How many dots appear on a pair of dice

• العربية • Aragonés • Asturianu • Basa Bali • বাংলা • Bân-lâm-gú • Български • Brezhoneg • Català • Čeština • Cymraeg • Dansk • Davvisámegiella • Deutsch • Eesti • Ελληνικά • Español • Esperanto • Euskara • فارسی • Français • Gaeilge • Galego • 한국어 • Հայերեն • Hrvatski • Bahasa Indonesia • Íslenska • Italiano • עברית • Jawa • Kiswahili • Latina • Latviešu • Lietuvių • Lingua Franca Nova • Magyar • मराठी • مصرى • Bahasa Melayu • 閩東語 / Mìng-dĕ̤ng-ngṳ̄ • Nederlands • 日本語 • Nordfriisk • Norsk bokmål • Norsk nynorsk • Occitan • Plattdüütsch • Polski • Português • Română • Runa Simi • Русский • Scots • Seeltersk • Sicilianu • Simple English • Slovenčina • Slovenščina • کوردی • Српски / srpski • Srpskohrvatski / српскохрватски • Sunda • Suomi • Svenska • Tagalog • தமிழ் • Taqbaylit • ไทย • Türkçe • Українська • اردو • Tiếng Việt • Võro • 文言 • 吴语 • 粵語 • 中文 Ban-sugoroku is similar to e-sugoroku is a Use [ ] Dice are thrown onto a surface either from the hand or from a container designed for this (such as a cup or tray). The face (or corner, in cases such as tetrahedral dice, or edge, for odd-numbered The result of a die roll is determined by the way it is thrown, according to the laws of One typical contemporary Construction [ ] Arrangement [ ] The pips on standard six-sided dice are arranged in specific patterns as shown. Asian style dice bear similar patterns to Western ones, but the pips are closer to the center of the face; in addition, the pips are differently sized on Asian ...

Suppose that the first die we roll comes up as a 1. The other die roll could be a 1, 2, 3, 4, 5, or 6. Now suppose that the first die is a 2. The other die roll again could be a 1, 2, 3, 4, 5, or 6. We have already found 12 potential outcomes, and have yet to exhaust all of the possibilities of the first die. The easiest way to solve this problem is to consult the table above. You will notice that in each row there is one dice roll where the sum of the two dice is equal to seven. Since there are six rows, there are six possible outcomes where the sum of the two dice is equal to seven. The number of total possible outcomes remains 36. Again, we find the probability by dividing the event frequency (6) by the size of the sample space (36), resulting in a probability of 1/6. In the previous problem, you may have noticed that the cells where the sum of the two dice is equal to seven form a diagonal. The same is true here, except in this case there are only two cells where the sum of the dice is three. That is because there are only two ways to get this outcome. You must roll a 1 and a 2 or you must roll a 2 and a 1. The combinations for rolling a sum of seven are much greater (1 and 6, 2 and 5, 3 and 4, and so on). To find the probability that the sum of the two dice is three, we can divide the event frequency (2) by the size of the sample space (36), resulting in a probability of 1/18. Again, we can easily solve this problem by consulting the table above. You will notice that ...

• • Quiz Categories • • • • • • • • • • • • • • • • • • Quiz Lab • • • • • Quizzes by... • • • • • • • Community • • • • • Badges • Playlists • Quests • Create • Showdown • Live 5 • Trivia Bingo • Activity • Puzzle Library • • Sporcle Events • • • • • • • • • • Videos • Blog • Mobile Apps • Remove Ads and Go Orange • Feedback • • • • • • • • • • • • • • • • Quiz Lab • Create • Community

What is the average number of rolls until a shooter "sevens out"? I know that a 7 will appear every 6 rolls, but with come-out 7-11s and craps, plus the possibility of shooters making multiple points, I think the average number of rolls may be higher than expected. Is there any mathematical reference material on this? Just a question about an Oriental dice game, where the players are supposed to guess which side of the die shows up. The players will first place their bets on 1,2,3,4,5,6 (like roulette) and then the "dealer" will roll 3 dice simultaneously. Payouts are 1:1 if the chosen numbers shows up once (on any of the 3 dice), 2:1 if the chosen no shows up twice, and 3:1 if the chosen number appears on all 3 dice. As the player can place any number of bets of the board, what will be the optimum number of bets to place? (assuming all my bets are equal in size) The probability of three matching is 1/216. The probability of two matching is 3*5/216. The probability of one matching is 25*5/216. The probability of 0 matching is 5*5*5/216. So the expected return is 3*(1/216)+2*(15/216)+1*(75/216)-1*(125/216)=-17/216=-7.87%. There is no optimal number of bets, you will give up an expected 7.87% of total money bet no matter what you do. These bets can be made in both The pair can be any one of 6 numbers. The other two singletons can be among the other five. So there are 6*combin(5,2)=60 combinations already. There are combin(4,2)=6 combinations of dice on which the pair can app...