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
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
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