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Forum index » Science and Technology » Math » num-analysis
Probability of 1234567890
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Stig Holmquist
science forum beginner


Joined: 30 Apr 2005
Posts: 48

PostPosted: Tue Jul 11, 2006 9:42 pm    Post subject: Probability of 1234567890 Reply with quote

What is the probability of drawing the digits
1-2-3-4-5-6-7-8-9-0 in any oder when sampling
ten times with replacement from an urn with ten
balls numbered 1-2-3-4-5-6-7-8-9-0 .

The probability of getting the same digit every
time would seem to be 1 per 10x10^10. So
to get the above set in any order must be
different. What formula should be used?

Stig Holmquist
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Dave Seaman
science forum Guru


Joined: 24 Mar 2005
Posts: 527

PostPosted: Tue Jul 11, 2006 10:27 pm    Post subject: Re: Probability of 1234567890 Reply with quote

On Tue, 11 Jul 2006 17:42:39 -0400, Stig Holmquist wrote:
Quote:
What is the probability of drawing the digits
1-2-3-4-5-6-7-8-9-0 in any oder when sampling
ten times with replacement from an urn with ten
balls numbered 1-2-3-4-5-6-7-8-9-0 .

The probability of getting the same digit every
time would seem to be 1 per 10x10^10. So
to get the above set in any order must be
different. What formula should be used?

The number of ways of drawing all ten digits, once each, is 10!. The total
number of possible draws is 10^10. The ratio, says Mathematica, is

In[1]:= 10!/10^10

567
Out[1]= -------
1562500

In[2]:= N[%]

Out[2]= 0.00036288



--
Dave Seaman
U.S. Court of Appeals to review three issues
concerning case of Mumia Abu-Jamal.
<http://www.mumia2000.org/>
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Andrew Poelstra
science forum beginner


Joined: 20 May 2006
Posts: 13

PostPosted: Wed Jul 12, 2006 2:41 am    Post subject: Re: Probability of 1234567890 Reply with quote

On 2006-07-11, Stig Holmquist <stigfjorden@hotmail.com> wrote:
Quote:
What is the probability of drawing the digits
1-2-3-4-5-6-7-8-9-0 in any oder when sampling
ten times with replacement from an urn with ten
balls numbered 1-2-3-4-5-6-7-8-9-0 .

The probability of getting the same digit every
time would seem to be 1 per 10x10^10. So
to get the above set in any order must be
different. What formula should be used?


Well, the number of permutations of 1234567890 is 10!

So, you have 10!/(10x10^10)

--
Andrew Poelstra <http://www.wpsoftware.net/projects/>
To email me, use "apoelstra" at the above domain.
"You people hate mathematics." -- James Harris
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Dave Seaman
science forum Guru


Joined: 24 Mar 2005
Posts: 527

PostPosted: Wed Jul 12, 2006 3:23 am    Post subject: Re: Probability of 1234567890 Reply with quote

On Wed, 12 Jul 2006 02:41:04 GMT, Andrew Poelstra wrote:
Quote:
On 2006-07-11, Stig Holmquist <stigfjorden@hotmail.com> wrote:
What is the probability of drawing the digits
1-2-3-4-5-6-7-8-9-0 in any oder when sampling
ten times with replacement from an urn with ten
balls numbered 1-2-3-4-5-6-7-8-9-0 .

The probability of getting the same digit every
time would seem to be 1 per 10x10^10. So
to get the above set in any order must be
different. What formula should be used?


Well, the number of permutations of 1234567890 is 10!

So, you have 10!/(10x10^10)

The number of ways of selecting 10 balls when sampling with replacement
is 10^10, not 10^11.



--
Dave Seaman
U.S. Court of Appeals to review three issues
concerning case of Mumia Abu-Jamal.
<http://www.mumia2000.org/>
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Andrew Poelstra
science forum beginner


Joined: 20 May 2006
Posts: 13

PostPosted: Wed Jul 12, 2006 5:30 am    Post subject: Re: Probability of 1234567890 Reply with quote

On 2006-07-12, Dave Seaman <dseaman@no.such.host> wrote:
Quote:
On Wed, 12 Jul 2006 02:41:04 GMT, Andrew Poelstra wrote:
On 2006-07-11, Stig Holmquist <stigfjorden@hotmail.com> wrote:
What is the probability of drawing the digits
1-2-3-4-5-6-7-8-9-0 in any oder when sampling
ten times with replacement from an urn with ten
balls numbered 1-2-3-4-5-6-7-8-9-0 .

The probability of getting the same digit every
time would seem to be 1 per 10x10^10. So
to get the above set in any order must be
different. What formula should be used?


Well, the number of permutations of 1234567890 is 10!

So, you have 10!/(10x10^10)

The number of ways of selecting 10 balls when sampling with replacement
is 10^10, not 10^11.


10*10^10 is 10^11. (I'm not sure if you factored that into your post or
not).

--
Andrew Poelstra <http://www.wpsoftware.net/projects/>
To email me, use "apoelstra" at the above domain.
"You people hate mathematics." -- James Harris
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Dave Seaman
science forum Guru


Joined: 24 Mar 2005
Posts: 527

PostPosted: Wed Jul 12, 2006 12:21 pm    Post subject: Re: Probability of 1234567890 Reply with quote

On Wed, 12 Jul 2006 05:30:22 GMT, Andrew Poelstra wrote:
Quote:
On 2006-07-12, Dave Seaman <dseaman@no.such.host> wrote:
On Wed, 12 Jul 2006 02:41:04 GMT, Andrew Poelstra wrote:
On 2006-07-11, Stig Holmquist <stigfjorden@hotmail.com> wrote:
What is the probability of drawing the digits
1-2-3-4-5-6-7-8-9-0 in any oder when sampling
ten times with replacement from an urn with ten
balls numbered 1-2-3-4-5-6-7-8-9-0 .

The probability of getting the same digit every
time would seem to be 1 per 10x10^10. So
to get the above set in any order must be
different. What formula should be used?


Well, the number of permutations of 1234567890 is 10!

So, you have 10!/(10x10^10)

The number of ways of selecting 10 balls when sampling with replacement
is 10^10, not 10^11.


10*10^10 is 10^11. (I'm not sure if you factored that into your post or
not).

The correct number of outcomes is 10^10, not 10^11. If you were drawing
11 times with replacement, then it would be 10^11. I don't care if you
write 10^11 as 10x10^10; it's still wrong.



--
Dave Seaman
U.S. Court of Appeals to review three issues
concerning case of Mumia Abu-Jamal.
<http://www.mumia2000.org/>
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