one person rolls the dice as many times as needed to obtain 12 consecutive rolls whereupon they obtain their point , their point being the number they get on the first roll coming out, after they have obtained their first point or rolled 7 or 11 on their first roll they proceed to roll the dice to obtain their second point or 7 or 11
216122
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.
622872
0.8037 or 80.37%\n\nThis is why:\n\nP(E)=1-P(not E); so the probability of at least 2 people having the same number is 1- the probability that each person has a unique number.\n\nThe probability that each person has a unique number:\n\nFirst person has 100 choices out of 100 numbers, so 100/100.\nSecond person can't choose the first person's number, so only 99 choices out of 100, so 99/100;\nThird only 98/100;\n. . .\n18th person, has (100-17)/100=84/100\n\nTherefore the probability that none of these people have the same number is (100/100)•(99/100)•(98/100)•(97/100)•. . .•(84/100) ≈19.63%\n\nSo the probability that at least two people have the same number is 100%-19.63%=80.37%.
190893
The number of combinations is determined by multiplying the number of possible numbers of the first by the number of possible numbers of the second, third and last. If the any of the numbers could be used more than once, the answer would be 4 X 4 X 4 X 4 = 256. Since this problem says that the number can only be used once, then the number of possible choices for the first is 4, the second is only 3, the third is 2, and the last is only 1. So the way to arrive at 24 is 4 X 3 X 2 X 1. Maybe an example to make sure it is clear. If the first digit was 3, then the second could only be one of 5, 8, or 9. If the second was chosen as 5, then the third could only be one of 8 or 9. The last then could only be the one not selected as the third.
178489
In game one between A and B, the outcomes are:  A wins, B wins, or they draw, equaling 3 outcomes.\n\nIn 14 games, there are 3^14 outcomes.
595089
If you don't spin, then there are five left.  Now since you said that the two bullets are next to each other and you didn't get shot the first time, then they both count as the same bullet, so that there are four possibilities left.  So that means the you have a 1 in 4 chance of getting shot.  You can also think of it this way: say you  have this situation.\n\n1 2 3 4 5 6\nX X X B B X\n\nso that there is a bullet in positions 4 and 5.  Since you shot once and got a blank, then that means you either were in the 1, 2, 3, or 6 position.  Out of those, only one (position 3) has a bullet the next time.  So the chance of getting shot is 1 in 4.\n\nIf you do spin, then you have a 2 in 6 chance of getting shot, since then you can consider that a repeat of the previous round.\n\nSince 1/4 is smaller than 2/6, I would go with not spinning.\n\n********************************************************************\n\nThis is a different result than if we couldn't assume whether the bullets are together or not.  Not spinning in that case would lead to a 2 in 5 chance of getting shot; spinning would lead to a 2 in 6 chance of getting shot.  In that case, spinning would be better.\n\nThis is an interesting question.  I'll need to keep it in mind for my students one day!
342063
Hmm, interesting question, actually, and a practical application of math.  Hey, all you folks who ask what this stuff is good for, here is a real-life situation where you can use it!\n\nGetting to the problem.  Obviously, the contestant to go last spins again if his score is lower than the best score.  If tied, there is a spin-off where his winning chance is 50%; so, if there is a tie, he should respin if he has 45 or less (giving him at least 11 out of 20 good numbers), not spin if he has 55 or more (giving him at most 9 out of 20 good numbers - better to go to the spin-off, where he has a 50% chance), while 50 is a "push".   \n\nNow, for the contestant to go 2nd to last: First, let's assume the first player busts (of course, the second player respins if his score is lower than the first player), which essentially makes this a two-player game.  Based on what number the second player stops on, this is the winnning percentage I calculate for the last player:\n\n100 - 4.875%\n95 - 14.375%\n90 - 23.375%\n85 - 31.875%\n80 - 39.875%\n75 - 47.375%\n70 - 54.375%\n65 - 60.875%  \n60 - 66.875%\n55 - 72.375%\n50 - 77.375%\n\nSo, you may think the second player should respin whenever his odds are under 50%, but that is actually incorrect.  For example, if the second player has 70, giving him a 45.625% chance of winning, respinning would give him a 70% chance of going bust - he is better off stopping with the 45.625% chance.  To determine whether or not respinning improves the second player's chances, a second calculation is needed that considers each of the 20 scores after the second spin - some of them bust the player but the others improve his chance of winning.  Based on that, here is a new table, giving the last player's chance of winning if the second player respins with the score given:   \n\n95 - 95.24375%\n90 - 90.9625%\n85 - 87.13125%\n80 - 83.725%\n75 - 80.71875%\n70 - 78.0875%\n65 - 75.80625%  \n60 - 73.85%\n55 - 72.19375%\n50 - 70.3125%\n45 - 69.68125%\n\nSo, comparing the two tables, the critical points are 55 and 60.  If the second player has 60, he should stop - the last player's chance of winning improves if the second player spins.  At 55, however, the second player should spin, improving his odds by about 0.2%.\n\n\nMoving on to the three player game... now it gets really messy.  The player to go first needs a score that can beat the other two players.  Does he stop on 55 again, hoping the other two players both bust?  He'll need a better score than that to have a 50% chance of winning, but, as was noted in the two-player analysis, stopping with a 30% chance can be better than making a second spin.  As for how to do a complete analysis, uh, beats me.  This may require a computer to do all the math.\n\n* * * * * *\n\nThinking about the three-player game, maybe it isn't as messy as I initially thought - the only additional consideration for the first player, as opposed to the second player, is that either player will stop if the first player's score is beaten.  This can be calculated from the first table.  For this table, listed is the chance of the first player losing (i.e. either player beats him) if he stands on the score given:\n\n100 - 9.51% \n95 - 26.68%\n90 - 41.29%\n85 - 53.59%\n80 - 63.85%\n75 - 72.31%\n70 - 79.18%\n65 - 84.69%\n\nSimilar to the two-player game, I can now calculate the odds if the first player decides to respin.  Since I’m getting a little tired crunching these numbers, for this last table I’ll only list the critical ones:\n\n70 - 83.4%\n65 - 82.3%\n\nIn other words, the first contestant should respin on 65, stop on 70.\n\nIn conclusion, I think the optimum strategies are.....\n\nFirst player: spin on 65, stop on 70\n\nSecond player: spin on 55, stop on 60 (assuming his score at least matches the first player)
455663
Total number of balls: 7\nLet's say the player draws each ball with each hand.\nThere can be 2 situations:\n1)Player draws 1 red ball with left hand and 1 black ball with right hand: (2/7 * 5/6)\n2)Player draws 1 red ball with right hand and 1 black ball with left hand: (5/7 * 2/6)\n\nTherefore, probability of drawing exactly 1 red ball is (2/7 * 5/6) + (5/7 * 2/6)= 10/21
167808
If you count the points on a cube, you end up with 8
667015
6 SIX\n\nLAST THREE ARE CONSECUTIVE
338826
Probability choose red the first time: 5/8\nAFter choosing red, have 4R and 5W. Therefore, probability of choosing Red again is 4/9. So, probability of choosing red and then red, is 5/8 * 4/9 = 5/18\n\nIf choosing white the first time: 3/8 chance of choosing white first. Then there are 7R and 2W, so probability of choosing red is 7/9. Probability of choosing white then red is 3/8 * 7/9= 7/24.\n\nAdd these two to find answer for first part. 5/18 + 7/24 = .56944\n\nFor second part, keep the proabailty of finding red and then red, and find the probabilty of finding white then white. Probabilty of finding white first time is 3/8. Then have 7R and 2W, so probabilty of finding white now is 2/9. Multiply the two 3/8 * 2/9 = 1/12.\n\nAdd 5/18 + 1/12 = .36111 as the answer for the second part.
303066
This is a binomial probability.\nP(success) = 1/5, P(failure) = 4/5.  There are 6 events here, so P(2 successs) = (1/5)(1/5)(4/5)(4/5)(4/5)(4/5).\nP(at least 2 successes) = P(2 successes) + P93 successes) + . . . . P(6 successes) or 1 - P(O successes) + P(1 success) + P(2 successes) which you already computed above.\nLet me know if you don't understand.
449571
The first die can be any of 6 numbers, the second any of 5, the third any of 4, so the total number of throws with all dice different is 6x5x4=120.  Total number of possible throws is 6x6x6=216.  So your answer is 120/216 = 5/9.
136914
You are dealing with probabilities here and your odds are going to be exponential. But here is what I have for you.\n\nOdds in rolling a:\n4 = 1/12\n5 = 1/9\n6 = 5/36\n7 or 11 = 2/9\n8 = 5/36\n9 = 1/9\n10 = 1/12\n\nNow for each roll after that you would take the odds and times it to the "n" power, where "n" is the number of the roll. So to roll any of these numbers 12 times in a row would mean taking the fraction to the 12th power. Below are those answers. (Hope you understand scientific notation.)\n\n(1/12)^12 = 1.13876838470415E+55\n(1/9)^12  = 1.13771030934265E+55\n(5/36)^12 = 5.1524055910299E-11\n(2/9)^12  = 1.14868734219891E+55\n\nIn other words you would be more likely to win the lottery several times before this would happen. Hope this helps.
336939
If the order is important you must use the formula 21P2 = 21!/19! or simply multiply first the 21 ways you can choose the first place times the 20 ways you can choose the second, since a student has already been chosen in the fist place. Answer = 420 ways.
474511
EDIT: This answer has been corrected.  See note below.\n\n\nIn the case of the dice, you have discrete values in the probability distribution.  It is not a "normal" (bell-curve) distribution.  But you can calculate the std dev. as follows:\n\nFor every possible value v of a throw i, compute P(vi).  In a large number of throws N, the expected number of times a value vi occurs, nvi = N*P(vi).  The mean value would then be m = (1/N)∑nv*v = (1/N)∑N*v*P(v) = ∑v*P(v).  The standard deviation would then be\n\nd^2 = [1/(N-1)]∑(xi - m)^2 =  [1/(N-1)]*{∑[N*vi*P(vi)]^2 - 2*N*m*∑v*P(vi) +N*m^2}\n\nd = √ [1/(N-1)]*{∑[N*vi*P(vi)]^2 - 2*N*m*∑vi*P(vi) +N*m^2}\n\n\nYou can't eliminate N (the total no of triails) but you can compute the result for a large N without actually rolling the dice.\n\nEDIT:  There is an error in the formula: the terms N and P(vi) should not be squared in the first term of the sum. The formula should be \n\nd = √ [1/(N-1)]*{∑N*P(vi)*vi^2 - 2*N*m*∑vi*P(vi) +N*m^2}\n\nThen, for large N, N-1≈1 and the N's cancel out to give\n\nd = √ [∑P(vi)*vi^2 - 2*m*∑vi*P(vi) +m^2]\n\nSo for all possible values of the dice (vi), compute the probabiity of that value P(vi), the mean value m = ∑vi*P(vi),  and sum over all value according to the above formula.\n\nPlugging the mean value into the formula gives the simplification:\n\nd = √ [∑P(vi)*vi^2 - 2*m^2 +m^2] = √ [∑P(vi)*vi^2 - m^2]\n\nFINAL:  Made final correction.  Above formula is now correct; the m^2 is not under the summation sign.\n\nCalculated for an ordinary pair of dice: m = 7, d = 2.415\nYou can see the calculation here http://img452.imageshack.us/img452/1734/dicestddevlk8.png
16573
2 people play each game so there are half the games in each subsequent round... so there are 16 games in the first round, 8 games in the second, 4 in the third, 2 in the fourth round, and 1 in the fifth and final round...\n16+8+4+2+1=31
303664
a. first draw doesnt matter since it's replaced; you're back at 3 in 10 chance\n\nb. two ways this could happen: first, 3/10 chance first ball is red, leaving you 2/9 chance then of second ball being red on second draw.  second, 7/10 chance you draw a white ball, leaving you with 3/9 chance of second ball being red.  (3/10*2/9)+(7/10*3/9) which leaves you with 3/10 chance. again.\n\n\nEDIT:\n\nsurprises me how many people can't do math.\n\nfor A), the second draw is completely independent from the first.  Doesn't matter if I took a dump in the bowl in the first draw; as long i fished it out, you're left again with the same start on the second draw-3 reds and 7 whites.  your probability doesn't change because of what happened in the first draw as the conditions are RESET.\n\nfor B), let's run through allll possible scenarios\nNumber the Balls\n\n1 - red\n2 - red\n3 - red\n4 - white\n5 - white\n6 - white\n7 - white\n8 - white\n9 - white\n10- white\n\nPossible draws (X denotes red second ball outcome)\n1,2 - X\n1,3 - X\n1,4\n1,5\n1,6\n1,7\n1,8\n1,9\n1,10\n(2 fav outcomes if ball 1 is drawn first)\n\n2,1 - X\n2,3 - X\n2,4 \n...\n2,10 \n(2 fav outcomes if ball 1 is drawn second)\n\n3,1 - X\n3,2 - X\n3,4 \n...\n3,10 \n(2 fav outcomes if ball 1 is drawn second)\n\n4,1 - X\n4,2 - X\n4,3 - X\n4,5 \n...\n4,10\n(3 fav outcomes if ball 4 is drawn first)\n\n...\n\n10,1 - X\n10,2 - X\n10,3 - X\n...\n10,8\n10,9\n(3 fav outcomes if balls 5-10 are drawn first)\n\n\nthere are 90 outcomes\nof which there are 2+2+2+3+3+3+3+3+3+3 favorable outcomes = 27.\n\n27/90 = 3/10\n\n\nSecond edit: guy below, 2*3 + 7*3 = 6 + 21 = 27, not 28