what is the probility of rolling a 1-6 dice and getting a 4.?
600596
There are 6 possible outcomes for the roll. There is only one outcome when the number is 4. Therfore the chances of getting a 4 are 1/6.
132305
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!
597041
Well first, come up with a story, then you could use random numbers for saving rolls, then use a basic hit point system.\nPlus, becuase you are doing it in vb6, do the entire game in plain text, since animation in vb6 is limited and slow in my opinion, and to make the player use thier imagination. \n\nthanks
704806
For discrete events, the basic procedure is to list all possible outcomes, and the probability of each (which must add up to 1). The probability of being in one of an identifiable smaller group of outcomes (called a subset) is simply the sum of the individual probabilities of the outcomes within that group.\n\nFor example, if you have two distinguishable dice, there are 36 different ways they can be rolled. If the dice are unweighted, the probability of each is 1/36. Of those, 6 add up to 7. That means the odds of getting 7 are 6/36=1/6.
627413
OK.  Here are the possibilities.  It is important to keep the order of the dice in mind, so you can think of this as rolling a die 5 times.\n\nSSSSD\nSSSDS\nSSDSS\nSDSSS\nDSSSS\n\nNow each of these will have the same probability, so if I can find the probability of one of them, I can take that and multiply by 5 since there are five combinations.\n\nSSSSD can also occur in different ways.  If S=1, then D could be 2, 3, 4, 5, or 6.  So if S=1, then the probability of this combination is (1/6)^4(5/6).\n\nNow if S=2, then D can be 1, 3, 4, 5, or 6.  So again, (1/6)^4(5/6).\n\nSo hopefully you can see that there are 6 ways that you can do this.  So the probability of SSSSD is 6(1/6)^4(5/6).\n\nSo overall, the probability of getting four of a kind (without getting five of a kind) is\n5*6*(1/6)^4(5/6) = 150/6^5.\n\nSo it looks like your method does work.  Good job; this isn't an easy on.  There is always more than one way to look at probabilities.
950716
The only way to fail to win at least one prize is to loose each time.\nThe probability that you will loose one time is 1 - 0.11 = 0.89.\nThe probability that you will loose 5 times is 0.89^5.\nSo the probability that you will win at least once, or fail to loose all 5 times, is 1 - 0.89^5 = 0.44
1152612
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.
107013
Maybe there's a trick to this, but if the probability is\n5%, then if (for example) you opened 100 boxes, you would\nhave "gotten something" 5 times (ie. 5% of 100 = 5).\n\nThat seems like you'd get something, on average, once\nevery 20 boxes.
435037
try this site for help on probabiliity\n\nhttp://www.futureaccountant.com/probability/
857483
You could do something like this:\n\nSuppose that we play a game where there is a token placed on a board that has spaces to the left and to the right.  The game will consist of rolling a die and subsequently moving the token left on odd numbered rolls and moving it right on even numbered rolls - that is, on roll #1 move to the left & on roll #2, move to the right, etc...  The object of such a game might be to see how many rolls it takes to end up at the same space where the token started.\n\nNow suppose that we modify the game in the following way:  We introduce two different collered dice - one red and one green - that are used alternatively on each roll.  Every time the red die is rolled, you move the token in the direction indicated by whether it is an odd numbered roll or whether it is an even numbered roll.  However, whenever the green die is rolled, you move the token in the direction OPPPOSITE of the rule indicated by the odd/even numbered roll.  \n\nOne possible outcome of playing the modified game would be that you start out rolling a 5 on the red die - and since this would be roll #1, you would move the token 5 places to the right.  On the next roll, you roll a 4 on the green die.  Since this is roll #2, you would normally move the token 4 to the left.  However, the green die changes the direction.  Therefore, the token would move 4 to the right.  The effect would be the same as if a 4 were rolled on the red die.\n\nLastly, associate the operation of addition with that of the direction indicated by even numbered rolls and the operation of subtraction with that of the direction indicated by odd numbered rolls.  Also, associate the concept of positive numbers with that of rolling a red die and the concept of negative numbers with that of rolling a green die.\n\nI don't know if this is what you are looking for.  I hope it helps!
647274
1/6 coz there r 6 sides on  a die  i dunno
236313
Wouldn't the probability than 5 of the 38 thought it was exciting be\n\n38C5(.79^5)(.21^33) = 6.63x10^-18 : basically impossible?  Even if you added in the probabilities of 4, 3, 2, 1, and 0, it's still tiny.\n\nSo I would infer that either the 79% is wrong or the sample was not independently chosen.
467797
The chance of not rolling a four is 5 out of 6. So five times:\n\n5/6 ^ 5 = 0.4019\n\nor about 2 in 5.
506414
I answered this one for you earlier...\n\nYou have X ~ Bin(6, 1/6)\n\nas you have 6 trials, with the proability of success (a four) being 1/6\n\nYou then simply need to stick this in the formula:\n\nP(X = x) = nCx × p^x × (1-p)^(n-x)\n\nIn this case you want the probability of 2 fours:\n\nP(X = 2) = 6C2 × (1/6)² × (5/6)^4 = 0.2009
332812
A fair dice is one in which the probability of getting one side is 1/6.\n\nSum of 6 or 11 can be obtained in the following ways: \n(1,5) (2,4) (3,3) (5,1) (4,2) (5,6) (6,5)\n\nThere are a total of seven elements here.\nTotal possible elements = (6 X 6) = 36\n\nHence probability = 7/36
554953
ddepends what the payout ratio is if its 5:1 then you get\n5$ per 1$ that you bet
51714
You have a 3/4 chance in the first level, which means 75% of the time you'll get to level 2.  In level 2 you have a 3/4 chance again, so your chances of getting to level 3 from the begining is (3/4)*(3/4) = 9/16 or 56.25%.  Level three is 3/4 again, so your chances of getting to level 4 are (3/4)*(3/4)*(3/4).  I'm not going to work out what that is, but you should be able to put it into a caculator easily enough.  In level 4 you have a 1/2 (i.e. 2/4) chance of winning so you have a (3/4)*(3/4)*(3/4)*(1/2) chance of getting to level 5.  If you don't see that pattern yet you probably shouldn't be taking whatever math class this is for.  I'm not going to finish it for you (you should do your own homework), but if you work out the rest of it and post your solution I'm sure someone can tell you if it's right.