A coin flip comes up heads 9 times in a row...you have to bet your life on the 10th flip - what do you pick?
453355
It doesn't matter.  It's still 50/50.
466586
Repeat the game and pay a lot of money!
405425
Probability of getting one head = 1/2\nProbability of getting two heads = (1/2) * 2 = 1/4\nYou have a 25% chance of getting a head two times...\nAnd 12.5% chance of getting a head 3 times...\nAnd 6.25% chance of getting a head 4 times...\n.\n.\n.\nAnd 0.09765625% chance of getting a head 10 times!!! :)
733920
a) There are 16 combinations and 4 of them include exactly 3 heads (HHHT, HHTH, HTHH, THHH).  So the probability is 4/16 or 1/4.\n\nb) This question results in the same probability (4/16 = 1/4), except it asks for it as 'odds' which is the probability of success divided by the probability of failure.  That is 1 to 3 odds.\n\nc) The odds against heads are just the reverse.  Probability of failure divided by probability of success.  That is 3 to 1 odds against.
572902
Take your choice. Even odds.
1112750
play with the stock market, but take to an advisor and play lotto..once a week you never know
26988
I think the odds would be 1 in infinite. That's just because you could just any number, and you only have one guess.
624825
Random Payouts. Psychologically your mind knows that the next hand, the next spin COULD be the BIG WIN. Some people get so into it that they are SURE they will hit big on the next spin, or the next hand, So sure that they can't put it down. That is when it becomes a habit like alcoholism or drugs. Gambling can be a gamblers cocaine!
1006793
there are lots of varial ... from body varial to varial kickflip....but i don`t know varial .... just varial flip :D
1058790
The gambler's fallacy applies to independent random events, like coin tossing or die rolling. It's the belief that the occurrence of independent random events in the past can help you predict whether those events will occur more or less frequently than expected in the future. It's a fallacy. There's still only once chance in two that a coin will land heads-up no matter how many times it's landed tails-up recently.\n\nBlackjack is a random game, with hidden information, but which cards will be drawn in the future is influenced by which cards have been drawn in the past. In one deck, for instance, there are only four jacks, so if a jack has already been drawn, the chance of another jack appearing before the next shuffle has decreased.\n\nDon't count on being able to count cards in a casino, though. They shuffle together several decks and reshuffle drawn cards into the decks before the decks are exhausted.
245215
There are five tosses with two choices each.\nEach toss is a 50% chance of coming up heads regardless of the results before it or after.\nThe probability of getting a certain number of heads would be calculated as such: number of heads/number of choices.\nWe are looking for two heads and there are ten choices (heads and tails of each coin)\nThe probability of it would be 2/10 or 20%\n\nIt is correct in that each toss is 50% chance, but in this case we are speaking of of the chances of a group of coins having a certain number turn up heads.  We are also speaking of the probability of this before any actions are taken.  The odds will change with each coin toss till the last one which will be 50%
463357
number 9 or number 10
1042197
50/50.  The odds don't change no matter how many times the coin is flipped.
1183362
I don't care you are just another statistic. I say flip a coin if its heads cut away, tails live another day.
323264
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.
232212
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.
809813
Seen this riddle at least 2 times already... getting old.... \n\nanswer is the other coin is the nickle and the first one is a quarter...