random number generator between 1-100 what is the probabiity of any number repeating at least once in 20 attempts?

I don't like to give flat out answers, so I'll just say to first simplify the question to see the logic. Answer this question: "What is the probability of any random number (1-100) repeating in two attempts?" So for any number out of one hundred, there is a .01 (1%) chance of hitting it. You have to run through the tree of possible outcomes: P(Miss-Miss) = .99 * .99 = .9801 P(Miss-Hit) = .99 * .01 = .0099 P(Hit-Miss) = .01 * .99 = .0099 P(Hit-Hit) = .01 * .01 = .0001 The chance for hitting it two attempts is .01 * .01. = .0001. So 1 in 1000. Now, "What is the possibility of any random number (1-100) repeating at least twice in three attempts?" In three "rolls", there are the following possible outcomes: P(Miss-Miss-Miss) = .99 * .99 * .99 = .970299 P(Miss-Miss-Hit) = .99 * .99 * .01 = .009801 P(Miss-Hit-Miss) = .99 * .01 * .99 = .009801 P(Miss-Hit-Hit) = .99 * .01 * .01 = .000099 P(Hit-Hit-Hit) = .01 * .01 * .01 = .000001 P(Hit-Hit-Miss) = .01 * .01 * .99 = .000099 P(Hit-Miss-Hit) = .01 * .01 * .99 = .000099 P(Hit-Miss-Miss) = .01 * .99 * .99 = .009801 Total probability of the above (all of P(outcome)) added together is 1.0. Therefore, our probability tree is good. So the probability in three attempts of hitting any number at least twice is: P(MHH) + P(HHH) + P(HHM) + P(HMH) = .000099 + .000001 + .000099 + .000099 = .000298 = 1 in 298,000 Now, there is a pattern here. I won't tell you how to solve it completely, but I will say that to answer the "What is the possibility of any random number (1-100) repeating at least twice in 20 attempts?"... start to think about it in the reverse: Total probability of all outcomes minus probability of hitting any random number only once in 20 attempts and minus the probability of not hitting it at all in 20 attempts: T(all P) - P(1 hit in 20 attempts) - P(0 hit in 20 attempts) = 1.0 - P(1 hit in 20 attempts) - P(MMMMMMMMMMMMMMMMMMMM) Now the question is... what is P(1 hit in 20 attempts)? Well, that would be P(HMMMMMMMMMMMMMMMMMMM) multipled by the 20 ways you can hit that outcome, right? I'll let you figure the rest out...

Let me take a stab at this. Probability that the number A, where 1<=A<=100, occurs in any attempt is p=1/100. Number of attempts = n=20. You want the number A to appear at least twice in the 20 attempts. Therefore, number of occurrences required is k>=2. Occurrence is also sometimes referred to as a success. Number of occurrences, K, is a random variable that follows a binomial distribution with parameters n=20 and p=1/100 and can take the values k=0,1,...,20. P(K>=2) = 1 - [P(K<2)] = 1 - [P(K=0) + P(K=1)]. P(K=0) and P(K=1) can be obtained using the probability mass function (pmf) for binomial distribution. Hope this is correct and it helps.

Prob of repeating at least once = 1 - (prob(it does not occur at all) + prob(it occurs exactly once)) For binomial distribution, here p = prob of success = 0.01, q = prob of failure = 0.99. Prob of getting exactly r successes in n trials = (nCr)*p^r*q^(n-r) So the twp probs are: (20C0)*0.01^0*0.99^20 = 1*1*0.99^20 = 0.8179 and (20C1)0.01^1*0.99^19 = 20*0.01*0.99^19 = 0.1652 So, answer is 1 - (0.8179 + 0.1652) = 0.0169

I won't just *give* someone an answer. I prefer for someone to at least do a little looking... follow the breadcrumb trail... and find the answer.

I’d like to take a stab at this, and I think it’s easiest to find the odds of the opposite happening and subtract that from one. The way I see it, we have 20 independent trials then and want the likelihood of 20 unique numbers. On the first trial the number will definitely be unique, so 100/100 odds; for the second any number but the one drawn in the first trial will do, so it’s 99/100; for the third 98/100; etc., and for the 20th it’s 81/100. Since the trials are independent, we just multiply these 20 factors through and get a ratio with 100^20 in the denominator. In the numerator we have 100x99x98…x81, which is basically 100!/80!. Plug it into a spreadsheet, and I get p = 1 – (100!/80!)/100^20 = 86.96% This better be right.:neutral:

First figure the probability of failure. The chance that the 2nd # does NOT match the first is 99/100. Then the 3rd # must avoid both of these, with a chance of 98/100, and so on. This looks like 100/100 * 99/100 *... * 81/100. Which is the same as the fraction 100 P 20 / 100^20, which is approximately equal to .13039. If that is the failure rate, the success rate is approximately .86961. Yes, this is the same as the math to the "birthday problem". There's also a way to estimate the answer. This might be needed because these large numbers are too big for many calculators. Consider, how many pairing occur in the 20? That is, how many ways are there to select 2 numbers out of 20? That's 20 C 2, or 190. So a decent estimate is .99^190 = .1481. 1 -.1481 = .8519. Not a bad estimate. (The P in 100 P 20 is for permutation, and that button is getting to be fairly common on calculators now. And 20 C 2 is for combinations, if your calculator does permutations it also does combinations, and the two buttons are probably right next to each other.)

Yeah, 100!/80! would be another way of getting the product of the integers 81 through 100. Good job!!

Thanks, those are very very large numbers actually, and I am sure there is a proper mathematical way to simplify that ratio. But Excel and even the Google search window apparently have no trouble with 160-digit numbers, so why bother haha. I couldn't crank 100! on my HP 12C calculator, it can handle up to 100 significant figures, now that I test it.

Since .99^100 is about 1/e, you can use the "e" button to estimate answers on the cheaper calculators. For example, the classic birthday problem says, "If 23 people are in a room, what is the chance that 2 share a birthday?" The exact way is 365 P 23 / 365^23. But the estimate way is: since 23 C 2 = 253. We could type (364/365)^253 = .4995, and 1 - .4995 = .5005, over 50%! Since 253/365 = .693, if a calculator can't handle this, e^-.693, (which is .50007) can still give a quick estimate. How do you do this with the google search window?

I was thinking there has to be some exact simplification for n!/k! that would have been used before Excel/Google calculator were widely available. Alas these supercomputing powers breed human laziness! As for the Google search window, just try typing expressions into it and see if you get a numeric answer. E.g., if I type "100!/80!/100^20", I get 0.130399502 from Google search. I mean Excel is more reliable, but Google search is great when you just need something very quickly, I use it for quick currency conversions all the time too!

For me, it wasn't the most worthwhile course, but it was very important. I think statistical modeling and analysis is the most useful and advanced mathematics that the ordinary person who doesn't want to be a physicist, chemist, or real (not computer programmers) engineer can get. My math background extends beyond just statistics, but for me... those concepts have been largely useless. Interesting maybe... but not really that useful.

a person without basic concepts or stats and probabilty, is it more likely he or she will make more wrong decisions/judgements than right in the long run?

I think so. Someone emailed me a newspaper article about the big difference in assets between couples who knew some math and those that did not. There were 3 pretty easy math questions, couples who went 3 for 3 had acquired assets of over a $ million, on average. Couples who did poorly, averaged assets of only about $ 200,000 for their retirement. I don't have the email or link anymore. But it supported what I've long suspected, which is that just a little mathematical understanding makes a huge difference over a long period of time.

Yes, it is more likely. The great thing about statistics is that it teaches someone that decisions can be made based on data. Rather than just with "your gut". Therefore, it enables people to make better decisions. That's why it is so valuable.

Just saw the thread, this (below) is what I was going to write, it seems the "easiest" way to look at the issue, I havent checked the final calculation but the methodology is exactly what I would have done. This is not necessarily a good thing, please note. Mind you, I do tell myself very often that I am very intelligent.