Probability with replacement: 256 cards numbered 1, 2, 3, ... , 256 are shuffled....

[Pindar Z]

256 cards have numbers on them 1,2,3,..,256 and the cards are shuffled.

Q1. Assuming 16 cards were selected at random with replacement, what is the probability that two cards out the 16 selections have the same number?
Q2. What is the possible number of selections to be performed in order to find two cards with the same number?

Thanks in advance .

 

[Deleted member 4993]

256 cards have numbers on them 1,2,3,..,256 and the cards are shuffled.

Q1. Assuming 16 cards were selected at random with replacement, what is the probability that two cards out the 16 selections have the same number?
Q2. What is the possible number of selections to be performed in order to find two cards with the same number?

Thanks in advance .

What are your thoughts?

Please share your work with us ...even if you know it is wrong

If you are stuck at the beginning tell us and we'll start with the definitions.

You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:

http://www.freemathhelp.com/forum/th...Before-Posting

 

[pka]

256 cards have numbers on them 1,2,3,..,256 and the cards are shuffled.
Q1. Assuming 16 cards were selected at random with replacement, what is the probability that two cards out the 16 selections have the same number?
Q2. What is the possible number of selections to be performed in order to find two cards with the same number?

I never cease to be amused at novices who makeup mathematics questions.
Do they have any idea how very large $\displaystyle \large 2^{2^7}$ is?

 

[pka]

Love writing simulation programs:
here, I got ~19/50. Sure surprisingly high: me goofed?

It is not a matter of goofing, but rather knowing mathematical test-speak.
The phrase "two cards out the 16 selections have the same number" means two and only two.
Not several pairs each having the same number. Not a number appearing more that once.
BUT it says "a single pair of matching numbers".
Now I agree that in most likely not the intended meaning, but nevertheless it is what it says.
It should have said "at least two equal numbers".

Now run a simulation for the meaning I propose.

 

[Pindar Z]

What are your thoughts?

Please share your work with us ...even if you know it is wrong

If you are stuck at the beginning tell us and we'll start with the definitions.

You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:

http://www.freemathhelp.com/forum/th...Before-Posting

My apologies. The problem is I have a very hard time understanding probability, so I thought if I can get an answer to the question I posted maybe I can begin my journey from there.

 

[Pindar Z]

It is not a matter of goofing, but rather knowing mathematical test-speak.
The phrase "two cards out the 16 selections have the same number" means two and only two.
Not several pairs each having the same number. Not a number appearing more that once.
BUT it says "a single pair of matching numbers".
Now I agree that in most likely not the intended meaning, but nevertheless it is what it says.
It should have said "at least two equal numbers".

Now run a simulation for the meaning I propose.

Thanks for the clarification. What actually meant was a number appearing more than once .

 

[pka]

Thanks for the clarification. What actually meant was a number appearing more than once.

You should have started by making that clear.
There are $\displaystyle \dfrac{256!}{(256-16)!}$ (or permutations of 256 taken 16 at a time) strings of length sixteen from two hundred and fifty six possible choices in which no two are the same value.

There are $\displaystyle 256^{16}-\dfrac{256!}{(256-16)!}$ strings of length sixteen from two hundred and fifty six possible choices in which at least two are the same value.

 

[Ishuda]

You should have started by making that clear.
There are $\displaystyle \dfrac{256!}{(256-16)!}$ (or permutations of 256 taken 16 at a time) strings of length sixteen from two hundred and fifty six possible choices in which no two are the same value.

There are $\displaystyle 256^{16}-\dfrac{256!}{(256-16)!}$ strings of length sixteen from two hundred and fifty six possible choices in which at least two are the same value.

Or, to put it another way: If we want at least 2 cards to be the same, that means 2, 3, 4, ...., 16 cards the same which is everything except 1 card the same and so we can solve our problem by getting the probability of 1 card the same, i.e. no duplicates, and subtracting that from one.

Numerator: Well if we need every thing different, the choices are 256 on the first round, 255 on the second, ..., 241 on the sixteenth round or the
₂₅₆P₁₆=$\displaystyle \dfrac{256!}{(256-16)!}$
given by pka.

Denominator: For the total number of possibilities we have a choice of 256 on the first round, 256 on the second, ..., 256 on the sixteenth round or 256¹⁶ possibilities.

So the probability of having all different is
$\displaystyle \dfrac{_{256}P_{16}}{256^{16}}$
and the probability of having at least two matching is
1 - $\displaystyle \dfrac{_{256}P_{16}}{256^{16}}$ = $\displaystyle \dfrac{256^{16}\, -\, _{256}P_{16}}{256^{16}}$

BTW: To avoid those large numbers, the calculation proceeds as
1 - $\displaystyle \dfrac{256}{256}\, \dfrac{255}{256}\, \dfrac{254}{256}\, ...\, \dfrac{241}{256}$ ~ 0.5696

 

[Pindar Z]

That's just as unclear...
Can you get someone to help you with your English?

Perhaps you should start with something simpler, like:
numbers from 1 to 8
5 draws with replacement

I totally agree. Thanks for the suggestion

 

[Pindar Z]

You should have started by making that clear.
There are $\displaystyle \dfrac{256!}{(256-16)!}$ (or permutations of 256 taken 16 at a time) strings of length sixteen from two hundred and fifty six possible choices in which no two are the same value.

There are $\displaystyle 256^{16}-\dfrac{256!}{(256-16)!}$ strings of length sixteen from two hundred and fifty six possible choices in which at least two are the same value.

Seems like we're dealing with a big very numbers. I think its best if I work with much smaller numbers. Thanks alot

 

[Ishuda]

Seems like we're dealing with a big very numbers. I think its best if I work with much smaller numbers. Thanks alot

What large numbers? In the actual calculation of the number, the largest number dealt with is 1 as shown in the earlier post.

Oh, I remember now, there are just as many numbers between 0 and 1 as between 0 and infinity so one must be very large

 

[Pindar Z]

What large numbers? In the actual calculation of the number, the largest number dealt with is 1 as shown in the earlier post.

Oh, I remember now, there are just as many numbers between 0 and 1 as between 0 and infinity so one must be very large

Thanks a lot sir. Really appreciate the assistance .