what are the odds of probability of getting your point or rolling 7 or 11 , twelve times in a row in craps?
476052
it's pretty complicated, and I'm too far removed from my math courses to keep it all straight.  But it varies based on what your point is, simply because some outcomes are more likely.\n\n4 has 3/36 odds of being the point\n5 has 4/36\n6 has 5/36\n8 has 5/36\n9 has 4/36\n10 has 3/36\n\nBut you can't roll 7.  So if we assume the point is 5, for each roll there is 32/36 odds that we don't roll 5, but 6/36 odds we roll a seven.  Reconcilling that involves equations that I've long since forgotten about.\n\nI think maybe [((8/9 - 1/6)^11)*(1/9)] would be the odds you'd roll 11 times without getting 7 or 5 and get a 5 on the 12th roll.  That'd be about .0214.  That you'd follow that up with a 7 or 11 would be .0214*(8/36), or .00535\n\nSite I link to be low discusses the outcomes a bit more formally than I'm cabaple of doing these days.
1120211
1)There are thirtysix possible combination of dice of which 10 are equal to or greater than 9. Therefore the probability is 10/36 or 27.8%.\n2) There is nine socks and only one with the correct number at that point. Therefore the odds are one in nine. \n3) x=-7/3\n4) x=-1\n5) x=-3
279356
This is always so much fun, isn't it.  This is probably why the previous answerer is now looking for satisfying employment in the food-service sector.\n\nBy the way, you forgot to mention if they have to specifically be in that order or in any order, though.  It's a very different answer.\n\nOkay, what's the probability of drawing one green marble each time you draw, if you put the marbles back each time?  It's the partial over the sum, or in other words 4 out of 6, or 2/3.  Similarly the probability of drawing a red marble is 2 of 6, or 1/3.\n\nSo if you specifically want the order to be green, green, red, then you just multiply the probabilites each time.  2/3 * 2/3 * 1/3 is  4/27.\n\nHowever, suppose it can be in any order?  Here's where it gets tricky.  First we figure how many ways there are to draw two greens and a red -- we can draw (g,g,r) or (g,r,g) or (r,g,g) -- so three ways.  The probability is the same for each time so we add them together, essentially 3 * 4/27 is 12/27, or a little less than 50%.\n\nPart b) is harder, since the probabilities change with each draw.  The first time we draw it's a 2/3 chance of getting green, but after that there are only 3 green marbles out of 5, so it's now 3/5 instead of 4/6.  So if we want to draw two green and one red IN THAT ORDER it's 2/3 * 3/5 * 1/2 = 1/5.\n\nIf you don't care the order, then figure out each of the ways to draw 2 greens and one red (same as above) and add the probabilities from each time.  I'll leave it to you for the answer.\n\nActually there are very few interesting people in the world who like statistics.  But even the boring ones can usually find very good jobs.  Also it's been a while since my stats classes so if I'm wrong, please don't come after me with a baseball bat, thanks  :)
894091
None. Two odds make an even
1049750
For any roll, the chances of a particular number on an N sided die coming up is 1 in N.  Therefore, generating a random number from 1-N for each roll would be the answer.
832564
The answer is one chance in 2^5  or 1 in 32.\n\nNotice that the chances are the same as flipping five heads in a row.
950716
The only way to fail to win at least one prize is to loose each time.\nThe probability that you will loose one time is 1 - 0.11 = 0.89.\nThe probability that you will loose 5 times is 0.89^5.\nSo the probability that you will win at least once, or fail to loose all 5 times, is 1 - 0.89^5 = 0.44
630810
a) probability of rolling a 2 is 1 out 6. for 5 balanced dice\n(1/6)(1/6)(1/6)(1/6)(1/6) = (1/7776)\n\nb) isnt this the same as part a?
249369
Is the term you are looking for "probability"?\n\nI hope this helps.
497428
It seems that a large proportion of people have answered your first point (and very well I might add). But regarding the '20 consecutive tails' part, strictly speaking the answer is that the chances of the coin coming down 'heads' is still 50-50.  However, the chances of a coin coming down 'tails' on 20 consecutive occasions is so remote, that we would have to consider the possibility that the coin has been tampered with!
901297
1/54 (for 6-sided dice of course)\nThere are 6*6*6 = 216 possible outcomes, and\n1,1,1\n1,1,2\n1,2,1\n2,1,1\nare the only outcomes with sum<5, so 4/216 = 1/54
981433
I don't think there's anyone reference for this, at least not an authoritative one, although you might look at "True Odds" by Walsh.\n\nIf you wanted to know the probability of particular events happening, I think it'd be easier to look them up on-line individually.  For instance:\n\nhttp://golf.about.com/od/faqs/f/holeinoneodds.htm
76964
Yes, it's true.  It happens in Las Vegas all the time.  Actually it's a common occurance in Statistics\n\nExample: Roll 2d6 and total the result.  You will get a result between 2 and 12.  Do this repeatedly and you will get a random collection of totals.  If you plot the number of occurances of each total, you get what's called, a Bell Curve (because of its shape).  The most common occurance is 7.  The least common is 2 and 12.\n\nMistWing SilverTail
97799
i don't believe the person above me got the second part right.  he is right about the beginning - the first probability is 2/5:\n\npossible numbers to draw: 2 4 6 11 13 15\n\nso you can choose any of these six out of the fifteen given (6/15 = 2/5).  the next number you draw must be also a "winning" number.  BUT, you have to have an even number less than 7 AND an odd number greater than 10.  this gives us two scenarios:\n\nIF the first number picked was even and <7, then the possible numbers you can pick from are ONLY: 11 13 15, since picking another even number <7 will not meet your requirements.  this gives us a probablity of 3/14, since there are 3 "winning" numbers out of the 14 numbers left.\n\n\nIF the first number picked was odd and >10, then the possible numbers you can pick from are ONLY: 2 4 6, since picking another odd number >10 will not meet your requirements.  this gives us a probablity of 3/14, since there are 3 "winning" numbers out of the 14 numbers left.\n\nSO, finally, suppose you go with the first scenario.  the probability would be: (6/15)*(3/14) = 3/35  going with the second scenario will give you the same thing, 3/35.  thus adding these two, since you can have either scenario occur and still meet your requirements, gives you: 3/35+3/35 = 6/35.\n\nAnd that would be the odds: 6/35.
1177996
The probability of winning n times is the ratio of combinations of winners and losers among your tickets to the combinations of tickets drawn:\n(winners and losers of your 5 tickets) / (total drawings)\nC(5, n) * C(1995, 20 - n) / C(2000, 20), or\n[ 5! / (n! * (5 - n)!) ] * [1995! / ((20 - n)! * (1975 + n)!)] / [2000! / (20! * 1980!)]\n\nThe probability of winning 0 times is approximately 95.094%, so the probability of winning at least once is about 4.906%.\n\nThe probabilty of winning 1 time is approximately 4.812%.\nThe probabilty of winning 2 times is approximately 0.093%.\nThe probabilty of winning 3 times is approximately 0.001%.\nThe probability of winning 4 or 5 times is miniscule.\n\n[Edited below to add comment]\nThe misconception many people are having is that it's a straight 2000 tickets drawn each time.  In a drawing like this, once the first of the 20 winning tickets is picked, there are only 1999 tickets left from which to draw.  After the second winner is chosen, the remaining tickets become 1998.  You have to use combinations to get the correct probability, here.
574995
The chance that it will be on the next draw is the same as it has always been 1/1000.  Draws are independent events and as much as our intuitive human nature tells us otherwise, a number is not "due" because it hasn't been picked in awhile.\n\nThe chance that a specific sequence wouldn't come up in 1 drawing is 999/1000.  For two drawings it is (999/1000)^2.  For 3 drawings (999/1000)^3, etc.\n\nFor 6588 drawings, the odds that it would have gone this long is (999/1000)^6588 or 0.00137226009.  This is very close to the probability that it will be drawn on the next draw (0.001).  They are both low probabilities but not out of the bounds of expectation.  If you check the draws, you'll probably find sequences of numbers that have gotten drawn more than a dozen times...\n\nI know you don't want to hear this but to end up with most likelihood of having money in the long run, the best odds are not to play at all.  The lottery odds always favor the state since the payouts average less than the expected number of tickets.  Perhaps not for a single draw, but odds will win out in the long run.
967397
You really think you can make that work?\n\nBut anyway. if you look at this in the eyes a pure statistician, than the more you play, the more you increase the odds that you win... the more squares you play, the more you increase the odds that you win.. as far as the odds of it actually paying off... god only knows