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# Select Unique Random Integers

Thu 17 Mar, 2011 12:32 pm
Suppose you have to select 22 random integers. what is the expected number of random integers to be selected to guarantee that you have 22 "different" random integers
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Type: Question • Score: 3 • Views: 2,332 • Replies: 11
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Ragman

1
Thu 17 Mar, 2011 12:47 pm
@sachisri,
Suppose you do the approporiate work on your own homework? Just think of how proud of yourself you'll be. The statistics on people becoming successful who do their own work is enormous.

I would never think of taking that away from you.
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Oylok

1
Thu 17 Mar, 2011 01:05 pm
@sachisri,
When is the assignment due?
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sachisri

1
Thu 17 Mar, 2011 01:14 pm
@sachisri,
I am trying to help another student with this assignment. so I am not sure about the due date. Would appreciate a reply ASAP
engineer

2
Thu 17 Mar, 2011 01:27 pm
@sachisri,
You can't answer this from what you've provided. For example, how big is the group of numbers you are selecting from?
Oylok

1
Thu 17 Mar, 2011 01:30 pm
@sachisri,
I'm not sure what the question is asking...

You have 22 numbers. You sample from them with replacement. You want to know the expected number of times you have to draw a number, before you end up drawing each of them at least once. Is that it? Or are you asking something else?

Have to go, sorry...
sachisri

1
Thu 17 Mar, 2011 01:35 pm
@Oylok,
Yes that is exactly what I am looking for.
Let me rephrase the question..
Ifyou want to select 22 "different" random integers then what is the expected number of random integers that you must select in order to guarantee that you have 22"different" random integers
sachisri

1
Thu 17 Mar, 2011 01:38 pm
@engineer,
If you want to select 22 "different" random integers . what is the expected number of random integers that you must select in order to guarantee that you have 22"different" random integers
The question does not give any parameters or tell us how big the group of numbers is.
Even a formula to find out N number of "different" random integers and the expected number of random integers you have to select to guarentee the above will be sufficient
thanks a lot
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Oylok

2
Thu 17 Mar, 2011 03:42 pm
@sachisri,
Oylok wrote:

You have 22 numbers. You sample from them with replacement. You want to know the expected number of times you have to draw a number, before you end up drawing each of them at least once. Is that it? Or are you asking something else?

sachisri wrote:

Yes that is exactly what I am looking for.

Well, I honestly don't see how the problem you keep giving us is the necessarily the same as the one I gave you just now, but I'll start talking about the problem that I posed for you.

We generally don't give homework answers here (I guess?), but I'll tell you what my approach was:

(1) Start by solving this problem instead: "You have n numbers. You sample from them with replacement. What is the expected number of times you have to draw before you end up drawing each of them at least once?"

(2) Let's give a name to the answer: call it e(n). So e(1) = 1. e(2) = 3, because it takes you one draw to come up with the first distinct number, and from that point it should take you 2 draws on average to get a second distinct number.

(3) Try to find a formula for e(n) in terms of e(n-1). (That's called a "recursive formula" for e(n).)

(4) Guess the formula for e(n) in terms of n, just by looking at the recursive formula, and figuring out creative ways of writing e(3), e(4), etc. If I'm right, the answer turns out to be n multiplied by the nth partial sum of a well-known infinite series.

There's probably a clever, more logical way of doing the problem. But that was my own "hack" approach.
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laughoutlood

2
Thu 17 Mar, 2011 07:22 pm
@sachisri,
If the integers were truly random then duplications could occur.

In sampling without replacement you only need a population of 22 integers from which to select (in a randomised fashion) and expect a guaranteed 22 different integers.
markr

1
Fri 18 Mar, 2011 01:06 am
@laughoutlood,
But if the selections are made from the set of all integers, then the probability of a duplicate selection is zero, and the expected number is 22.
oolongteasup

1
Fri 18 Mar, 2011 01:17 am
@markr,
I expected 22, you too.

If it can happen it will so it's only approximately zero, i can guarantee it.
0 Replies

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