Thread: Combinatorial probability.

It is a long time since I have worked on probabilities and to my dismay I can’tquite get my head around how many permutations there are in the formularequired.

I tried the old p (1) X p (2) X p (3) X ... X etc. . . . but I don’t think Iquite have it ..

This is my issue:-

Day one there are 57 spaces and the 57 numbers for these spaces are placedin a hat and drawn. Obviously the probability of drawing a seat is 1 in 57 and seattwo one in 56 as there is one less every time and so on.

What need to know is on Day two what is the probability of the draw repeatingitself exactly as day one?, as in the guy that pulled seat 1 pulls it also fromthe bag day two and all the way through that the guy will pull from the bag thesame seat as the day before ?

Can someone give me the answer and the formula before I get an injury doing this?


Thanks

Tim

Originally Posted by timpe

Day one there are 57 spaces and the 57 numbers for these spaces are placedin a hat and drawn. Obviously the probability of drawing a seat is 1 in 57 and seattwo one in 56 as there is one less every time and so on.[/FONT][/SIZE]

What need to know is on Day two what is the probability of the draw repeatingitself exactly as day one?, as in the guy that pulled seat 1 pulls it also fromthe bag day two and all the way through that the guy will pull from the bag thesame seat as the day before ?

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Lets say you have five numbers in the hat.
There are ways to draw those out of hat.
Say is the order on day one.
The is only one way in 120 for that exact string to appear on day two.
Is that your question?

If not, please try to clarify the question.

Just about

this is an actual event where 57 people draw 57 numbers from a hat, now on day two we repeat this. a member of our society has objected due to the possibility of all 57 people being able to draw the exact same seat number from the hat on day two as day one..

I used to be good at this and thats why i am banging my head to get around it.

Thanks for your time

Tim

Originally Posted by timpe

this is an actual event where 57 people draw 57 numbers from a hat, now on day two we repeat this. a member of our society has objected due to the possibility of all 57 people being able to draw the exact same seat number from the hat on day two as day one.

The probability that every member will get the same seat on day two is . That is almost zero.

Now here is a different question.
The probability that some member will get the same seat on the second day is a bit more that .

so that would be 0.0175438596 so if i was to quote this as the odds then it would be over a million to one or? how do you put a number to that ..

pacience for an old guy (40 ish) that thought he was still good at math... if you dont use it you lose it i am afraid..

Thanks

Tim

Originally Posted by timpe

so that would be 0.0175438596

That is not correct.
57!=4.05269195048772E+76 is

so that is odds of ?? to 1 ?

Tim

Originally Posted by timpe

so that is odds of ?? to 1 ?

A 4 followed by 75 digits. That is a 76 digit number.

Originally Posted by Plato

A 4 followed by 75 digits. That is a 76 digit number.

Yes I had a cobweb removed and got there..

Thanks for your help.

Tim