Probability

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jyua015

Hi Everyone, please help me solve this

20 people are invited to pick a random number between 1 and 100. Fin
the probability that, out of the 20 random numbers selected, that ther
are exactly 18 different numbers.

Thank you so much for solving i
 
Here's my attempt, but I'm no mathamatician...

The first person picking a number will have a probability of 100/100 (1/1)
that they will pick a number nobody else has as they are the first. The
second person has a probability of 99/100 that they will pick another number
not chosen previously. The third is then 98/100.. etc.

I guess if you do this through 18 iterations, you have your 18 different
numbers exactly. The last two people then have an 18/100 chance of picking
the same number as the previous people (assuming you are allowing both of
the final people to pick the same number, if not then it would be 18/100 for
the 19th person and 17/100 for the last person)

You probably then have to multiply all the probabilities together:

100/100 x 99/100 x 98/100 x 97/100 x 96/100 x 95/100 x 94/100 x 93/100 x
92/100 x 91/100 x 90/100 x 89/100 x 88/100 x 87/100 x 86/100 x 85/100 x
84/100x 83/100 x 18/100 x 18/100

= 6.36x10^37 / 1x10^40

Take out the exponants to the lowest common denominator is

= 6.36 / 1x10^3, which is 6.36/1000

So the probability would be 157.23 to 1 against that out of 20 people,
exactly 18 will pick different numbers with 2 picking numbers that have
previously been chosen.

Can anyone out there tell me even if I'm on the right track here?

MGS
 
addendum...

It seems that the low probability is mainly the problem of the 2 people who
have to pick only 18 numbers out of 100 rather than a selection of 90 or so.
They've pushed it out so far.

The probability depends on the number of people chosing different numbers
and the number of numbers they can chose from. For instance, if all 20
people had to chose different numbers then the probability would be:

7.69 to 1

If they all have to pick the same number then it would be:

10,000,000,000,000,000,000,000,000,000,000,000,000,000 to 1

This obviously doesn't take into account any psychological issues such as
the fact that a person is less likely to pick the number 1 or 100, single
digit numbers, round numbers, etc. Hence why the birthday problem is more
random.

MGS (again)


MGS shared with us the following:
 
You have calculated the probability of (in order) 18 unique numbers
followed by two duplicates. Why couldn't the result occur in some other
order?

Jerry
 
Should still be the same. There will still be 18 different numbers chosen at
the end which the 2 people have to duplicate regardless of when they pick
them. As the rest of the numbers are unique they will still have the same
probability in a different order:

100, 99, 98, 18, 97, 96, 95, 94, 93, 92, 18...etc

would be the same as

100, 99, 98, 97, 96, 95, 18, 18, 94, 93... etc

This assumes that nobody knows what number the other person has chosen until
the end. If this wasn't the case then the probabilities would be much lower
as everyone would know which numbers they wouldn't have to chose.

If we assume that everyone chooses at the same time, you might say that the
probability would be:

((82^18) x 18 x18) / 100^20

....however this suggests that each person only has 82 numbers to chose from,
not 100.

Jerry W. Lewis shared with us the following:
 
No, there is a mutiplicative combinatoric term that allows for the
reordering. Since this is almost certainly a homework problem, I will
let jyua015 work out the details.

Jerry
 
Good Call

As I said, I'm no mathematician - just testing my mind and trying to think
back to school.

No idea what a multiplicative combinatronic term is :-)

MGS


Jerry W. Lewis shared with us the following:
 
Simplify to see the problem:

3 people are invited to pick a random number between 1 and 3. Find the
probability that, out of the 3 random numbers selected, that there are
exactly 2 different numbers.

Your solution would then be 3/3 x 2/3 x 2/3 = 12/27

But, of the 3^3 = 27 possible outcomes, 18 rather than 12 have exactly 2
different numbers:

112, 113, 121, 131, 211, 311,
221, 223, 212, 232, 122, 322,
331, 332, 313, 323, 133, 233

Jerry
 
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