Is winning just pure luck?\nDo the odds increase playing in a group? If so, how many?\nIs there mathematical reasoning involved?
125228
It is just pure luck.  The best way to win is to be rich enough to buy tickets with every one of the possible number combinations.
375916
diidy, how could the odds get worse when you buy more tickets?\n\nI think there are actually two answers, depending on whether each ticket is independent of the others or not.\n\nIf each ticket is independent, and there's no guarantee as to how many tickets there are or how many winners there will be, then think of it this way.  The odds of any ticket NOT winning are 299:300.  Therefore the odds of NO ticket winning are 299^n:300^n, where n is the number of tickets.  The odds of at least one ticket winning are 1 - the odds of no ticket winning, or 1 - 299^n:300^n, or (300^n - 299^n):300^n.\n\nIf there are 300 tickets of which 1 is a winner, then the odds of winning are n:300.
528604
I watched a special on big lottery winners recently, and YES you can win too much money.. I think it depends on a personality type, but when you start putting solid gold statues by your pool?  Seems a bit over the top!
478855
That is essentially correct, HOWEVER...\nthere are other factors.\n\nYou have in effect calculated the expected value of your ticket if they only rewarded the top prize.  There are other prizes for fewer numbers hit as well, and these are often set values which do not change as the prize "pool" increases.  These would have to be taken into account. \nAlso, there is the slight but very possible chance that someone else hits the big prize simultaneously, (especially when the pool becomes relatively large compared with the odds:  the number of people buying tickets rises much greater than linearly) , in which case you split it.  While this would be taken into account in both lotteries, the differential between the "value" of the two would be smaller.\nNow if the odds of a lottery were, say 10,000,000 : 1, and the payout for the top prize were 20,000,000, then it might actually be worth spending $10,000,000 to purchase every ticket, guaranteeing a win!  (Not only do you win the top prize, but you have hundreds of smaller prizes too.  But again, this would depend heavily on the number of tickets in circulation).  In fact, a few years ago an international consortium did just that and purchased ~96% of possible powerball ticket combinations when the payout exceeded 4 times the (inverse of the) probability.  (They won by the way)\nIf you go for the one with the higher probability as another answerer has suggested, then by continuation of this hypothesis you should just forget about the lottery and flip coins against your neighbor for $1.\nIn any event,  the lottery is usually just a tax on people who are bad at math.
504113
When you are computing the probability of a given event such as 2 heads, 2 tails, or one of each, you are essentially counting the number of ways a given event can occur and dividing this number by the number of all possible outcomes.\n\nIf you toss 2 coins once ( or one coin twice) there are 4 possible outcomes:\n\nHH = both heads\nHT = 1st heads, 2nd tails\nTH = 1st tails, 2nd heads\nTT = both tails\n\nAssuming the coins are "fair" each outcome is equally likely.\n\nThere is one way in four that two heads (or 2 tails) can happen so the probability for each of these outcomes is 1/4\nBut there are two ways you can have one head and one tail and so the probability of this happening is 2/4 =1/2. The sum of all the probabilities of the distinct outcomes will always equal 1. \n1/4 +1/4 + 1/2 =1\n\nHow about if you have 3 tosses? The 8 possible outcomes are as follows:\n\nHHH \nHHT\nHTH\nTHH\nHTT\nTHT\nTTH\nTTT\nWe see there is only one way to have 3 heads (or 3 tails) and so the probability of 3 heads (or 3 tails) = 1/8\nHowever there are 3 ways you can have 2 heads and one tail, so the probability of this happening is 3/8. And there are also 3 ways you can have one head and 2 tails, so 3/8 is the probability for this too.
52279
Hearts is a game involving chance but also considerable skill. If someone were very proficient at keeping track of the cards that had been played, they could probably win 30+ times in a row. Because of this, I don't think it would be able to calculate a probability. But congrats, though, 13 times is tres awesome!
90201
I believe the result is the same, assuming replacement.  This is because every event (drawing a ball) is independent of every other and it does not matter if you look at all 1000 results or some 100 of them.\nGood question.
217004
There are a lot of sports fans out there that also gamble. So it is like watching a sports event because they use stradegy and cunning just like in other sports and people have dreams of winning all the money. And for me it is to gain tips on how to improve my game.
406665
theres no such thing as lucky numbers it's just a superstition...good luck :P
128914
Shuffling is pretty complex as you probably mean a riffle shuffle or something like that.  But let's assume you're shuffling by spreading out all the cards on the table, and randomly picking them.\n\nAssuming you want the cards in a specific order, then for each card, you'd have to randomly select the "right" card.  For the first card, you have 52 to choose from, and a 1/52 chance of getting it right.  For the second card, you now have 51 to choose from and thus a 1/51 chance of getting it right...on down to the last card.\n\nSo your odds are:\n\n1/52 * 1/51 * 1/50 * 1/49 * ... * 1/3 * 1/2 * 1/1\n\nor 1/52! or about 1/8*10^69.  To put that in perspective, lets say all 6 billion people on earth are shuffling one trillion times a second (man is that fast shuffling!) for a billion years, you're still not even in the ball park of being close (1/10^30) of having one person get it right.
10512
Sometimes, it's easier to find the probability of the opposite case you are looking for.  This is one of those times.  Try to find the probability that you don't get the right one.\n\nAfter taking you two chances, one of the three objects is left.  What are the chances that this one that is left is the right one? The one that is left is one of three, so there is a 33.3% chance that it is left behind- which means there is a 33.3% chance that you DID NOT get the right one.  Therefore, there is a 66.7% chance that you WILL get the right one. \n\nIf you are still not convinced, it is easy to show that your reasoning that got you to 83% is wrong.  What if you used the same reasoning with 4 objects and 3 chances to get the "right" one.  Your logic would say there is a 25% chance the first time, a 33% chance the second time, and a 50% chance the third time.  If you just added them (as you did to get 83%), you would find that 25% + 33% + 50% = 108%.  Clearly you can't have a percentage above 100%, which illustrates the problem with that reasoning.\n\nHere's why that reasoning dosn't work; 33% of the times you play this game, your second try has a 0% probability of getting the right one because you already got it on the first try.  The other 66% of the times you play the game, your second try has a 50% probability.  So the AVERAGE probaility of the second try is (33.3% x 0%) + (66.7% x 50%), or 33.6%.  Add this 33.3% to the 33.3% probability of the first try, and you get the correct answer of 66.7%.\n\nIf you still aren't convinced, come over some time for a "friendly" game of dice.
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
435608
Using probability, you can show that it is better to switch.  It does raise your chance of winning a good prize.\n\nLet's think of this in terms of having two doors with a goat and one door with a car.  Thus, the probability that you originally picked the car is 1/3 and probability that you origianally picked a goat is 2/3.\n\nIf you picked the car (1/3 probability), then one of the two doors with a goat would be openned.  If you switch, then you get a goat automatically.  If you picked one of the goats (2/3 probability), then the other goat would be shown.  If you switch, you would win the car.  Thus, if you switch, you have a 1/3 probability of winning a goat and a 2/3 probability of winning the car.\n\nThink about if you kept your door.  The original probabilties would still hold.  1/3 probability of getting a car; 2/3 probabilty of getting a goat.\n\nWikipedia has a good discussion of this problem.  I especially like the tree diagram that is shown there.
464574
The odds of getting all six are 13,983,816 to 1\n\nNote: the above answer forgets to divide by through the possible combinations of the six numbers. \n\nIf we multiply the numbers 49 x 48 x 47 x 46 x 45 x 44 we get 10,068,347,520. However, each possible group of six numbers (combination) can be drawn in different ways depending on which number in the group was drawn first, which was drawn second, and so on. There are 6 choices for the first, 5 for the second, 4 for the third, 3 for the fourth, 2 for the fifth, and 1 for the sixth. Multiply these numbers out to arrive at 6 x 5 x 4 x 3 x 2 x 1 = 720. We then need to divide 10,068,347,520 by 720 to arrive at the figure 13,983,816 as the number of different groups of six numbers.\n\nEdit: Yes, well the above answer has now changed their response. Well done and all that.
267784
1 in 176 million for for each ticket you buy the Mega MIllions game.\n\nSuperObots idea that it is about 1 in 39 shows the level of innumeracy that people have that allows this "voluntary tax on the non-thinker" to thrive as it does.\n\nYou wanted to see "work":\n\nThere are 56 possible number for each of the first five choices in this game (56 balls number 1 thru 56)\n\nSo the chances of getting the first number right are 1 in 56.  Second number on its own is also 1 in 56.  But cahnces of getting both right are 1 in (56 x 56)\n\nChances of getting all five right are 1 in (56 x 56 x 56 x 56 x 56)\n\nIn this game there is a sixth ball chosen from 45 balls, so the chances of getting them all are 1 in (56 x 56 x 56 x 56 x 56 x 45)\n\nBIG NUMBER.\n\nFat chance.\n\nOdds are much, much much better that you will be struck by lightning (about 1 in 800,000)\n\nSo you are about 200 times more likely to be struck by lightning than to win this lottery! (Around 1000 folks a year get struck by lightning, and about 10% of them die!)
144364
the meaning is simple... play lottery
377600
I am scratching my head on this one!  Why are some of these people saying that you have no chance?  \n\nFirst off...this is FANTASY football and any good manager should know that on any given week no team is unbeatable.  Sooo...those saying you have no chance whatsoever are already showing that they know very little about fantasy games.\n\nWhile on paper your opponent has the edge this week....the odds are still at worse 40/60 of you pulling out a win.  Good luck!