Statistics of Dice Throw Statistics of Dice Throw The probababilities of different numbers obtained by the throw of two dice offer a good introduction to the ideas of probability . For the throw of a single die, all outcomes are equally probable. But in the throw of two dice, the different possibilities for the total of the two dice are not equally probable because there are more ways to get some numbers than others. There are six ways to get a total of 7, but only one way to get 2, so the "odds" of getting a 7 are six times those for getting "snake eyes". This simple example raises the idea of distinguishable states. For example, throwing a 3 is twice as likely as throwing a 2 because there are two distinguishable ways to get a 3. The probability of getting a given value for the total on the dice may be calculated by taking the total number of ways that value can be produced and dividing it by the total number of distinguishable outcomes. So the probability of a 7 on the dice is 1/6 because it can be produced in 6 ways out of a total of 36 possible outcomes.

Go Back Of Dice and the Binomial Distribution The throw of a die or the picking of a card out of a deck are perhaps the most visible examples of the statistics of random events. Most of the conceptual tasks in probability for these kind of events can be handled with the binomial distribution . The binomial distribution can serve as a kind of "multi-tool" for common statistical questions. Given one throw of a die, what is the probability of throwing a 2? What is the probability of throwing two 2's in a row? What is the probability that in six throws of the die you will not throw any twos? What is the probability that you will throw at least one 2 in six throws? What is the average number of 2's that you would throw with a total of six throws of the die? The average number for a given outcome is the number of trials times the probability for that outcome. So the average is = np = 6(1/6) =1. What is the probability that you will throw exactly one 2 in six throws? Activity: An Experiment with Dice Many people think that one of these cubes is called "a dice". But no! The plural is dice, but the singular is die: i.e.

1 die, 2 dice. The common die has six faces: We usually call the faces 1, 2, 3, 4, 5 and 6. Throwing Two Dice and Adding the Scores ... Example: when one die shows 2 and the other shows 6 the total score is 2 + 6 = 8 Question: Can you get a total of 8 any other way? What about 6 + 2 = 8 (the other way around), is that a different way? Yes! Because the two dice are different. Example: imagine one die is colored red and the other is colored blue. There are two possibilities: So 2 + 6 and 6 + 2 are different. And you can get 8 with other numbers, such as 3 + 5 = 8 and 4 + 4 = 8 High, Low, and Most Likely Before we start, let's think about what might happen. Question: If you throw 2 dice together and add the two scores: 1. What is the least possible total score? 2. What is the greatest possible total score? 3. What do you think is the most likely total score?

The first two questions are quite easy to answer: 1. The least possible total score must be 1 + 1 = 2 2. The greatest possible total score must be 6 + 6 = 12 3. The most likely total score is ... ??? Are they all just as likely? Or will some happen more often? To help answer the third question let us try an experiment. The Experiment Throw two dice together 108 times, add the scores together each time, record the scores in a tally table. Why 108? That seems a strange number to choose. I will explain later. You can record the results in this table using tally marks : Added Can you see the Symmetry in this table? 2 and 12 have the same number of ways = 1 each 3 and 11 have the same number of ways = 2 each 4 and 10 have the same number of ways = 3 each 5 and 9 have the same number of ways = 4 each 6 and 8 have the same number of ways = 5 each 108 Throws OK, why 108 throws?

Well, 36 throws isn't enough for good results, 360 throws is be great but takes a long time. So 108 (which is 3 lots of 36) seems just right. So let's multiply all these numbers by 3 to match our total of 108: Total Those are the theoretical values, as opposed to the experimental ones you got from your experiment. The theoretical values look like this in a bar graph: How do these theoretical results compare with your experimental results? This graph and your graph should be quite similar, but they are not likely to be exactly the same, as your experiment relied on chance, and the number of times you did it was fairly small. If you did the experiment a very large number of times, you should get results much closer to the theoretical ones. And, by the way, we've now answered the question from near the beginning of the experiment: What is the most likely total score? 7 has the highest bar, so 7 is the most likely total score. Hey, is that why people talk about Lucky 7 ... ?

Probability On the page Probability you will find a formula: Probability of an event happening = Number of ways it can happen Total number of outcomes Example: Probability of a total of 2 We know there are 36 possible outcomes. And there is only 1 way to get a total score of 2. So the probability of getting 2 is: Probability of a 2 = 1 36 Doing that for each score gets us: Total Probabilities for Rolling Two Dice By Courtney Taylor Updated April 20, 2016. One popular probability problem is to roll a die . A standard die has six sides with numbers 1, 2, 3, 4, 5 and 6. If the die is fair (and we will assume that all of them are), then each of these outcomes are equally likely. Since there are six possible outcomes, the probability of obtaining any side of the die is 1/6. Thus the probability of rolling a 1 is 1/6, the probability of rolling a 2 is 1/6 and so on for 3, 4, 5 and 6. But what happens if we add another die? What are the probabilities for rolling two dice?

What Not to Do To correctly determine the probability of an event we need to know two things. First, how often the event occurs. Then second divide the number of outcomes in the event by the total number of outcomes in the sample space . Where most go wrong is to miscalculate the sample space. Their reasoning runs something like this: “We know that each die has six sides. We have rolled two dice , and so the total number of possible outcomes must be 6 + 6 = 12.” continue reading below our video 4 Tips for Improving Test Performance Although this explanation was straightforward, it is unfortunately incorrect. It’s plausible that going from one die to two should cause us to add six to itself and get 12, but this comes from not thinking carefully about the problem. A Second Attempt Rolling two fair dice more than doubles the difficulty of calculating probabilities. This is because rolling one die is independent of rolling a second one. One roll has no effect on the other one. When dealing with independent events we use the multiplication rule . The use of a tree diagram demonstrates that there are really 6 x 6 =36 outcomes from rolling two dice.

To think about this, suppose that the first die we roll comes up as a 1. The other die could be either a 1, 2, 3, 4, 5 or 6. Now suppose that the first die is a 2. The other die again could be either 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. A table of all 36 of the outcomes are in the table below. Sample Problems With this knowledge we can calculate all sorts of two dice probability problems. A few follow: Two fair six-sided dice are rolled. What is the probability that the sum of the two dice is seven? Two fair six-sided dice are rolled. What is the probability that the sum of the two dice is three? Two fair six-sided dice are rolled. What is the probability that the numbers on the dice are different? Three (Or More) Dice The same principle applies if we are working on problems involving three dice . We multiply and see that there are 6 x 6 x 6 = 216 outcomes.

As it gets cumbersome to write the repeated multiplication, we can use exponents to simplify our work. For two dice there are 62 outcomes. For three dice there are 63 outcomes. In general, if we roll n dice, then there are a total of 6n outcomes. Outcomes for Two Dice Probability - Dice - Wizard of Odds Home › Ask The Wizard › Probability - Dice Probability - Dice If you are rolling 6 six-sided standard dice what are the odds of rolling six of a kind? Jeff B. from Miami, Florida The answer is 6*(1/6)6 = 6/46,656 = 1/7,776 =~ 0.0001286 . 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? Grshooter from Kansas City, Missouri The average number of rolls per shooter is 8.525510.

For the probability of exactly 2 to 200 rolls, please see my craps probability of survival page . What are the odds (and frequency) in 100,000 rolls of the dice (craps/dont pass line) of losing a DP bet 2x’s.3x’s 4x’s 5x’s 6x’s 7x’s 8x’s or 9x’s in a row. Derick from Minneapolis, USA My craps appendix shows how to work out the odds for any one bet. There you will see the probability of losing the don’t pass bet is 2928/5940. The probability of losing n bets in a row is (2928/5940)n. The frequency in 100,000 of losing exactly n can be closely approximated as 100,000 * (2928/5940)n+2. What are the odds of rolling the same number with six dice in one roll?

Kyle Hill from Colstrip, Montana The odds of getting six of the same number with six dice is 6*(1/6)6=1/7776 =~ 0.01286%. How often can you roll a pair of dice 28 times without getting a 7? How do you figure this? Congratulation on your site, it’s great. Arturo G. from Mexico City, Mexico Thanks for the compliment. I take it you mean what is the probability of rolling a pair of dice 28 times without getting a 7. The probability of not rolling a 7 on any one roll is 5/6. The probability of not rolling a 7 in 28 rolls is (5/6)28 = 0.006066, or about 1 in 165. 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) Jansen from Toronto, Canada 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 sic bo and chuck a luck . Mr.

Wizard, what is the probability of rolling two pair when rolling four dice? Brian from St. Catherines, Canada There are combin(6,2)=15 different sets of pairs possible. There are combin (4,2)=6 ways the dice can roll any specific two pair. There are 6^4=1296 ways to roll four dice. So the probability is 90/1296=6.9444%. If I roll a die, my probability of rolling a six is 1/6. If I roll two dice, does my probability of rolling a six on one of them increase, or does it stay at 1/6? Mike R. from Rosemount If you rolled x dice the probability of getting at least one 6 is 1-(5/6)2. In the case of two dice this is 30.56%. How many different ways are there of rolling 3 ones using 6 dice? Jamie from Croydon, England First there are combin (6,3)=20 ways you can choose three dice out of 6 for the three ones. Then each of the other three can be any of five numbers.

So, the total ways are 20×53=2500. The total number of ways to throw all the dice are 66=46,656, so the probability of rolling exactly three ones is 2500/46656=0.0536. For help with the combin function see my probabilities in poker section. What is the probability of any one die showing ONE when 3 dice are throw together. My understanding is it should be 50% chance (1/6+1/6+1/6=1/2 -->50%) But in your odds table it shown 34.72%. Please help. John C. from Singapore The probability of rolling exactly one one with three dice is 3*(5/6)2*(1/6) = 75/216 = 34.72%. What is the probability of rolling a "pair" when tossing 4 dice? Anthony from Toronto, Canada 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 appear. The two singletons can be arrange in two ways. So there are 60*12=720 ways to throw a pair. The total number of all ways to throw the dice is 64=1296. So the probability is 720/1296 =~ 55.56%. My friend owns a bar and has a "shake of the day" where there are ten dice in a Tupperware container, what are the odds of matching 8 out of the 10 in one shake. Thank you for your time. August from Oshkosh, USA The probability that if you roll 10 dice and exactly 8 numbers are the same is 6*combin(10,8)*(1/6)8*(5/6)2 = 1/8957.952.

The probability of matching at least 8 is 6*[combin(10,8)*(1/6)8*(5/6)2 + combin(10,9)*(1/6)9*(5/6) + (1/6)10] = 1/8569.469. I recently rolled, during a game of backgammon, double sixes four consecutive times. What are the odds of this happening again? David from Sunland, USA With every new roll the probability the next four rolls will be all double sixes is (1/36)4 = 1 in 1679616. What is the probability of getting a straight by a single throw of 5 dice? Teodoro C. Deocares from Dagupan, Philippines There are two possible spans: 1 to 5 and 2 to 6. Each of these spans can be ordered in 5!=120 ways. There are 65 = 7776 ways to roll five dice. So the probability is 2*120/7776 = 3.09%.

The probability of this seems to be much higher right after I put mark 0 for large straight during a game of Yahtzee. A fair sided die is rolled 30 times. What is the expected number of times that number 1 will come up? What is the probability that number 1 will come up it’s expected number of times? Anonymous The expected number of ones is 30*(1/6) = 5. The probability of exactly 5 ones is combin(30,5)*(1/6)5*(5/6)25 = 19.21%. If I have any give number of dice what is the probability that if I roll them all of them at least one will land on a one? Anonymous The probability that all the dice will not be a one is (5/6)n. So the probability of at least one 1 is 1-(5/6)n. Let’s take an example of five dice. The answer would be 1-(5/6)5 = 59.81%. If I throw 36 dice what is the probability of getting at least one six?

Anonymous 1-(5/6)36 = 99.86% If I kept throwing and removed all the sixes each time, how would I predict the theoretical number of dice remaining after a particular number of throws? Anonymous Each roll the expectation is that 5/6 of the dice will remain. So the expected number of dice remaining after n throws would be 36*(5/6)n. For example after 10 throws you would have 5.81 dice left, on average. If I roll three dice, what is the probability of getting at least two numbers the same? Anonymous The probability all numbers will be different is (5/6)*(4/6)=20/36. So the probability at least two numbers will be the same is 1-(20/36) = 16/36 = 44.44%. If two people throw a pair of dice, What is the probability that it is the same number? Is there a formula to figure this out? Anonymous Yes. You simply run through all total from 2 to 12 and determine the probability of rolling each twice.

So the answer would be (1/36)2+(2/36)2+(3/36)2+(4/36)2+(5/36)2+(6/36)2+(5/36)2+(4/36)2+(3/36)2+(2/36)2+(1/36)2 = 11.27%. I recently attended a hospital fete. There was a new car as a prize if 7 dice produced 7 sixes in one throw. £1.00 a go. Odds on this must be high but how high? Anonymous The probability of throwing seven sixes with seven dice is (1/6)7 = 1 in 279,936. So the car would have to have a value of £279,936 or more for this to be a good bet. Even your average Rolls Royce is not worth this much, so I would say that was a terrible bet. [Bluejay adds: Uh, yeah, but I think the point was that it was for charity.

What’s more fun: Donating £1.00 to charity and getting nothing back but the good feeling of helping out, or donating £1.00 and getting the good feeling plus the longshot chance of winning a car?] What are the probabilities for a 5 of a kind, 4 of a kind, 3 of a kind, full house, 2 pair, pair, straight, and nothing with the roll of five dice? Anonymous Five of a kind: 6/65 = 0.08% (obvious) Four of a kind: 5*6*5 = 1.93% (five possible positions for the singleton * 6 ranks for the four of a kind * 5 ranks for the singleton). Full house: combin(5,3)*6*5/65 = 3.86% (combin(5,3) positions for the three of a kind * 6 ranks for the three of a kind * 2 ranks for the pair). Three of a kind: COMBIN(5,3)*COMBIN(2,1)*6*COMBIN(5,2) / 65 = 15.43%.

(combin(5,3) positions for the three of a kind * combin(2,1) positions for the larger of the singletons * 6 ranks of the three of a kind * combin(5,2) ranks for the two singletons. Two pair: COMBIN(5,2)*COMBIN(3,2)*COMBIN(6,2)*4 / 65 = 23.15% (combin(5,2) positions for the higher pair * combin(3,2) positions for the lower pair * combin(6,4) ranks for the two pair * 4 ranks for the singleton. Pair: COMBIN(5,2)*fact(3)*6*combin(5,3) / 65 = 46.30% (combin(5,2) positions for the pair * fact(3) positions for the three singletons * 6 ranks for the pair * combin(5,3) ranks for the singletons.

Straight: 2*fact(5) / 65 = 3.09% (2 spans for the straight {1-5 or 2-6} * fact(5) ways to arrange the order). Nothing: ((COMBIN(6,5)-2)*FACT(5)) / 65 = 6.17% (combin(6,5) ways to choose 5 ranks out of six, less 2 for the straights, * fact(5) ways to arrange the order. Hi Wizard, I wanted to know if you can answer this. In a popular gambling game in 17th century France, a player would roll a pair of dice 24 times. He would win his bet if at least one of these rolls was a double six. There was a debate at the time over whether the probability of winning was above or below an even 50%. Can you help me? Anonymous Sure, this is easy. The probability of rolling at least one 12 in 24 rolls is 1-(35/36)24 = 49.14%. So the odds favor betting against a 12.

This is a clever bet because the expected number of twelves in 24 rolls is 2/3. However that does not mean the probability of a 12 is 2/3, because sometimes there will be more than one 12, and the player betting on 12 doesn’t win any more for extra twelves after the first one. If the probability of winning any given trial is p, the number of trials is n, and the probability of at least one win is w then solving for n in terms of p and w gives us... w=1-(1-p)n log(1-w) = n*log(1-p) n= log(1-w)/log(1-p) So in your example n = log(1-.5) / log(1-(1/36)) = log(0.5) / log(35/36) = 24.6051. So if the probability of success is 50% in 24.6 rolls it must be slightly less in 24 rolls. What is the probability of rolling 1,2,3,4,5,6 with six dice, six times in a row?

Anonymous The probability of rolling 123456 with six dice in a single roll can be expressed as prob(second die does not match first die) * prob(third die does not match first or second die) * ... = 1*(5/6)*(4/6)*(3/6)*(2/6)*(1/6) = 0.015432. So the probability of doing this six times in a row is 0.0154326 = 1 in 74,037,208,411. If I roll a single die 6 times, what is the probability of getting a "2" exactly 4 times? Anonymous Combin(6,2)*(1/6)4*(5/6)2 = 0.008037551. What is the probability of rolling 13 or more with 3, 4, and 5 dice, if you are allowed to keep the highest three dice in your roll? Anonymous 4 dice: 48.77% 5 dice: 66.13%. Suppose we roll three fair six-sided dice.

What’s the conditional probability that the first dice shows 4, given that the sum of the three numbers showing is 12? Shikha from North Ryde The probability of A given B is the probability of A and B divided by the probability of B. In this case the probability of rolling a 4 on the first die and then a total of 8 on the other two is (1/6)*(5/36) = 5/216. The probability of rolling any total of 12 with 3 dice is 25/216, as shown in my sic bo section. So the answer is (5/216)/(25/216) = 5/25 = 20%. In a recent programming exercise myself and other students were asked to describe a six-sided die in code, and then use our dice to determine play simple game. The object of the game was to roll the dice until the sum of the tosses reached exactly 100. Any toss that put the total over 100 would not be added and merely added to statistics.

Quickly it was determined that 17 throws would be the least amount of throws needed to reach 100. However calculating the odds of that occurring has proved elusive. Calculating the odds of a specific sequence of throws is rather straight forward, but how might one factor in both non-specific ordering of throws, and the different ways of reaching 100 in 17 throws (16*6 + 1*4 and 15*6 + 2*5)? from The two ways you mention are the only ways to throw a total of 100 in 17 throws. The probability of throwing 16 sixes and one four is 17*(1/6)17. There are 17 possible positions of the 4 and each sequence has a probability of (1/6)*(1/6)*...*(1/6) with 17 terms. The number of ways to get 15 sixes and 2 fives is combin (17,2) = 136. So the probability of 15 sixes and 2 fives is 136*(1/6)17.

So the total probability is (17+136)*(1/6)17. = 1 in 110,631,761,077. There are 3 dice, 2 are proper six sided dice, while one is a die with all sides containing a six. All the dice are in my pocket. I randomly take out a die and throw it. The result is a 6. What is the probability that the die was one of the proper dice with 6 different values? Annojh from Toronto Let A = Choosing the normal die Let B = Rolling a 6 with randomly chosen die Answer = Pr(A given B) = Pr(A and B)/pr(B) = ((2/3)*(1/6))/((2/3)*(1/6)+(1/3)*1) = (2/18)/((2/18)+(6/18)) = 1/4. If you can roll six dice only once, what is the probability of rolling 6,6,6,6,1, and 4 in any order?

Aubrey from Kokomo There are 6!/(4!*1!*1!) = 30 ways to arrange these numbers in any order. Another way to look at it is there are 6 positions to put the 1, and 5 left to put the 4, so 6*5=30. The probability of getting 666614 in exactly that order is 1 in 66 = 1 in 46656. Multiply that by 30 for the 30 possible orders and the answer is 30/46656 = 0.0643%, or 1 in 1552.2. My understanding of "wait time" for an event is the reciprocal of the probability of that event. I’m interested in calculating the wait time to roll consecutive 2s using one die. In a simulation I get 42 rolls on average. How do I make the connection with the probability of rolling consecutive 2s? Lee from Andover It is true that for single events if the probability is p then the average wait time is 1/p. However it gets more complicated with consecutive events.

Let x be the state that the last roll was not a two. This is also the state at the beginning. Let y be the state that the last roll was a two. After the first roll there is a 5/6 chance we will still be in state x, and 1/6 chance we will be in state y. Let Ex(x) be the expected number of rolls from state x, and Ex(y) the expected number from state y. Then... Ex(x) = 1 + (5/6)*ex(x) + (1/6)*ex(y), and Ex(y) = 1 + (5/6)*ex(x) Ex(x) = 36*(7/6) = 42 So the average wait time for two consecutive twos is 42 rolls. I have the same type of problem, only the expected flips to get two heads, in my site of math problems , see problem 128. Can you tell me the odds of rolling two of the same number with two dice, three dice, and four dice? I am wondering how many dice would one have to roll at one time so that the odds are on the side of the person rolling the dice.

(It does not make any difference which number is doubled.) Mary from Minneapolis, MN Here is the probability of getting at least one number more than once according to the number of rolls: Probability of a Pair or More Rolls 6 rolls 98.46% So if you were to book this you should offer the yes with 3 rolls or the no at 4 rolls. What is the chance of getting a sum over 100, when rolling 20 dice? Kind regards Terje from Stockholm I started to use the Normal approximation to solve this, but the probability of over 100 points is too low for that method to be accurate. So I did a random simulation of 8.25 million trials and the number of trails that were 101 points or more was 127. So the probability is about 1 in 65,000. Wizard, could you please describe the equivalent odds of the California SuperLotto Plus (1 in 41.4 Million), in terms of number of consecutive times of rolling 7 or 11? I heard it somewhere before. Most people cannot comprehend the lottery odds. But, the rolling of dice -- they can relate.

Tim from Belmont, CA Let n be your answer. The probability of rolling a 7 or 11 is 8/36. To solve for n: (8/36)n = 1/41,400,000 n = -7.617 / -0.65321 n = 11.6608 So there you go, the probability of hitting the SuperLotto is the same as rolling a seven or eleven 11.66 times in a row. For those who can’t comprehend a partial throw I would rephrase as the probability falls between 11 and 12 consecutive rolls. We are in a disagreement between workers. there is a bar down the street that has a shake a day. which is you must throw five dice at once and all five must end up being the same "like yahtzee" but he gives you three chance at it. but you must pick up all the dice all three times. so the questions is what’s the odds to do it in one shake and what’s the odds to do it in the three shakes allowed. Thanks , if you already answered this before i am sorry but i couldt find it.

Dan and co workers at maple island from Forest Lake The probability of a five-of-a-kind on one throw is 6*(1/6)5 = 1/1,296. This is because there are six different five-of-a-kinds (one to six) and the probability each die will be that number is (1/6). The probability of not getting a five-of-a-kind is 1-(1/1,296)=1,295/1,296. The probability of going three attempts without a three of a kind is (1,295/1,296)3=99.77%. So the probability of getting at least one five-of-a-kind in three tries is 100%-99.77% = 0.23%. What is the expected number of tosses required in order to obtain at least one of each of the possible outcomes on an unbiased 6 sided dice? Michael from Melbourne If the probability of something is p then on average it will take 1/p trials for it to happen the first time. Obviously, on the first roll you’ll cross off one number.

The probability of rolling one of the other five numbers next is 5/6. So it will take on average 1/(5/6)=6/5=1.2 rolls for that to happen. Following that reasoning to the end, the expected number of rolls is (6/6)+(6/5)+(6/4)+(6/3)+(6/2)+(6/1) = 14.7. What is the classical probability of getting a total of 12 when 5 balanced dice are rolled? Kennith H. from Winters I hope you’re happy, I just added a new section answering questions such as this for 1 to 25 dice. As the five-dice table shows, the probability of rolling a total of 12 is 0.039223251028807. Somebody bet he that he could roll a total of 6 and 8, with two dice, before I could roll a total of seven twice. This seemed like a good bet because seven is the most common total. However I lost $2500 doing this over and over. What are the odds?

Anthony from Indiana I’m afraid you had the square side of this bet. The probability of rolling two sevens before a six and eight is 45.44%. Here are all the possible outcomes. The first column is the order of petintent rolls to the outcome of the bet, ignoring all others. Two Sevens before Six and Eight Bet Relavant Rolls (6/16)*(5/16)*(6/11) Win Basically, the reason the 6 and 8 is the better side is you can hit those numbers in either order: 6 then 8, or 8 then 6. With two sevens there is only one order, a 7 and then another 7. Two questions, please: 1) What is the probability of rolling 6,6,6,6,6,6 @ one time, with (6)6-sided die? 2)What is the probability of rolling 1,2,3,4,5,6 @ one time with (6) 6-sided die? Thanks! It’s killing me! Heather from Petaluma The probability of six sixes is (1/6)6 = 1 in 46656.

The probability of rolling 1,2,3,4,5,6 with six dice is 6 ! /66 = 1 in 64.8 What is the probability of getting any given number more than once if you roll a die ten times? Ryan from Silay 1-(5/6)10-10 × (1/6) × (5/6)9 = 51.55%. In a game called Taxation and Evasion, a player rolls a pair of dice. On any roll if the sum is 7,11, or 12, the player gets audited; any other sum they avoid taxes. If a player rolls the pair 5 times, what is the probability that he avoids taxes? George P. from Stevens Point, WI The probability of a 7, 11, or 12 is (6+2+1)/36 = 9/36 = 1/4. See my section on dice probability basics for how I arrived at that figure. The probability of rolling anything else is 3/4.

The probability of going five rolls without rolling a 7, 11, or 12 is (3/4)5 = 23.73%. How many throws of a die does it take before it is likely that you have thrown a 1, 2, 3, 4, 5, and 6 at least once each? Any ideas on generalizing this for an n-sided die? Asif from Columbia, SC Not that you asked, but let me address the mean first. For a six-sided die, the expected number of throws to get each face at least once is (6/6)+(6/5)+(6/4)+(6/3)+(6/2)+(6/1) = 14.7. For an n-sided die the expected throws is (n/n) + (n/(n-1)) + (n/(n-2)) + ... + n. The median number of throws required is 13. The probability of taking 13 rolls or less is 51.4%, and 13 rolls or more is 56.21%. I know you’re skeptical of dice control.

I have been practicing dice setting and controlled shooting for 3 months. What is the probability of throwing 78 sevens over 655 throws randomly? Thanks for the help :) Eric B. from Boston, MA For large numbers of throws we can use the Gaussian Curve approximation. The expected number of sevens in 655 throws is 655 × (1/6) = 109.1667. The variance is 655 × (1/6) × (5/6) = 90.9722. The standard deviation is sqr(90.9722) = 9.5379. Your 78 sevens is 109.1667 − 78 = 31.1667 less than expectation. This is (31.1667 - 0.5)/9.5379 = 3.22 standard deviations below expectation. The probability of falling 3.22 or more standard deviations south of expectations is 0.000641, or 1 in 1,560.

I got this figure in Excel, using the formula, normsdist(-3.22). This is about controlling the dice at Craps. You previously discussed the Stanford Wong Experiment , stating, "The terms of the bet were whether precision shooters could roll fewer than 79.5 sevens in 500 rolls of the dice. The expected number in a random game would be 83.33. The probability of rolling 79 or fewer sevens in 500 random rolls is 32.66%.... The probability of rolling 74 or fewer sevens in 500 random rolls is 14.41%." The question I have about this bet is that 14.41% still isn’t "statistically significant" [ i.e. p < 0.05 ] , which is usually taken to mean greater than two Standard Deviations from the Mean -- or a probability of less than a *combined* 5% of the event happening randomly on EITHER end of the series.

How many Sevens would have to be rolled in 500 rolls before you could say that there is a less than 2.5% chance that the outcome was entirely random (i.e. that the outcome was statistically significant) ? Many Thanks & BTW , yours is ABSOLUTELY the BEST web site on the subject of gambling odds & probabilities that I’ve found .... keep up the good work !!! Plexus from Warwick, Rhode Island Thank you for the kind words. You should not state the probability that the throws were non-random is p. The way it should be phrased is the probability that a random game would produce such a result is p. Nobody expected 500 rolls to prove or disprove anything. It wasn’t I who set the line at 79.5 sevens, but I doubt it was chosen to be statistically significant; but rather, I suspect the it was a point at which both parties would agree to the bet. The 2.5% level of significance is 1.96 standard deviations from expectations. This can be found with the formula =normsinv(0.025) in Excel.

The standard deviation of 500 rolls is sqr(500*(1/6)*(5/6)) = 8.333. So 1.96 standard deviations is 1.96 * 8.333 = 16.333 rolls south of expectations. The expected number of sevens in 500 throws is 500*(1/6) = 83.333. So 1.96 standard deviations south of that is 83.333 − 16.333 = 67. Checking this using the binomial distribution, the exact probability of 67 or fewer sevens is 2.627%. What is the expected number of rolls needed to get a Yahtzee? Ian F. from Provo Assuming the player always holds the most represented number, the average is 11.09. Here is a table showing the distribution of the number of rolls over a random simulation of 82.6 million trials. Yahtzee Experiment 82600000 1 Consider a hypothetical game based on the roll of a die.

If die lands on 1, the player loses $1 and the game ends. If the die lands on anything else, the player’s wins $1. At this point the player may let it ride, or quit. The player may keep playing, doubling each bet, until he loses or quits. What is the best strategy? Byron P. from Newington, CT Speaking only in terms of maximizing expected value, the player should play forever. While the probability is 1 that the player will eventually lose, at any given decision point the expected value always favors going again. It seems like a paradox. The answer lies in the fact that some events have a probability of 1, but still may not happen. For example, if you threw a dart at a number line from 0 to 10, the probability of not hitting pi exactly is 1, but it still could happen. However, for practical purposes, there is some stopping point. This is because the happiness money brings is not proportional to the amount. While it is commonly accepted that more money brings more happiness, the richer you get, the less happiness each additional dollar brings you. I believe a good way to answer this question is to apply the Kelly Criterion to the problem.

According to Kelly, the player should make every decision with the goal of maximizing the expected log of his bankroll after the wager. To cut to the end of this (I cut out a lot of math), the player should keep doubling until the wager amount exceeds 96.5948% of his total wealth. Wealth should be defined as the sum of the amount won plus whatever money the player had before he made the first wager. For example, if the player had $100,000 to start with, he should keep doubling up to 23 times, to a win of $4,194,304. At that point the player’s total wealth will be $4,294,304. He will be asked to wager 4,194,304/4,294,304 = 96.67% of his total wealth, which is greater than the 96.5948% stopping point, so he should quit. Players A and B throw a pair of dice.

Player A wins if he throws a total of 6 before B throws a toal of 7, and B wins if he throws 7 before A throws 6. If A begins, show that A’s chances of winning are 30/61. Sangeeta from Mumbai, India Let the answer to this question be called p. The probability of rolling a total of six is 5/36, and the probability of rolling a total of seven is 6/36. If you don’t understand why, please see my section on dice probability basics . We can define p as: p = Prob(6 on first roll) + Prob(no 6 on first roll)*Prob(no 7 on second roll)*p. This is because, if neither player wins after the first two rolls, the game is back to the original state, and the probability of player A winning remains the same. So, we have: 98280 Credit to Alan Tucker, author of Applied Combinatorics . Can you calculate what the probability is of two numbers coming up behind each other in a roll of the dice?

Meaning what is the probability of two 4’s or two 6’s or two 7’s back to back? I realize that the past cannot predict the future but is there a way to calculate 7/36 X 7/36 happening back to back? I hope that makes sence. James from Birmingham Sure. That would be Pr(2)2 + Pr(3)2 + ... + Pr(12)2 = (1/36)2 + (2/36)2 + (3/36)2 + (4/36)2 + (5/36)2 + (6/36)2 + (5/36)2 + (4/36)2 + (3/36)2 + (2/36)2 + (1/36)2 = 11.27%. In 180 consecutive rolls of the dice, how many times can I expect to see the following: Two sevens in a row? Three sevens in a row? Four sevens in a row? Thanks for your time :-). Melanie D.

from Elizabeth City, NC I can’t think of any useful reason to know this information, but I get asked this kind of thing a lot, so I’ll humor you. It is a little easier getting a specified sequence of sevens starting with the first roll, or ending with the last, because the sequence is bounded on one side. Specifically, the probability of getting a sequence of s sevens, starting with the first roll, or ending with the last, is (1/6)s × (5/6). The 5/6 term is because you have to get a non-7 at the open end of the sequence. The probability of starting a sequence of s sevens at any point in the middle of the sequence is (1/6)s × (5/6)2. We square the 5/6 term, because the player must get a non-7 on both ends of the sequence. If there are r rolls, there will be 2 places for an inside sequence, and r-n-1 places for a run of n sevens. Putting these equations in a table, here is the expected number of runs of sevens, from 1 to 10.

The "inside" column is 2*(5/6)*(1/6)r, and the "outside" column is (179-r)*(5/6)2*(1/6)r, where r is the number of sevens in the run. So, we can expect 3.46 runs of two sevens, 0.57 runs of three sevens, and 0.10 runs of four sevens. Expected Runs of Sevens in 180 Rolls Run 0.000002 0.000002 Two dice are rolled until either a total of 12 or two consecutive totals of 7. What is the probability the 12 is rolled first? Anonymous The answer and solution can be found on my companion site, mathproblems.info , problem 201. I’m a tabletop gamer, and was having some discussion with my friends about non-cubical platonic solid dice (If you’re a big enough nerd, that means d4, d8, d12, and d20). They argued that those would be the only ones that would be demonstratively fair.

I argued that manufacturing them to be fair would be entirely too difficult. Also, the only games would be craps variants rendered overly cumbersome due to the number of extra outcomes. Has any casino ever had a game that used non-traditional six sided dice? Bayani from Carnagie, PA This is Lisa Furman, the model from my M casino review . When I tried to impress her by saying that the balloon figure on the left is a truncated icosahedron , she just smiled and rolled her eyes. Don’t you dare challenge my math nerd credentials! When I was a high school sophomore, I constructed not only all the platonic solids with poster board and electricians tape, but all the Archimedean solids as well. If you limit yourself to the regular polygons, and want every face to have the same probability, then you are limited to the platonic solids. However, if you can lift the regular polygon requirement, then you can add the 13 Catalan solids as well. To answer your other question, no, I have never seen a game actually in a casino that used any dice other than cubes.

About ten years ago I saw a game demonstrated at a gaming show in Atlantic City that I think used a Rhombic triacontahedron , one of the Catalan solids, but I don’t think it ever made it to a casino floor. There is a game I see year after year at the Global Gaming Expo that uses a spinning top (like a dreidel), but alas, I’ve never seen that in a casino either. If I roll three six-sided dice, what are the odds of rolling a straight and, also, what are the odds of rolling a three of a kind? Mark from Fargo, ND There are 63=216 ways to roll three dice. Six of those combinations will result in a three of a kind (1-1-1 to 6-6-6). So the probability of a three of a kind is 6/216 = 1/36. There are four possible spans for a straight (1-2-3 to 4-5-6). There are also 3!=6 ways to arrange the three dice in a straight. So, there are 4*6=24 straight combinations.

Thus the probability of a straight is 24/216 = 1/9. What is the average sum when rolling four six-sided dice after subtracting the lowest result (known as 4d6-L)? What is the standard deviation for this roll? Aaron from New York 1296 The mean result is 12.2446, and the standard deviation is 2.8468. My question is based on dice odds. I know that there are six ways to get 7 and one way to get 12, but what are the chances of getting six 7’s before one 12? Are they even, and if not, how many twelves should be added to the equation to make it an even proposition? nick The probability of rolling a 7 is 1/6, and the probability of rolling a 12 is 1/36. The probability of rolling a 7, given that a roll is a 7 or 12 is (1/6)/((1/6)+(1/36)) = 6/7.

So the probability that the first six times a 6 or 12 is rolled it is a 6 every time is (6/7)6 = 39.66%. If you rephrase the question to be what is the probability of rolling five 6’s before a 12, then the answer is (6/7)5 = 46.27%. With four rolls it is (6/7)4 = 53.98%. So there is no number of 7’s before a 12 that is exactly 50/50. If you’re looking for a good sucker bet, suggest you can either roll four 7’s before a 12, or a 12 before five 7’s. This question was raised and discussed in the forum of my companion site Wizard of Vegas . Is there an easy way to calculate the probability of throwing a total of t with d 6-sided dice? Anon E. Mouse Here is a handy trick, courtesy of Robert Goodhand of Somerset, UK.

First put on a row six ones surrounded by five zeros on either side, as follows: One-Die Probabilities 0 0 To get the probability of any given total, divide the number of combinations of that total by the total number of combinations. In the case of three dice, the sum is 216, which also easily found as 63. For example, the probability of rolling a total of 13 with three dice is 21/216 = 9.72%. So for d dice, you’ll need to work your way up through 1 to d-1 dice. This is very easily accomplished in any spreadsheet. How many rolls of two dice would it take to have a 50/50 chance of rolling at least one 12? Maff That is a classic problem in the history of the field of probability. Many people incorrectly think the answer is 18, because the probability of a 12 is 1 in 36, and 18×(1/36)=50%. However, by that logic, the probability of getting a 12 in 36 rolls would be 100%, which clearly it isn’t.

Here is the correct solution. Let r be the number of rolls. The probability a roll isn’t a 12 is 35/36. The probability there are 0 12s in r rolls is (35/36)r. So we need to solve for r in the following equation: (35/36)r = 0.5 r = log(0.5)/log(35/36) r = 24.6051 So there isn’t a round answer. The probability of rolling a 12 in 24 rolls is 1-(35/36)24 = 49.14%. The probability of rolling a 12 in 25 rolls is 1-(35/36)25 = 50.55%. If you want to make a bet on this, say you can roll a 12 in 25 rolls, or somebody else can’t in 24 rolls. Either way you’ll have an advantage at even money. In Dice Wars , what is the probability of success for any given number of attacking and defending dice? As an attacker, what ratio has the greatest expected gain?

Anonymous For those unfamiliar with the game, both the attacker and defender will roll 1 to 8 dice, according to how many armies they each have at that point in a battle. The higher total shall win. A tie goes to the defender. If the attacker loses, he will still retain one army in the territory where he initiated the attack. For this reason, he must have at least two armies to attack, so if he wins one can inhabit the conquered territory and one can stay behind. The following table shows the probability of an attacker victory according to all 64 combinations of total dice. Probability of Attacker Win What is the probability of forming a Yahtzee with up to n rolls of the dice? Anonymous For the benefit of other readers, a Yahtzee is a five of a kind with five dice. In the game of Yahtzee the player may hold any dice he wishes and re-roll the rest. He can do this up to three rolls. The player may re-roll previously held dice, if he wishes.

For example, if the player's first roll is 3-3-4-5-6 and he holds the threes and then has 3-3-5-5-5 after the second roll he may keep the fives and re-roll the threes on his third roll. The following table shows the maximum number of dice of the same face for 1 to 20 rolls. The table shows the probability of getting a Yahtzee within three rolls is about 4.6%. Yahtzee Probabilities Rolling the Dice - Free Math Help Rolling the dice... What are the most likely outcomes from rolling a pair of dice? Assuming we have a standard six-sided die, the odds of rolling a particular value are 1/6. There is an equal probability of rolling each of the numbers 1-6. But, when we have two dice, the odds are not as simple. For example, there's only one way to roll a two (snake eyes), but there's a lot of ways to roll a seven (1+6, 2+5, 3+4).

Let's count how many ways there are to get each value, 2 through 12: Outcome Dice Probability - Wizard of Odds Dice Probability The Field Bet Example Before you play any dice game it is good to know the probability of any given total to be thrown. First lets look at the possibilities of the total of two dice. The table below shows the six possibilities for die 1 along the left column and the six possibilities for die 2 along the top column. The body of the table shows the sum of die 1 and die 2. Two Dice Totals 11 12 The colors of the body of the table illustrate the number of ways to throw each total. The probability of throwing any given total is the number of ways to throw that total divided by the total number of combinations (36). In the following table the specific number of ways to throw each total and the probability of throwing that total is shown. Total 36 100% The following shows the probability of throwing each total in a chart format. As the chart shows the closer the total is to 7 the greater is the probability of it being thrown.

The Field Bet Example Now that we understand the probability of throwing each total we can apply this information to the dice games in the casinos to calculate the house edge. For example consider the field bet in craps . This bet pays 1:1 (even money) if the next throw is a 3, 4, 9, 10, or 11, 2:1 (double the bet) on the 2, and 3:1 (triple the bet) on the 12. Note that there are 7 totals that win and only 4 that lose which might cause someone who didn't know better to think it was a good gamble. The player's return can be defined as the sum of the products of the probability of each event and the net return of that event. The following table shows each possible total, the net return, the probability of throwing that total, and the average return. The average return is the product of the net return and the probability. The player's return is the sum of the average returns. Total Probability - throwing dice Single dice Click here for how to write a probability. Throwing dice is more complicated than tossing coins , as there are more than 2 values.

If you throw a single dice, then it can fall six ways, each of which is equally likely if the dice is true. So the probability of getting one particular value is 1/6. If you want either of two values it is 2/6 or 1/3, and so on. Sum of two dice It gets more interesting when you have two dice. One thing that you can do is work out what the total of the dice is. The dice experiment allows you to simulate throwing pairs of dice and see what the result is. This is a good introduction to probability, since you can see which combinations are more likely. But the real world, or even a simulated real world, never matches completely with calculated probability. So how do we calculate it? The first thing is to work out what the range is. You can't have a total less than 2 (both dice being 1) and you can't have a total more than 12 (both dice being 6). The easiest way to see what the probabilities is to write out the possible totals. There are 36 of them in all (6 x 6).

Total on dice 6+6 1/36 = 3% The percentages above are rounded. Compare these with the dice experiment . It won't match exactly, although the more throws you do, the closer it will get. Other conditions with two dice There are plenty of other conditions that you can have with two dice. Often, the easiest way to get the probability of a condition is to list all possible throws, and count the ones that fit the condition. The best way to make sure you have remembered all throws is to list them in a 6 x 6 table, like below. Imagine two dice, red and green. This table shows all possible throws of these dice, with red numbers for the red dice, and green for the green dice. The background is black if the condition is satisfied. You can change the numbers and the condition to see what happens. You find the probability of this condition by counting the black squares, and dividing by 36 (all possible throws). First number: See below for an explanation Either dice is a particular number Return to interactive exercise for conditions The probability of one dice being a particular number is 1/6.

You would assume that it would be twice as likely that either of two dice being a particular number, or 1/3, but this would be wrong. If you select the first condition above, you will see why. While there are six conditions where one dice is a number, and six conditions where the other dice is that number, there is one condition in both. So the probability is (12-1)/36 or 11/36. There is a better way to calculate this. If you click on the second condition, then back on the first, you will find that they are opposites of each other. That means that their probabilities add up to one. The second condition is easier to calculate directly (see below), as 25/36. So this probability is (36-25)/36 = 11/36. Neither dice is a particular number Return to interactive exercise for conditions The probability of one dice not being a particular number is 5/6. The probability of two dice not being a that number is 5/6 x 5/6 = 25/36.

You can do this as they don't overlap as all (the correct term is that they are independent events). Both dice are the same particular number Return to interactive exercise for conditions The probability of one dice being a particular number is 1/6. The probability of two dice being the same particular number is 1/6 x 1/6 = 1/36. This is not the same as saying that both dice are the same number. There are six different possible numbers, so that would be 6/36 or 1/6. At least one dice isn't a particular number This is the opposite of both dice being the same particular number, so the probabilities will add up to one. So the probability of at least one dice isn't a particular number = 1 - (1/36) = 35/36. The highest dice is a particular number Return to interactive exercise for conditions This is a fun conditions, but a little strange to work out! Try experimenting with different numbers, and you will get different collections of throws fitting the conditions, and so different probabilities. To take an example, considering one dice, if the particular number is 5 then there are 5 throws which fit the condition.

There are another 5 for the other dice, and there is one overlap, so the result is (5 + 5 - 1)/36 = 9/36 or 1/4. But this varies for other numbers. The lowest dice is a particular number The probabilities vary considering what the number is. But the probability of the lowest dice being a particular number is the same as the probability of the highest dice being six minus that particular number. The dice have two particular numbers Return to interactive exercise for conditions Like the first condition, there are two ways to work this out. The first is doing it directly. Each dice has 6 throws for each number, but there are four overlaps. So the probability is (4 x 6 - 4)/36 = 20/36. An easier way is to realise that this condition is the opposite of the next one, dice are not either number. That probability is 16/36 so this one is 1 - 16/36 = 20/36 or 5/9.

The dice are not either particular number Return to interactive exercise for conditions The probability of one dice not being either of two numbers is 4/6, so the probability of both dice fitting this condition is 4/6 x 4/6 = 16/36. We can do this as the conditions don't overlap, so they are independent events. The dice are two particular numbers There are only two ways that two dice can have two values (try the exercise above to see why), so the probability is 2/36 or 1/18. More than two dice As the number of dice increase, then the different conditions can become more complicated. You can work out some simply. The easiest is the probability of a number of dice being a particular number. For n dice, this is 1/(6n). For example, the probability for throwing 5 dice and getting them all sixes is 1/(65) = 1/(6x6x6x6x6) = 1/7776 = 0.000128 or 7775 to 1.

You can work out what is the chance of getting various numbers when throwing several dice (or one dice several times), but you do it in a back-to-front way, as above . For example, if you throw three dice, what is the chance of one of the dice getting less than five? This means that one of the dice (at least) will be a one, two, three or four. Now this is quite hard to work out directly. What we can do is work out the opposite. The chance of one dice not getting less than five means the chance of it getting a five or six, and this is 2/6, or 1/3. But we need all dice to get this. This is 1/3x1/3x1/3 or 1/27. This was the opposite of what we want, so we must take it away from one. 1 - 1/27 or 26/27 which is about 96%. Die or dice People occasionally point out to me that the singular of 'dice' is 'die'. Yes, I know that. However, a lot of people don't, and I prefer to use 'dice' since then my meaning is clear.

It stops them wondering why I'm talking about dying! However, if you think that I should be accurate rather than comprehensible, I'm afraid that the authorities are not on your side. The Oxford English Dictionary says "The form dice (used as pl. and sing.) is of much more frequent occurrence in gaming and related senses than the singular die." I think that probability is related to gaming! It gives a quote: 1474 CAXTON Chesse 132 "He caste thre dyse and on eche dyse was a sise." (Translation - "He casts three dice and on each dice was a six.") Dice Probabilities - Rolling 2 Six-Sided Dice 12 - 2.78% As you can see, 7 is the most common roll with 2 six-sided dice . It's six times more likely that you'll roll a 7 than a 2 or a 12 -- a huge difference. And it's twice as likely that you'll roll a 7 than a 4 or a 10. However, it's only 1.2 times more likely that you'll roll a 7 than a 6 or an 8.

Another way of looking at these numbers is that, over time, you will roll one 4 or 10 for every two 7s rolled. You'll see six 7s for every 2 or 12. Of course, dice have a bad habit of defying expectations. Don't rely on probabilities to work out precisely according to a chart like this -- you'll often find yourself on the losing end of a game if you do.

Question: What is the most likely total to result from a throw of a pair of dice?
Answer: Seven