being (n-1)!?
    Last edited: Aug 15, 2013
  2. jcsd
  3. Aug 15, 2013 #2


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    Hey cdux.

    The key thing you have to specify is whether the probability changes after a ball is released (or a specific bowl is hit).

    If the probabilities don't change at all with respect to the above attributes then probability is always the same and under a uniform distribution (all probabilities are likely) then the probability is 1/n for all bowls.

    If the assumptions are different then you will get a different distribution (the above is the simplest case with the easiest assumptions).
  4. Aug 15, 2013 #3
    The question didn't specify so they are probably equiprobable. Being from the beginning of a book, I guess it's certain.

    PS. I'm mainly interested in the logic of (n-1)!/n! (that led to 1/n) rather than going directly to 1/n.
  5. Aug 15, 2013 #4

    D H

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    When placing the kth ball, you have to put it in one of the k remaining empty bowls. You can't drop it on the floor or put it in your pocket. That places a constraint on the placement. You have k-1 choices rather than k choices as to where to place that kth ball. If you don't put it into one of the first k-1 empty bowls you have no choice but to drop it into the last empty bowl.
  6. Aug 24, 2013 #5


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    Try this: if ball x falls in bowl x, how many ways can the other n-1 balls be placed? What fraction is that of the total number of ways of placing n balls?
  7. Aug 24, 2013 #6
    OK I have a mental image of a base of containers shuffling around that nets a result of n! states. Then if you hold into place k ball into container k then you can shuffle the balls only in (n-1)! ways which produces our desired (n-1)!/n! result for classical probability N(A)/N(Ω).

    But there may be a missing link in the logic here because I suspect I was biased by knowing the answer when reaching that interpretation.

    Is there really a concrete way of looking at probability that involves 'holding the desired result into place' while all the rest outcomes 'shuffle' around it?

    It sounds impressive for a movie but I wonder if it has holes. i.e. exceptions.

    The logic of subtracting elements from the numerator group for more successes appears to be correct since the answer is also (n-l)!/n! for l successes.

    Going to the extreme of n successes appears to make it easier to understand since that would net 0!/n! which is what one would expect for the chance of getting an n! combination right.

    If you're going to have 2 failures in matching, it doubles the probability of the contraption, and 3 failures can shuffle around in 6 ways making the probability of that X 6 of the initial. So it might be another case of thinking of failures instead of successes might be more direct.

    ..And you don't want to see the answer for "at least l successes..".
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