November 14th, 2009 at 9:05:55 AM
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At the airport, at the luggage carousel. If I have 1 bag my average wait is n/2 bags if there are n bags on the flight.

If I take 2 bags, my wait is going to be longer and with 3, longer still.

I have manged to work out my average wait for 1,2 or 3 bags using a simulation in Visual Basic but don't know how to produce a formula to express the average wait.

Can anyone help, please?

If I take 2 bags, my wait is going to be longer and with 3, longer still.

I have manged to work out my average wait for 1,2 or 3 bags using a simulation in Visual Basic but don't know how to produce a formula to express the average wait.

Can anyone help, please?

November 14th, 2009 at 9:49:37 AM
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Good question. Makes for a good integration problem. As the number of bags approaches infinity, if your number of bags is n, then the expected wait time until your last bag is n/(n+1) times the wait until the last bag. For example, if it takes all bags 15 minutes to come out, and you have 3 bags, then your expected wait time until your last bag is (3/4)*15 = 11.25 minutes.

I think I'll make an "ask the wizard" question about of this one, thanks.

I think I'll make an "ask the wizard" question about of this one, thanks.

It's not whether you win or lose; it's whether or not you had a good bet.

November 15th, 2009 at 9:42:02 AM
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Thanks for the response. That was just what I was looking for.

I appreciate that you qualified the answer with "as n approaches infinity" but it bothers me slightly that the n/(n+1) formula means that, if every bag on the flight is mine, I manage to leave the baggage claim slightly before my final bag appears.

Still, I've had less credible things happen to me at airports.

Thanks.

I appreciate that you qualified the answer with "as n approaches infinity" but it bothers me slightly that the n/(n+1) formula means that, if every bag on the flight is mine, I manage to leave the baggage claim slightly before my final bag appears.

Still, I've had less credible things happen to me at airports.

Thanks.

November 15th, 2009 at 10:04:00 AM
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Yes, I had to assume an infinite number of bags. Integration doesn’t work with discrete distributions. To adjust for the discrete distribution I suggest adding 1/(b+1), where b is the total number of bags. It makes sense to add something, because you have to wait for your bag to come all the way out of the chute. So my revised formula t*[n/(n+1) + 1/(b+1)], t = total time, n = number of your bags, b=total bags. It may not be exactly right, but should be very close.

It's not whether you win or lose; it's whether or not you had a good bet.

January 3rd, 2010 at 3:01:59 PM
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Quote:WizardYes, I had to assume an infinite number of bags. Integration doesn’t work with discrete distributions. To adjust for the discrete distribution I suggest adding 1/(b+1), where b is the total number of bags. It makes sense to add something, because you have to wait for your bag to come all the way out of the chute. So my revised formula t*[n/(n+1) + 1/(b+1)], t = total time, n = number of your bags, b=total bags. It may not be exactly right, but should be very close.

After working on this some more, my answer is [b*combin(n,b)-(sum for i=n to b-1 of combin(i,b)]/combin(n,b)/n

It's not whether you win or lose; it's whether or not you had a good bet.