Consider a purely probabilistic game with potential outcomes x1, x2, ..., xn (each of which is a specified gain or loss of some number of dollars) which occur with the probabilities p1, p2, ..., pn respectively (where, p1 + p2 + ... + pn = 1.0). Assume that x1, x2, ..., xn and p1, p2, ..., pn are known. Theoretically speaking, how would you define or measure the risk of playing this game?
207443
This question is asking the EXPECTATION for this game, which can be found by summing the product of each outcome with it's respective probability:\n\np1x1 + p2x2 + ... + pnxn\n\nThis gives the theoretical winnings or losings for the game.  If it's a negative number (loss) it's high risk, while a positive number (gain) would mean you're expected to win.  Of course, with any probability problem, the actual results will match up with the theoretical results only over the long term.
665816
The probability is zero if u take into account everything and everything is random. The previous sentence simply means, given no information AT ALL, u cannot guess anything.\n\nProbability is not the law that governs the universe. It's one of the models that describes some aspects of this universe.... uncertainty. Incomplete information.\n\nIt's a philosophical thing to argue if information can ever be complete. Can uncertainty be completely removed? Let's skip this.\n\nThe reason u have to estimate the probabilities is because u have no know model to predict which sperm would reach the egg. no model to predict when, why, etc, of why ppl want sex. No model to predict decisions. u have no model to describe about a lot of things. Even with models, there r so many variables, the probability would be very very small if u wanna take into account EVERYTHING.\n\nFor example, I said to u, pick any positive integer. Can I predict what u will pick? take 0 to be positive integer, your possible choices are, 0,1,2,3,4,5,....9999999999999999999999999..... infinitely many choices. So, without much info about anything, the probability of any choice is 0. But, i know u r a human. You probably won't pick a number that require u 1 month to write it down, or recite it to me. Maybe some psychologists did research and found out people love 3 digit numbers, or people love the number 9, or whatever. These can all be utilised to form opinions of probabilities.\n\nSo, probability is subjective. The probability distribution of a certain event depends on the assessor's information. What's the probability of the coin landing on its head? The ideal case is when we assume the coin is perfect and ignoring other events. Say, the coin roll off under the couch and u cannot observe it... this throw would not be counted, etc. This probability is definitely different if I know the coin is biased. Or I take into account the various events that could forbid me from observing the results, say, a comet crash into my roof, destroying everything, including the coin, etc etc. there are infinitely many possibilities and taking these into account, these all affect the probability.\n\nSo, there is no violation or anything. Probability is only meaningful when they r defined properly... u lay out the Sample Space, define the subset in suitable form(the sigma field), define some mapping from this field to the real number field(the random variable X), construct some measure on them(probability measure).\n\nSo, for your question... it all depends on how much u know about the.. everything. if u have knowledge of everything perfectly, u can then say for sure whether an event will 100% happen, or 100% not happen. And if u know nothing at all, u cannot be sure of ANYTHING because there are infinite kind of events that can happen and each with equal probability of infinitely small. Even if u know some information, and u know certain event are more likely than other, if u have infinite events to consider, the probability can approach 0 too. So, with infinite possibilities, only with infinitely precise information can u access the probability of events.\n\n** information is gained. A few hours ago, my information is different. If someone asked me if u will ask this question, all I can say is the probability based upon my knowledge and my opinions, etc then. But a few minutes ago, I gained information that u wrote this question. With that information, my assessment of the probability of this event happening is 100%(kinda like cheating, i know LOL). So, if u want less unpredicted accident happening to u, start KNOWING more.
635894
1) If no one is injured, then no subs ever get to play.  The probability is (.9)^11 = .3138106 or about 31.38%.\n\n2) (.9)^9 * (.1)^2 = .003874205 or about 0.39%\n\n3) "At least" is best calculated as the complement of "no substitutes were needed."  In the first calculation we determined the probability of not needing any substitutes.  So the probability of needing at least one sub would be .6861894 or about 68.62%\n\n4) The trick here is that the two injured could both be players, or one could be a player and the other could be the sub.  (They could not both be subs, however.)  The calculation in #2 (above) is part of this--the probability that two of the players were injured.  For the other part we need to know the probability that one player is injured (while 10 are not) and that the sub is also injured.  .1 * (.9)^10 * .15 = .0052302.  So the overall result for either of these outcomes (add this result with the result from #2) is 0.00910438 or about 0.91%.
2901
That would be a "random variable". So I guess other words of the puzzle might be wrong.
390713
CASE 1:\nThe probability that 2 green balls are selected from box A:\n1/2 (chance of picking box A)\nChance of picking first ball green (3/6)\nChance of picking second ball green (2/5)\n\nSo the combined odds are: 1/2 x 3/6 x 2/5 = 1/4 x 2/5 = 1/10\n\nCASE 2:\nThe probability that 2 red balls are selected from box B:\n1/2 (chance of picking box B)\nChance of picking first ball red (4/9)\nChance of picking second ball red (3/8)\n\nSo the combined odds are: 1/2 x 4/9 x 3/8 = 12/72 = 1/6
127418
The probability of requiring exactly k stages to reduce the number of black balls by 1 is:\n\np^(k-1)*(1-p)\n\nso the expectation value of k is:\n\n1*(1-p) + 2*(1-p)p + 3*(1-p)p^2 +... = 1/(1-p)\n\nThis has to be done m times so the expectation value for the entire process is:\n\nm/(1-p)
392048
Random chances are the same as probabilities. If you are so interested, you should look into Quantum Theory. It describes it in depth.
35666
F- even and odds.  There are 3 even sides and 3 odd sides. So the likelihood is the same for both.
285500
Your logic is flawed. Say you always pick card #2. The first time, you'll have 14.2857...% (that's the same as saying one seventh) of the chances of winning the car.\n\nThe second time, with a new set of cards, your chances that the card #2 is the winner are again 14.2857...%, and so will every new time, because your chances are always 1 out of 7. You are picking one card out of seven cards that you are given.\n\nYour reasoning would be right if you could be assured that the car card will always be a different one (that for example, it will be in the first case #4, then #6, and so on). In that case, after the 7th. guess (if not earlier), you'd get the car. But the way you explained the game, nothing prevents the car from being always for example at card #5. If that is the case, always trying card #2 will make you fail every time.
322670
you multiply the number of teams by the number of games... 10 x 2 = 20 possible outcomes.\n\nThe different variations is the mutilples of all the options... for example, the number of sandwich combinations?&#92;Options, wheat, white or rye... (3)\nchedder or swiss cheese.... (2)\nham, turkey, or roast beef... (3)\n\nThere are 18 different sandwiches.  (3 x 2 x 3)
45354
If you have two events "A" and "B"\n\nThe probability that\n\na) Both would occur = P(A intersection B)\nb) Only one would occur = P (A intersection B') + P (B intersection A')\nc) At least one of them would occur = P (A union B)\nd) None of them would occur = P (A' intersection B')\n\nTry this site for a good explanation on probability\n\n\nhttp://www.futureaccountant.com/probability/\nhttp://www.schoolingkids.com/
427734
Not every problem should be solved with an elephant gun.\nJust ask yourself a different question. What is the probability that x2 is closer to x1 than to x3. since the answer to that question is the same as the question you ask and the only 2 possibilities are certain\nand mutually exclusive, the answer should be the same, 1/2.\n\nWhat about if x2 is exactly in the middle you ask?\n\nIt's like asking what is the probability that a random number in an interval is the midpoint. 1/infinity=0
94826
If a coin is flipped several times, you expect that it will turn heads half (.5) of the times and tails half (.5) of  the times. This expectation, is the same if you flip it only once, that is, there is a 50% chance it will be tails and 50% chance it will be heads (even though you'll either end up with 100% of the single flip being heads/tails and 0% being tails/heads.\n \nSo, supposing the bet is of $1.00 (i.e., you get to keep the coin if you win), you receive $1.00 if you call the out come correcetly and zero otherwise. The expected pay off of a flip is\n \nE[pay off] = $1.00*.5 + $0.00*.5 = $0.50\n \nNow, the "how much would you pay" question should be changed to "what is the most you can pay and still expect a profit". The expected profit is given by:\n \nE[profit] = expected pay off - amount payed to enter the bet\nE[profit] = E[pay off] - amount payed to enter the bet\nE[profit] = $0.50 - X\n \nIf you set the expected profit to be the least possible but still more than zero, E[profit] becomes $0.01 (one cent). Now you solve for X.\n \n0.01 <= 0.50 - X or X <= 0.49 (where <= should be read as less than or equal).\n\nNow it should be observed that this answers the question: "what is the most you can pay and still expect a profit?". The question "How much would you pay to enter this bet?", is far more comlex and depends on something called "risk preference".\n \nIf you are "indeferent to risk", the solution above is still correct but, if you are "averse to risk", you may prefer not to enter a game even if it has a positive expected profit. For examle, you may not be willing to pay 0.20 to enter this bet even though its expeceted profit is 0.30, because there is still a chance that you'll lose all of your 0.20 and have a profit of -0.20.\n \nOn the other hand, if you have an attraction to risk, you might pay 0.60 to enter the game and have an expected profit of -0.10 because you might make a profit of 0.40. (When you enter a bet at a cassino or lottery, the odds are always against you, i.e., your E[profit] < 0.)\n \nHope this helps.
579526
Okay, your possible outcomes number 16. They are:\n\n1+1=2\n1+2=3\n1+3=4\n1+4=5\n2+1=3\n2+2=4\n2+3=5\n2+4=6\n3+1=4\n3+2=5\n3+3=6\n3+4=7\n4+1=5\n4+2=6\n4+3=7\n4+4=8\n\nYour probabilites are:\n2: 1/16\n3: 2/16\n4: 3/16\n5: 4/16\n6: 3/16\n7: 2/16\n8: 1/16\n\n2,3,5 and 7 are prime, thus the probability of rolling a prime is (1+2+4+2)/16 or 9/16 exactly.
429350
Experimental probability is obtained by carrying out some trials to see what fraction of the time the event you are looking for occurs.  For example, if you want to know the probability of scoring a total of 8 when you throw a pair of dice, one way of doing it would be to throw a pair of dice five times and see how many times you get 8.  If it happened twice, you would say the experimental probability is \n2/5, or 0.4\nNot a very good result, though, too few trials.  If you did it 100 times, and 8 showed up 15 of those times, the experimental probability would be \n15/100, or 0.15\n\nTheoretical probability is found by considering how many equally likely outcomes there are, and in how many of those outcomes the event you are looking for occurs.  In the example I just gave, there are 36 different ways for the dice to land, and the total is 8 in 5 of those ways, so the theoretical probability is\n5/36, or 0.13888...\n\nWe expect a very small number of trials will not give a very good picture of the real probability.  Choose your answer from that.
656018
If you're accepting the thought that every hand is random and that the probability of winning any hand is 1/2, then you're correct.\n\nThis would be a binomial distribution.  The probability of winning at least one hand = 1 - (the probability of losing all four).\n1 - (1/2)^4 = 1 - 1/16 = 15/16.\n\nThe only fallacy to the actual process is that the probability of winning subsequent hands after the first isn't going to be exactly 1/2, as the cards that were used are gone from the deck and change the odds for the rest of the deck.  But if you're looking for a good ballpark figure, you're on the right track.
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