Suppose we have a list of 21 numbers and out of those 21 numbers - the number 4 shows up 1 in 3. Now I want to analyze how often the number 4 comes up after not coming up once. For example if we have this set of numbers 4, 5, 8, 3, 4, 2, 9, 4, 3, 8, 3, 2, 4, 3, 4, 7, 2, 8, 4, 2, 4, We see that 4 came up 7 times out of 21 possible times. Which means that it has a probability of 1 in 3 (assuming numbers from 1-10 do not have an equal probability of showing up for some anonymous reason & these numbers are a reflection of the actual probability of each number showing up) Now if I want to analyze the probability of 4 showing up after not coming up (or missing) once I get this result 4 Didn't come up 3 times (5,8,3) 4 Didn't come up twice 4 Didn't come up 4 times (3,8,3,2) 4 Didn't come up once (3) 4 Didn't come up three times (7,2,8) 4 Didn't come up twice (2,9) 4 So probability of 4 showing up after one time not showing up would be 1 in 6 -- correct ? Or is it less than 1 in 6 since 4 initially had a probability of coming up 1 in 3?

Can you say that all in Chinese please? Or was it said in Chinese actually? Use this link and stay away from this forum for a while... Code: [URL]http://easycalculation.com/statistics/probability.php[/URL] I stopped doing even mental arithmetic after I left school and purchased a simple calculator to do the job. I am very happy now.

Wouldn't it just be 1/3 still, because each number's value is independent of the others, so it doesn't matter how many times 4 did or didn't appear before the number? Just like flipping coins...

Yes. It happened 7 times in 21 tries, so that's the final and the only correct answer. But he keeps adding so many things after that and that confuses the mind... that's not in the spirit of mathematics...

It depends on whether you are trying to figure out the probability for that particular sequence if it were rescrambled, ie if all those same numbers have to come up again but can come up in a different order or if you're just doing another completely random spin of numbers but saying that for some reason the number 4 has a weighted probability of coming up 3 times as much as any other number.

Download the .rar file which contains this Excel file: probability of madness lol.xlsx Input your number (between 1 and 9) in the green highlighted box (G2), and press F9 when you wish to recalculate. Look at corresponding cells to find what you want... The headings are at the bottom of the range of cells. Disclaimer: The program may contain 0 to infinite errors or mistakes! No responsibility taken or given... blah, blah, blah... If this program helps you, and if possible, then please donate whatever amount, or/and give good food, to the poor near your home and wish them Merry Christmas! Thanks! View attachment probability of madness lol.rar VT scan... Clean! [TR="class: collapsable hide"] [/TR] File name: probability of madness lol.rar File type: RAR Detection ratio: 0 / 46 Analysis date: 2012-12-23 10:13:08 UTC ref.: Code: [URL]https://www.virustotal.com/file/4555c0000f1be5766b4bb66219de9c92f624c5030d53d9faffb96208ba368183/analysis/1356257588/[/URL]

The second is more correct because you need to leave the first occurrence. Or it becomes 2 times out of 6 if you include the set1 also as it already has a 4 as the last item in set1. More confusion???! . For both, I would say "insufficient data to calculate the right answer" . How would you know for sure when that "1 in 6" set/pattern would come up again? You can calculate "exact" probability only using the known data IMHO. You can arrive to an approximate value if you take 10000s of data sets (e.g., 21 sets like you have, then keep generating such 21 random-numbers (1 ~ 9) sets as many times per hour / per day / per month, and see if the randomness has a definite pattern of producing result!!! IMHO, that's impossible for 9 unique number sets. It may be possible to find the probability for less unique numbers IN a set (e.g., set1 = numbers 1 to 3, or set1 = numbers 1 to 4). For me, this is the end of it! I am off to enjoy my non-mathematical-based free life! If you find the right and the only answer, then please tell us... P.S.: I would love to solve such riddles based on probability though: Example: A gambler goes to bet. The dealer has 3 dice, which are fair, meaning that the chance that each face shows up is exactly 1/6. The dealer says: "You can choose your bet on a number, any number from 1 to 6. Then I'll roll the 3 dice. If none show the number you bet, you'll lose $1. If one shows the number you bet, you'll win $1. If two or three dice show the number you bet, you'll win $3 or $5, respectively." Is it a fair game? Hint What will happen if there are 6 gamblers, each of whom bet on a different number? Answer (Select the whole text below) It's a fair game. If there are 6 gamblers, each of whom bet on a different number, the dealer will neither win nor lose on each deal. If he rolls 3 different numbers, e.g. 1, 2, 3, the three gamblers who bet 1, 2, 3 each wins $1 while the three gamblers who bet 4, 5, 6 each loses $1. If two of the dice he rolls show the same number, e.g. 1, 1, 2, the gambler who bet 1 wins $3, the gambler who bet 2 wins $1, and the other 4 gamblers each loses $1. If all 3 dice show the same number, e.g. 1, 1, 1, the gambler who bet 1 wins $5, and the other 5 gamblers each loses $1. In each case, the dealer neither wins nor loses. Hence it's a fair game.

So, what we have here i what we call a game of luck - we have a pool of numbers (1-10) and we pick one each time. The distribution of probabilities for those numbers can be anything (in your case: 4 has 33% of getting picked and the rest whatever). Now, the critical part - which we don't know from your information - is if the results of each game depend on the results of the previous ones - for example, in games like the roulette or lotto, it does not. If somehow the probability distribution changes with each games, it does (so in that case 4 wouldn't always have 33%, it might have 30% in one game, 10% in another, 50% in an another etc.) I will assume it 's the first case (the previous games do not change the probability distribution for the future ones). Given that assumption, the answer is obviously 33% Examples: Suppose you have 1,5, X -> 33% of getting a 4 Suppose you have 1,5,8,2,3,1,1,9,6,3,3,3,3,X -> 33% of getting a 4