# math answers fast

**Number of times the coin was flipped 200 times: 5,000,000**

** Probability of a single event occurring: .5 (a fair coin)**

** Number of total times we will flip this coin: 200**

** Number of consecutive runs of heads we are looking for: 5**

** Number of times out of the total games played we saw our specified event occur: 4,829,647**

** The formula for the above is easy. For example, with three dice you do it this way: Multiply the chances of NOT**

** getting a one or a five (4/6) on the first die by the chances of NOT getting a one or a five on the second die (4/6) by the chances of NOT getting a one or a five on the third die, and the subtract the whole thing from 1.**

** 1) The probability the number on the 5th die equals the number on the 7th die is 1/6 or 16.666% (The number on the 5th die has an equal chance of being 1 through 6. likewise with the 7th dice.)**

** 2) There are 16 different “results,” all equally likely, after four coins are flipped:**

** 2) Exactly four? 76,545 / 279,936 which reduces to 35/128 or 27.34375%.**

** I shouldn’t be doing your homework for you!**

** 2-1, 2-2, 2-3, 2-4 , 2-5, 2-6, 2-7, 2-8, 2-9**

** 3-1, 3-2, 3-3, 3-4, 3-5, 3-6, 3-7, 3-8, 3-9**

** 4-1, 4-2 , 4-3, 4-4, 4-5, 4-6, 4-7, 4-8, 4-9**

** 2-1, 2-2, 2-3, 2-4, 2-5, 2-6, 2-7, 2-8, 2-9**

** 3-1, 3-2 , 3-3, 3-4, 3-5, 3-6, 3-7, 3-8, 3-9**

** 6 5 5 6 5 6 6 6 5 6 6 6**

** That’s the most incorrect statement I’ve read in this forum in awhile.**

** Team A loses 9 games out of 100 -$100 loss x 9 games = -$900**

** 20/24 x 19/23 x 18/22 x 17/21 = .45595. about 45.59%.**

** What is the probability that none of them are yellow?**

** – Jake always has to use the bathroom so he must sit on one of the ends.**

** – Justin Jane don’t get along and need to have at least two seats between them.**

** – Jane Julie are best friends and insist on sitting next to each other.**

** – Jesse has a crush on Jessica and wants to sit next to her, but he can’t sit next to Jordan (after all, Jordan Jessica did just break up.)**

** – Joey needs to sit in one of the two middle seats.**

** 1 5 6 8 2 7 4 3 1 5 6 8 4 2 7 3 1 5 6 8 4 7 2 3 1 5 6 8 7 2 4 3**

** 1 6 5 8 2 7 4 3 1 6 5 8 7 2 4 3 2 7 4 1 8 5 6 3 2 7 4 1 8 6 5 3**

** 2 7 4 8 1 5 6 3 2 7 4 8 1 6 5 3 2 7 4 8 6 5 1 3 2 7 5 6 8 1 4 3**

** 2 7 5 6 8 4 1 3 2 7 6 5 8 1 4 3 3 1 4 8 2 7 5 6 3 1 4 8 2 7 6 5**

** 3 1 4 8 6 5 2 7 3 1 4 8 6 5 7 2 3 1 4 8 7 2 5 6 3 1 4 8 7 2 6 5**

** 3 1 5 6 8 2 7 4 3 1 5 6 8 4 2 7 3 1 5 6 8 4 7 2 3 1 5 6 8 7 2 4**

** 3 1 6 5 8 2 7 4 3 1 6 5 8 7 2 4 3 2 7 4 8 1 5 6 3 2 7 4 8 1 6 5**

** 3 2 7 4 8 6 5 1 3 2 7 8 4 1 6 5 3 2 7 8 5 6 1 4 3 4 1 8 2 7 5 6**

** 3 4 1 8 2 7 6 5 3 4 1 8 5 6 2 7 3 4 1 8 5 6 7 2 3 4 1 8 6 5 2 7**

** 3 4 1 8 6 5 7 2 3 4 1 8 7 2 5 6 3 4 1 8 7 2 6 5 3 4 2 7 8 1 5 6**

** 3 4 2 7 8 1 6 5 3 4 2 7 8 5 6 1 3 4 2 7 8 6 5 1 3 4 7 2 8 1 5 6**

** 3 4 7 2 8 1 6 5 3 4 7 2 8 5 6 1 3 4 7 2 8 6 5 1 3 5 6 1 8 2 7 4**

** 3 5 6 1 8 4 2 7 3 5 6 1 8 4 7 2 3 5 6 1 8 7 2 4 3 5 6 8 1 4 2 7**

** 3 5 6 8 1 4 7 2 3 5 6 8 2 7 4 1 3 5 6 8 7 2 4 1 3 6 5 1 8 2 7 4**

** 3 6 5 1 8 4 2 7 3 6 5 1 8 4 7 2 3 6 5 1 8 7 2 4 3 6 5 8 1 4 2 7**

** 3 6 5 8 1 4 7 2 3 6 5 8 2 7 4 1 3 6 5 8 7 2 4 1 3 7 2 4 8 1 5 6**

** 3 7 2 4 8 1 6 5 3 7 2 4 8 6 5 1 3 7 2 8 4 1 6 5 3 7 2 8 5 6 1 4**

** 4 1 6 5 8 2 7 3 4 1 6 5 8 7 2 3 4 2 7 8 1 5 6 3 4 2 7 8 1 6 5 3**

** 4 2 7 8 5 6 1 3 4 2 7 8 6 5 1 3 4 7 2 8 1 5 6 3 4 7 2 8 1 6 5 3**

** 4 7 2 8 5 6 1 3 4 7 2 8 6 5 1 3 5 6 1 4 8 2 7 3 5 6 1 4 8 7 2 3**

** 5 6 1 8 2 7 4 3 5 6 1 8 4 2 7 3 5 6 1 8 4 7 2 3 5 6 1 8 7 2 4 3**

** 5 6 2 7 8 1 4 3 5 6 2 7 8 4 1 3 5 6 7 2 8 1 4 3 5 6 7 2 8 4 1 3**

** 6 5 1 8 2 7 4 3 6 5 1 8 4 2 7 3 6 5 1 8 4 7 2 3 6 5 1 8 7 2 4 3**

** 6 5 2 7 8 1 4 3 6 5 2 7 8 4 1 3 6 5 7 2 8 1 4 3 6 5 7 2 8 4 1 3**

** 7 2 4 1 8 5 6 3 7 2 4 1 8 6 5 3 7 2 4 8 1 5 6 3 7 2 4 8 1 6 5 3**

** 7 2 4 8 6 5 1 3 7 2 5 6 8 1 4 3 7 2 5 6 8 4 1 3 7 2 6 5 8 1 4 3**

** 2 1 5 2 5 1 3 1 4 3 4 1**

** 4 1 3 4 3 1 5 1 2 5 2 1**

** 1 – .2953782 = .7046218.**

** 34.15% of the time you will lose $2,000**

** Sorry for not getting back to you earlier. I’ve been busy in a chess tournament all weekend, out of town. I just got home.**

** EVERYONE, at one time or another comes up with this idea / system, yours truly included. Often it’s for Roulette, or Blackjack, but it can be used for any game, including Craps.**

** Losing rolls: 1-3, 3-1, 2-2**

** As far as systems go, it’s no better or no worse than any other type of system. But there is no way to bet, or manner of betting that will put the odds in your favor. In your case, you will break even in the long run. again, because there is no house edge, when laying 2 to 1 on a true 2 to 1 proposition.**

** The one problem with the above, and correct me if I’m wrong, is that these bets are made AFTER a point is established. But you’re not allowed to just walk up to a table and wait for a point to be established and do this. You have to first bet on the pass or don’t pass bet. Only then can you bet on the 4 point making it or not, and taking the odds. (True odds, in this case.) And that amount that you can bet on these rolls after a point is established is limited to that particular casino. (That’s the triple odds and quadruple odds and ten times odds and stuff I can recall seeing on marquees.)**

** Hope this helps. at least a little bit. Good luck in your Crap playing. May the dice be with you. 🙂**

**Assume the following:**

**A total of ten, twelve-sided dice are rolled. seven attacking dice and three defending dice.**

**In order for a kill to occur, three (or more) 12s must be rolled from the seven attacking dice. However, for every 1 that is rolled on the attacking dice, this cancels out a 12.**

**For every 12 that is rolled on the defending dice, this also cancels out a 12 on the attacking dice.**

**To clarify, there must be at least three 12s remaining on the seven attacking dice, after subtracting one 12 for every 1 that is rolled on the seven attacking dice, and for every 12 that is rolled on the three defending dice.**

** The probability is exactly 0.00877970848425242. This is verified by first a computer simulation, over 1 billion trials, and then simply by examining all possible ways to roll ten, 12-sided dice.**

**There are exactly 61,917,364,224 ways to roll ten 12-sided dice. (12^10) According to a program I wrote that examines each roll and determines if a kill occurred, (that takes several minutes to run), exactly 543,616,408 of these rolls have three or more 12s remaining. 543,616,408 divided by 61,917,364,224 is exactly .00877970848425242, which verifies the simulation.**

**What is the probability of the sum of a pair of six-sided dice equaling either 2, 3, 4, 9, 10, 11, or 12 at least once in ten rolls?**

**There are 36 different ways two, 6-sided dice can be rolled. There a total of 16 suitable outcomes and 20 non-suitable outcomes.**

**The only way I know of to solve this problem is to write a program to loop through all possible two and three digit numbers, multiplying them together, looking for the given criteria. I did so and the program returns the following:**

** 27 x 594 = 16,038**

** 36 x 495 = 17,820**

** 39 x 402 = 15,678**

** 45 x 396 = 17,820**

** 46 x 715 = 32,890**

** 52 x 367 = 19,084**

** 54 x 297 = 16,038**

** 63 x 927 = 58,401**

** 78 x 345 = 26,910**

** Notice only one of the products is odd. That might make for a good math puzzle.**

** I came across your site while looking for some help to a statistics assignment that has me all stressed out. It is not all that difficult but I am having a hard time. I am taking virtual classes for college, this is my last semester to graduate, but my first doing online classes, since I just moved and I needed the flexibility. I miss having a teacher to ask questions to. The system here is pretty weird – they just don’t seem to care and when you have doubts it is kind of your own problem. So here I am, reaching to you to see if maybe you can help.**

** This is the assignment:**

** Imagine yourself at a fair playing one of the midway games. Pick a game and calculate the expected value and post your results along with how you calculated them. For example, you may decide to throw a basketball to try to win a $10 bear. You paid $2.00 for three shots. What is the expected value? (Please do not use this example in your answer)**

** And so I came up with this example (which if you honestly think it sucks, if you can help me change it, if needed, it would be great).**

** I would like to thank you in advance.**

** I hope to hear from you soon!**

** Expected Value = (Cost to play the game x the chance of losing) + (the gain if you win x the chance of winning)**

** Here’s further proof the formula works. Maybe you’re familiar with the casino game of roulette. A quick Google search will tell you the house percentage for roulette is .0526. That’s the house edge, so expressed differently, the percentage against the player is – .0526.**

** I said earlier I would explain more about why I didn’t like the two games in your e-mail. (The game you proposed and the game in the assignment example. There is no way to determine the Expected Value. There is no way to determine the exact odds of winning.**