Say i have 100 people who secretly chose a number between 1 and 60 each, how do i work out the probability that a particular number will not be picked, and the likely number of numbers that will not be picked (is this latter one even possible to predict) TIA
Prob. of one person not picking a particular number is 59/60 or 0.9833rec Therefore Prob. of 100 people not picking a particular number is (59/60)/100 or 0.009833rec
If 100 people independently choose a single number between 1 and 60 the chances are out of every 1 draw (of numbers 1 - 60) 1.66 people will have chosen that number.......I think 3 draws gives 5 winners Most draws go the other way ,60 people choosing from 1 - 100 that way there is a .6 (60%) chance of a winner - 5 draws gives 3 winners. I could be wrong
Well if 100 people are picking 60 numbers, I'd say they can pick the same number twice otherwise 40 people will be without a number. You're not blond are you?
The above replies are mostly incorrect. Suppose that you want to find out the probability that the number 1 ball will not be picked. The probability that the first person does not pick that ball is 59/60. The probability that a second person does not pick that ball is also 59/60. The probability that the third person does not pick that ball is also 59/60 and so on. The probability that none of the 100 people pick the number 1 ball is (59/60)*(59/60)*(59/60)*.... otherwise written as (59/60)^100=0.1862. I'm a Master of Mathematics, literally.
The second part of your question is quite obvious since there are more people than numbers. The most likely number of balls that will not be picked is 0. I'll try and explain why. We already know that the probability of everyone not picking one ball is 0.1862. The probability that everyone will not pick two particular balls is (58/60)^100=0.0337 The probability that everyone will not pick three particular balls is (57/60)^100=0.00592 The probabilities are quickly tending to 0 and since the sum of probability always equals 1 we have that 1-0.1862-0.337-0.00592=0.77 approximately. Therefore, the most likely number of balls not to be picked is 0.
Thanks Blivy. I understand your answer. It hadnt clicked that i had to do it from the 'not picking' angle rather than the picking angle
another variation, not sure if your second answer covered it. Given there is a 18.62% chance that a particular number does not get picked, what is the probably number of unpicked balls at the end. Is it simply 18.62% of the total number of balls (so 11 out of 60), or is it something totally different, like zero!
My other answer does cover it but you haven't quite grasped the meaning. There is a different probability that at the end, there will be 0 balls left unpicked, 1 ball left unpicked, 2 balls left unpicked etc. If you want to know what the probable number of balls that are left unpicked at the end is, what you're asking for is which of the above has the highest probability which we've already worked out. This means that there's a 0.77 chance that no balls will be left unpicked. There's a 0.1862 chance that 1 ball will be left unpicked. There's a 0.0337 chance that 2 balls will be left unpicked. There's a 0.00592 chance that 3 balls will be left unpicked. This means that the most likely outcome is that NO balls will be left unpicked since it has the highest probability, meaning that all balls ARE picked the majority of the time. This will happen about 3 times for every 4 runs that you do. The probability that 11 of the 60 balls are left unpicked is (49/60)^100=0.00000000016 which is very unlikely.
Almost correct But I bet that because these are humans that are picking, the choices will not be purely random and trends will be seen. That'll bugger up all the maths. People will allus bugger up a fine theory.
But more to the point How on earth has this situation arisen where you find yourself laying odds on a hundred number pickers? What are these 'numbers' and why do you want to get your hands on a particular one?
Re: Almost correct Yep. It's all based on the assumption that the picking of the numbers is uniformly distributed and that if you ran several simulations they were all independent and identically distributed.
Re: But more to the point It wasnt relating to a real-life event. I was wanting to get something straight in my head and it was more the theory that i was after rather than the actual numbers. I had done the calcs at very low levels - 5/6 people but couldnt see the formula for taking it higher. So i just came up with a an example to give it some context.
