on the roll of one dice you win $5 if 1 or 2 shows up, $10 if 3 or 4 show up, $0 if a 5 how much should you?
965173
You're almost there.  What you want is for the average amount of money per roll to end up being zero.  Thus you need to add up the amounts from all six possible rolls, using a variable for the one you don't know, then solve for that variable, like this:\n2(5) + 2(10) + 0 - x = 0\n10 + 20 - x = 0\nx = 30\n\nYou need to lose $30 if you roll a six.
981521
A dice has 6 faces.\nProb that the two dice show equal = 1/6.\nSo Prob that the two dice show unequal = 5/6.\nSo the expected value of X is\n\nE(X) = 10*5/6 + 100*1/6 = 150/6 = 25.
598521
What were the odds?\n\n4 to 1?\n\n8 to 1?\n\n20 to 1?\n\n**********************************\nEDIT\n\nIf the odds were 6 to 1, then you won $120.
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?
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.
977563
You need to give us a problem, but let's say the problem was:\n\nYou have ten coins (dimes or nickels) and they add up to 65 cents...\n\nYou could guess that you have 5 dimes and 5 nickels:\nGuess:\n5 dimes, 5 nickels\nCheck\n75 cents <-- too high\n\nSo try guessing 4 dimes and 6 nickels:\nGuess:\n4 dimes, 6 nickels\nCheck:\n70 cents <-- still too high\n\nFinally try guessing 3 dimes and 7 nickels:\nGuess:\n3 dimes, 7 nickels:\nCheck\n65 cents <-- correct answer.
718899
The facts in this riddle are clear: There is an initial $30 charge. It should have been $25, so $5 must be returned and accounted for. $3 is given to the 3 friends, $2 is kept by the bellhop - there you have the $5. The trick to this riddle is that the addition and subtraction are done at the wrong times to misdirect your thinking - and quite successfully for most. Each of the 3 friends did indeed pay $9, not $10, and as far as the friends are concerned, they paid $27 for the night. But we know that the clerk will tell us that they were charged only $25 and when you add the $3 returned with the $2 kept by the bellhop, you come up with $30.
568434
Players - There are 22 players, each with a box numbered between 1 and 22.  Each day a player is chosen at random to play the game.  That person's space is filled the following day, and each person remains in the game as a box holder until they are selected to play the game themselves.\n\nThe Game - The player for that day does not know how much is in their box.  The board shows the amounts of money in play.  They are 1p, 10p, 50p, £1, £5, £10, £50, £100, £250, £500, £750, £1000, £3000, £5000, £10000, £15000, £20000, £35000, £50000, £75000, £100000, and £250000.\nThe player chooses boxes one at a time, and their values are eliminated from the game board to show the possible amounts available still to win.\nAfter 5 boxes, the phone will ring and the banker will offer an amount of money based on the possibilities still remaining.  Following this, the banker will make an offer after every 3 boxes have been chosen.\nAfter each offer, the player must decide whether to accept this offer, or eliminate more boxes in the hope that he will knock out the low amounts, and therefore generate a higher offer from the banker.  Ideally, he will have the £250000 in from of him, meaning it cannot be eliminated, and will generate a high offer to the end of the game.\nThe question "Deal Or No Deal" is asked each time, hence the title of the show.\nIf the player accepts the offer at any time, that is the amount they go home with.  The game is then played to it's conclusion to see how much they would have won if they had carried on to the end.  The offers continue to be made by the banker as if the player had not made a deal at an earlier time.
955894
x1*p(x1) + x2*p(x2)\n\nx1 = 1 for  event dots> 3,\nx2 = 4 for event  dots<=3.  \nThese are equally probable because there are 6 dots on the die.  \n\n1*(1/2) + 4*(1/2)
921908
1\n\njust do a little bit of trial you will see that the smaller number you use it nears 1.
585757
The odds are based on the number of numbers you can select from.  For example, if the winning number is five numbers out of a total of 70, you have a chance of 1 in (70 x 69 x 68 x 67 x 66), since this is probability without replacement.  So the calculated odds are 1 in 1,452,361,680 in this example.\n\nAnd as for the answer that made fun of your use of commas, feel free to point out to him that his use of apostrophes was erroneous as well.  "comma's" was the word he used, and it really should have been "commas." :)
978546
The expected value you get for rolling a normal die is 3.5, but there is no 3.5 on a die.
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%
882956
Probability is 3 out of 5
263161
This is a good probability problem -- I LOVE probability problems.  This is really asking 4 different questions.  First, what are the chances of the first pull being black.  (9 total socks, 5 are black ...).  Then what are the chances of the second one being blue (Keep in mind that now you only have EIGHT socks in the drawer, unless you put the last one back - so 1 blue and eight total socks).  Then chances of a brown sock with only 7 socks left (3 brown, seven total socks).  \n\nWhen you get all those probabilities, in order to see what the chances of this AND this AND this happening, you have to multiply the fractions (if it was "What are the chances of this OR this happening, you'd add them).\n\nGood luck.  This stuff is FUN!  (Especially if you ever want to go to VEGAS!)
1056545
Find the probability of each hand and add them up.  The hands would be: straight, flush, full house, 4 of a kind (straight flush would be included in straights AND flushes, don't count them twice).\n\n\nThere are 52C5 = 2598960 total hands, that's the bottom number.\n\nThere are 4x4x4x4x4 = 1024 ways to get an A-2-3-4-5 straight\nThere are the same for each of the ten possible straights up to 10-J-Q-K-A so 10x1024 = 10240 possible straights.  40 of these would be flushes too.\n\nDo this for the other hands then add them up. Put the sum over the total.
569858
I'll assume that he is starting with the coin.\n\nPossible scenarios: Let H be getting a head on a coin flip, T be getting a tail on a coin flip, and N be getting a nonsix on a die roll.\n\nH (Meaning he got a head on the first flip)\nTNH (Meaning he got a tail, then a nonsix, then a head)\nTNTNH\nTNTNTNH\n...\n\nSo P(H before S) = P(H or TNH or TNTNH or ...)\n= P(H) + P(TNH) + P(TNTNH) + ... since all are mutually exclusive or disjoint.\n= 1/2 + (1/2)(5/6)(1/2) + (1/2)(5/6)(1/2)(5/6)(1/2) + ...\n= 1/2(1 + (1/2)(5/6) + (1/2)^2(5/6)^2 + (1/2)^3(5/6)^3 + ...\n= 1/2((5/12)^0 + (5/12)^1 + (5/12)^2 + (5/12)^3 + ...\n= 1/2*sum(i = 0 to inf)(5/12)^i\n= 1/2(1/(1-5/12)) (since |5/12| < 1)\n= 1/2(12/7)\n= 6/7\n\nNow if he rolls the die first then possible scenarios:\n\nNH\nNTNH\nNTNTNH\n...\n\nP(NH or NTNH or NTNTNH or ...)\n= P(NH) + P(NTNH) + P(NTNTNH) + ...\n= P(N)P(H) + P(N)P(TNH) + P(N)P(TNTNH) + ...\n= P(N)(P(H) + P(TNH) + P(TNTNH) + ...)\n= P(N)(6/7)\n= (5/6)(6/7)\n= 5/7\n\nSo it depends on what was flipped first.