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    Random numbers




    Hello, let me go direct to point:

    Let's say i have a total of 60 numbers (1-60) and i can generate 6 random numbers among them, in this case: 5, 17, 22, 35, 39, 51.
    Now, if i do it again i'll get 6 new random numbers, cool!
    What's the probability that 0 of the first 6 numbers will come out? what about 1 of 6
    How can i calculate that?

    Another question, why is there less possibility of getting 2-3-4-5 than 2-7-12-16? just an example! is there a name for that? i know logically, but can't figure out why!

    Thanks in advance!

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    Re: Random numbers

    Quote Originally Posted by msw View Post
    Let's say i have a total of 60 numbers (1-60) and i can generate 6 random numbers among them, in this case: 5, 17, 22, 35, 39, 51.
    I assume the pool of 60 numbers consists of distinct numbers—that is, there are no duplicate numbers among the 60. I also assume that the selection is without replacement—that is, if you selected, say, 13, it is removed from the pool of numbers and so 13 is no longer available to be drawn. These are important factors and constraints to know in calculating the required probabilities.


    Quote Originally Posted by msw View Post
    What's the probability that 0 of the first 6 numbers will come out?
    In other words, you want that none of six specific numbers should be drawn, which means the first number for the second draw can be any one of 54 out of 60, the second number any one of 53 out of 59, …, and the sixth number any one of 49 out of 55. The probability that none of those numbers is drawn is thus P = (54/60)×(53/59)×(52/58)×(51/57)×(50/56)×(49/55) ≈ 0.5159.


    Quote Originally Posted by msw View Post
    what about 1 of 6
    Here it gets a bit more complicated. Suppose the first number matches one of the six previously drawn. The probability of this happening is six out of 60. The second number must not match any of those previously drawn, and so can be any one of 54 out of 59. Similarly for the third and subsequent ones. Thus, the probability of drawing a match first, followed by five non-matches is P = (6/60)×(54/59)×(53/58)×(52/57)×(51/56)×(50/55) ≈ 0.06317. However, we have only covered the case when the first number drawn is a match. Using a similar analysis and assuming that the second number drawn is the one that matches, the same probability P ≈ 0.06317 is found. In the cases where it’s the third, fourth, fifth or sixth number that matches, the same probability is again found in each case. Therefore, there are six different equiprobable ways of ending up with exactly one match, and so the overall probability is P = 6×(6/60)×(54/59)×(53/58)×(52/57)×(51/56)×(50/55) ≈ 0.3790.

    (For two matches, P = (6|2)×(6/60)×(5/59)×(54/58)×(53/57)×(52/56)×(51/55) ≈ 0.09475 where “(6|2)” means “6 choose 2” = 6×5/2 = 15, which is the number of ways two items can be selected from six. Can you see how we get this answer for two matches?)


    Quote Originally Posted by msw View Post
    Another question, why is there less possibility of getting 2-3-4-5 than 2-7-12-16?
    There is no difference in their probabilities. All four-number sequences have exactly the same probability of being drawn at random. The fact that the first one has an obvious pattern does not make it less likely.

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    Re: Random numbers

    Thank you for the explanatory answers!
    Still can't figure out why it seems to have a probability difference for n, n+1 and n, n, maybe because of the pattern itself! but still..

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    Re: Random numbers

    Quote Originally Posted by msw View Post
    Still can't figure out why it seems to have a probability difference for n, n+1 and n, n, maybe because of the pattern itself! but still..
    I don't understand what you are trying to say.

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    Re: Random numbers

    Sorry,

    Still can't figure out why it seems to have a probability difference for getting 1-2-3 and 3-34-57, maybe because of the pattern itself! but still..

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    Re: Random numbers

    I would guess it’s a purely psychological effect. We are surprised when we find a pattern where we don’t expect one.

    If at the beginning you specify the order in which the numbers are to be drawn, e.g., 1 first, 2 next, then 3, etc., then the probabilities become considerably smaller—but again this is equally true for sequences without any pattern.

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    Re: Random numbers

    I see.

    Another point following my current game analysis, every number has 1,67% probability of being drawn, let's assume draw # 1 was 6 random numbers, but none of them "15", how to calculate the probabilities of a number being drawn every time it doesn't.
    Assuming every draw is a new set of 6(selection without replacement) from 1-60, draw 1 = 1,2,3,4,5,6, those numbers are not excluded for draw 2.

    I'm doing a lottery study, is it okay to continue every time i have a question regarding the matter? thanks!
    Last edited by msw; 06-04-2015 at 07:10 PM.

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    Re: Random numbers

    Submit as many questions as you like.

    Basics, when you say random selection every number has the same chance. Once you pick the first selection, every number has the same chance still but the denominator gets smaller and probability gets bigger because of that. More or less there are fewer to select from.
    Stop cowardice, ban guns!

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    Re: Random numbers

    Quote Originally Posted by hlsmith View Post
    Submit as many questions as you like.

    Basics, when you say random selection every number has the same chance. Once you pick the first selection, every number has the same chance still but the denominator gets smaller and probability gets bigger because of that. More or less there are fewer to select from.
    Thanks, i got that!

    To be more clear, assume draw #1 is 1,2,3,4,5,6
    There were 54 numbers not drawn.
    Somehow the probability of these 54 numbers being drawn in the next draw goes up, right? Until the point a number must be drawn, not accurately i guess, but in a range.
    So, let's say the number 37 is one of them, he wasn't drawn in draw #1, what's the probability he gets drawn in draw #2, if not, draw #3 etc..

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    Re: Random numbers

    No, because each draw of six numbers from 60 is an independent event. If you drew 13 as part of the first series of six numbers, then 13 does not somehow become less likely to appear in the next series. Thinking that it is less likely is known as the “gambler’s fallacy.” If you think of flipping a coin, if you just got a Head, getting a Tail on the next flip is still a 50% probability.

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    Re: Random numbers

    Really nice, it's quite interesting!

    So, the probability of 3 heads in a fair coin is 0,5^3 = 12,5%, which means it's less likely to occur(prior to the first flip), since you have a 50% chance for heads/tails, yes you can still get 3 heads or even 20 heads, but an increasing equal subsequent result is less likely(again, prior), else doing 10.000 coin flip wouldnt be ≈ 50%.

    Based on that, if i'm not completely wrong, every time i get tails, i'll expect heads, since it's 50%, ok that might be inconsistent to what i've written, i realize i'd expect heads because i flipped tails but still..
    I mean, you get 20 tails, you seriously wouldn't guess next flip being heads? even though the probability is - by the fallacy definition - still 50%! (I would guess it’s a purely psychological effect again? if so, it must be explained somehow)

    So this logic should apply to my game, every number has 1,7% chance, 6 numbers = 10%, which means in 10 draws a number must come out, which is the average of a number being drawn by my tests.

    I do not have sufficient knowledge to understand this completely.
    I'm not even sure what i'm thinking.
    My mind is blown... lol
    Last edited by msw; 06-05-2015 at 04:41 AM.

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    Re: Random numbers

    If we use the coin-flipping analogy, the probability of getting a specified sequence such as HHTHTHTHHTTTHHTTTHHT is just as likely as getting 20 Heads. That may seem completely counterintuitive, but it is nonetheless true, assuming a fair coin. The important point to note is that each different 20-flip sequence―and there are 2^{20} = 1\ 048\ 576 unique ones―has the same probability of occurring as any other 20-flip sequence.

    Quote Originally Posted by msw View Post
    So this logic should apply to my game, every number has 1,7% chance, 6 numbers = 10%, which means in 10 draws a number must come out, which is the average of a number being drawn by my tests.
    No, it is a big mistake to think that if the probability of something is 10%, then it will occur once in ten trials. If that logic were correct, flipping a coin would always produce a Tail after a Head and a Head after a Tail, and all sequences would be HTHTHTHTHTHTHTHT... What 10% probability means is that if you run a huge number of trials (theoretically, an infinite number) then 10% of the trials will have the property for which the probability is 10%.
    Last edited by Con-Tester; 06-05-2015 at 04:53 AM. Reason: Added info.

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    Re: Random numbers


    You have two sets of random numbers 2,3,4,5 and 2,7,12,16 and you maintain the one is more likely than the other, but consider this: At the first event you get some number that is just as likely as any other numer. Say you get 2 as above. Now any second number that you get is just as likely as any other number, so 2,3 is as likely as 2,7 and so you can go on to show that 2,3,4,5 is as likely as 2,7,12,16
    And also holds for a sequence (given by,say, tossing a coin 10 times). HHHHHHHHHH is as likely as HTHTHTHTHT. This is because the coin "does not know" that it fell heads last time, so heads or tails is equally likely at the next throw. BUT if you ask how many equiprobable sequences there are with 10 heads in and how many equiprobable sequences there are with 5 heads in the answer is very different. There are 10!/(10!0!) sequences with 10 heads in and 10!/(5!5!) sequences with 5 heads in. So you have 1 sequence with 10 heads in and 252 sequences with 5 heads in and all the sequences are equiprobable. The probability of getting a sequence with 5 heads in is 252 times that of getting a sequence with 10 heads in. I HOPE MY MATHS IS CORRECT. I used the counting rule that says: The number of distinguishable arrangements of n items (10 in this case) of which n1 are of one kind and n2 are of another kind is n!/(n1!n2!)
    Last edited by Swayseker; 07-22-2015 at 10:22 AM.

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