dice - objects used in games of chance wih 6 sides

Free 1 Die Roll Calculator - Calculates the probability for the following events in the roll of one fair dice (1 dice roll calculator or 1 die roll calculator):
* Probability of any total from (1-6)
* Probability of the total being less than, less than or equal to, greater than, or greater than or equal to (1-6)
* The total being even
* The total being odd
* The total being a prime number
* The total being a non-prime number
* Rolling a list of numbers i.e. (2,5,6)
* Simulate (n) Monte Carlo die simulations.
1 die calculator

2 dice are rolled what is the probability that doubles are rolled less than 11 List out the doubles with a sum less than 11: (1, 1) (2, 2) (3, 3) (4, 4) (5, 5) The probability of each double is 1/6 * 1/6 = 1/36. We have 5 of them, so we have 5*1/36 = [B]5/36[/B]

Free 2 dice roll Calculator - Calculates the probability for the following events in a pair of fair dice rolls:
* Probability of any sum from (2-12)
* Probability of the sum being less than, less than or equal to, greater than, or greater than or equal to (2-12)
* The sum being even
* The sum being odd
* The sum being a prime number
* The sum being a non-prime number
* Rolling a list of numbers i.e. (2,5,6,12)
* Simulate (n) Monte Carlo two die simulations. 2 dice calculator

2 fair sided die are rolled. How many ways can the dice be rolled to sum exactly 6? [URL=' our 2 dice calculator[/URL], we get the following options: [LIST] [*]2,4 [*]3,3 [*]4,2 [*]1,5 [*]5,1 [/LIST] The probability of rolling a sum of 6 is [B]5/36[/B]

6 sided die probability to roll a odd number or a number less than 6 First, we'll find the set of rolling an odd number. [URL=' this dice calculator[/URL], we get: Odd = {1, 3, 5} Next, we'll find the set of rolling less than a 6. [URL=' this dice calculator[/URL], we get: Less than a 6 = {1, 2, 3, 4, 5} The question asks for [B]or[/B]. Which means a Union: {1, 3, 5} U {1, 2, 3, 4, 5} = {1, 2, 3, 4, 5} This probability is [B]5/6[/B]

A 6-sided die is rolled once. What is the probability of rolling a number less than 4? Using our [URL=' dice calculator[/URL], we get: P(N < 4) = [B]1/2[/B]

A blue dice and a red dice are tossed what is the probability that a 6 will appear on both dice Each event is independent. P(Blue dice 6) = 1/6 P(Red Dice 6) = 1/6 P(Blue 6, Red 6) = 1/6 * 1/6 = [B]1/36[/B]

A coin is tossed and a die is rolled. Find the probability pf getting a head and a number greater than 4. Since each event is independent, we multiply the probabilities of each event. P(H) = 0.5 or 1/2 P(Dice > 4) = P(5) + P(6) = 1/6 + 1/6 = 2/6 = 1/3 P(H) AND P(Dice > 4) = 1/2 * 1/3 = [B]1/6 [MEDIA=youtube]ofsbmHmQmjs[/MEDIA][/B]

A dice has six sides. The dice is rolled once. What is the probability that a six will be the result. P(6) = [B]1/6[/B]

A pair of dice are rolled. Find the probability for P(Not 2 or Not 12). P(Not 2 or Not 12) = 1 - P(2) - P(12) P(Not 2 or Not 12) = 1 - [URL=' - [URL=' P(Not 2 or Not 12) = 34/36 [URL='(Not 2 or Not 12)[/URL] = [B]17/18[/B]

A pair of dice is cast. what is the probablitly that the sum is less than 5? Using our [URL=' dice calculator[/URL], we get 1/6 or 16.67%

A pair of dice is rolled. Find the probability of rolling a sum of not less than 5. The phrase [I]not less than[/I] also means greater than or equal to. So we [URL=' our 2 dice calculator for a sum roll of 5 or greater[/URL] and we get: [B]5/6[/B]

A pair of standard dice is rolled, how many possible outcomes are there? We want the number of outcomes in the sample space. The first die has 6 possibilities 1-6. The second die has 6 possibilities 1-6. Our sample space count is 6 x 6 = [B]36 different outcomes [/B] [LIST=1] [*](1, 1) [*](1, 2) [*](1, 3) [*](1, 4) [*](1, 5) [*](1, 6) [*](2, 1) [*](2, 2) [*](2, 3) [*](2, 4) [*](2, 5) [*](2, 6) [*](3, 1) [*](3, 2) [*](3, 3) [*](3, 4) [*](3, 5) [*](3, 6) [*](4, 1) [*](4, 2) [*](4, 3) [*](4, 4) [*](4, 5) [*](4, 6) [*](5, 1) [*](5, 2) [*](5, 3) [*](5, 4) [*](5, 5) [*](5, 6) [*](6, 1) [*](6, 2) [*](6, 3) [*](6, 4) [*](6, 5) [*](6, 6) [/LIST]

A Pairs of fair dice is tossed. What is the probability of not getting a sum 7 or 8? Not a 7 or 8 means 2 - 6 or 9 - 12 [URL=' our 2-dice calculator[/URL], P(2 - 6) = 5/12 [URL=' our 2-dice calculator[/URL], P(9 - 12) = 5/18 Since the sum could be either of these, we add probabilities: P(Not a 7 or 8) = P(2 - 6) + P(9 - 12) P(Not a 7 or 8) = 5/12 + 5/18 [URL='(Not a 7 or 8) [/URL]= [B]25/36[/B]

A standard die is rolled. Find the probability that the number rolled is greater than 3. Using our [URL=' calculator[/URL], the probability is [B]1/2 or 0.5[/B]

An ordinary fair die is rolled twice. The face value of the rolls is added together. Compute the probability of the following events: Event A: The sum is greater than 6. Event B: The sum is divisible by 5 or 6 or both. [URL=' greater than 6[/URL] = [B]7/12[/B] Sum is divisible by 5 or 6 or both This means a sum of 5, a sum of 6, a sum of 10, or a sum of 12. [URL=' of 5[/URL] = 1/9 or 4/36 [URL=' of 6[/URL] = 5/36 [URL=' of 10[/URL] = 1/12 or 3/36 [URL=' of 12[/URL] = 1/36 Adding all these up, we get: (4 + 5 + 3 + 1)/36 [B]13/36[/B]

Chuck-a-luck is an old game, played mostly in carnivals and county fairs. To play chuck-a-luck you place a bet, say $1, on one of the numbers 1 through 6. Say that you bet on the number 4. You then roll three dice (presumably honest). If you roll three 4s, you win $3.00; If you roll just two 4s, you win $2; if you roll just one 4, you win $1 (and, in all of these cases you get your original $1 back). If you roll no 4s, you lose your $1. Compute the expected payoff for chuck-a-luck. Expected payoff for each event = Event Probability * Event Payoff Expected payoff for 3 matches: 3(1/6 * 1/6 * 1/6) = 3/216 = 1/72 Expected payoff for 2 matches: 2(1/6 * 1/6 * 5/6) = 10/216 = 5/108 Expected payoff for 1 match: 1(1/6 * 5/6 * 5/6) = 25/216 Expected payoff for 0 matches: -1(5/6 * 5/6 * 5/6) = 125/216 Add all these up: (3 + 10 + 25 - 125)/216 -87/216 ~ [B]-0.40[/B]

Farah rolls a fair dice and flips a fair coin. What is the probability of obtaining a 5 and a head? Give your answer in its simplest form. Probability of a 5 is 1/6 Probability of a head is 1/2 Since each event is independent, we get the total probability by multiplying both together: P(5,H) = 1/6 * 1/2 P(5,H) = [B]1/12[/B]

How many possible outcomes of rolling 4 dice Each die has 6 faces, so we have: 6 * 6 * 6 * 6 = [B]1,296 possible outcomes[/B]

How many ways can you rolls two dice and get a sum less than 4? Answer is [B]1/12[/B] using our [URL=' dice calculator[/URL].

If two standard dice are rolled, what is the probability that the sum is 3? [URL=' our 2 dice calculator[/URL], we get: [B]1/18[/B]

Jerry rolls a dice 300 times what is the estimated numbers the dice rolls on 6 Expected Value = Rolls * Probability Since a 6 has a probability of 1/6, we have: Expected Value = 300 * 1/6 Expected Value = [B]50[/B]

Larry is rolling two dice. His dad told him that he can skip doing the dishes that night unless he rolls double sixes. What is the probability that Larry will be able to skip doing the dishes? P(6, 6) = 1/6 * 1/6 = 1/36 P(Not 6,6) = 1 - 1/36 = [B]35/36[/B]

Probability of getting 4 or 6 when rolling a dice P(4 or 6) = P(4) + P(6) P(4 or 6) = 1/6 + 1/6 P(4 or 6) = 2/6 We can simplify this. We [URL=' this fraction into our search engine, choose simplify[/URL], and we get: P(4 or 6) = [B]1/3[/B]

Sum of 8 equal to 5/36 shown [URL=' At least one 4 means one of three scenarios: [LIST=1] [*](4, not 4) = 1/6 * 5/6 = 5/36 [*](not 4, 4) = 5/6 * 1/6 = 5/36 [*](4, 4) = 1/6 * 1/6 = 1/36 [/LIST] The phrase "or", means we add both probabilities (sum of 8) and (at least one 4): 5/36 + (5/36 + 5/36 + 1/36) 16/36 Simplify by dividing each part of the fraction by 4 [B]4/9[/B]

Free Rational Exponents - Fractional Indices Calculator - This calculator evaluates and simplifies a rational exponent expression in the form a^b/c where a is any integer or any variable [a-z] while b and c are integers. Also evaluates the product of rational exponents

Use our [URL=' dice calculator[/URL], you have 2 ways to roll an 11 [LIST=1] [*](5, 6) --> P(5, 6) = 1/36 [*](6, 5)--> P(6, 5) = 1/36 [/LIST] [U]We want P(5, 6) + P(6, 5):[/U] P(5, 6) + P(6, 5) = 1/36 + 1/36 P(5, 6) + P(6, 5) = 2/36 P(5, 6) + P(6, 5) = [B]1/18[/B]

Sarah rolls 2 fair dice and adds the results from each. Work out the probability of getting a total that is a factor of 6. Factors of 6 are {6, 12} [URL='(Roll a 6)[/URL] = 5/36 [URL='(Roll a 12)[/URL] = 1/36 P(Roll a 6 or Roll a 12) = P(Roll a 6) + P(Roll a 12) P(Roll a 6 or Roll a 12) = 5/36 + 1/36 P(Roll a 6 or Roll a 12) = 6/36 Using our [URL=' simplifier[/URL], we see that: P(Roll a 6 or Roll a 12) = [B]1/6[/B]

Free Set Notation Calculator - Given two number sets A and B, this determines the following:
* Union of A and B, denoted A U B
* Intersection of A and B, denoted A ∩ B
* Elements in A not in B, denoted A - B
* Elements in B not in A, denoted B - A
* Symmetric Difference A Δ B
* The Concatenation A · B
* The Cartesian Product A x B
* Cardinality of A = |A|
* Cardinality of B = |B|
* Jaccard Index J(A,B)
* Jaccard Distance J_σ(A,B)
* Dice's Coefficient
* If A is a subset of B
* If B is a subset of A

The singular form of the word "dice" is "die". Tom was throwing a six-sided die. The first time he threw, he got a three; the second time he threw, he got a three again. What's the probability of getting a three at the third time? Since all trials are independent: 1/6 * 1/6 * 1/6 = [B]1/216[/B]

The sum is greater than 7, the sum is divisible by 2 2 dice sum greater than 7 means 8, 9, 10, 11, 12. Now take this set, and intersect it with sums divisible by 2. [B]8, 10, 12[/B]

Three ordinary dice are rolled. What is the probability that the results are all less than 5 Calculate individual die probabilities: [LIST] [*]Die 1 P(x < 5) = 4/6 = 2/3 [*]Die 2 P(x < 5) = 4/6 = 2/3 [*]Die 3 P(x < 5) = 4/6 = 2/3 [/LIST] Since each roll is independent, we have: P(Die 1 < 5, Die 2 < 5, Die 3 < 5) = 2/3 * 2/3 * 2/3 P(Die 1 < 5, Die 2 < 5, Die 3 < 5) = [B]8/27[/B]

Two dice are rolled. Enter the size of the set that corresponds to the event that both dice are odd. If dice 1 is odd, then we have the following face values: {1, 3, 5} If dice 2 is odd, then we have the following face values: {1, 3, 5} [URL=' this 2 dice odds face link[/URL], we see that the size of the set is 9. [LIST=1] [*]{1, 1} [*]{1, 3} [*]{1, 5} [*]{3, 1} [*]{3, 3} [*]{3, 5} [*]{5, 1} [*]{5, 3} [*]{5, 5} [/LIST]

two unbiased dice are thrown. find the probability that the total number on the dice is greater than 10 [URL=' our 2 dice calculator[/URL]: We have (5,6),(6,5),(6,6) P(Sum) > 10 is [B]1/12[/B]

The only way you can roll a 12 with two dice is 6 and 6. Since each die roll is independent, we have: [LIST] [*]P(12) = P(6) * P(6) [*]P(12) = 1/6 * 1/6 [*]P(12) = 1/36. [/LIST] Now, what is the probability we roll a 12 five times in a row? The same rules apply, each new roll is independent of the last, so we multiply: [LIST] [*]P(12, 12, 12, 12, 12) = 1/36 * 1/36 * 1/36 * 1/36 * /36 [*]P(12, 12, 12, 12, 12) = [B]1/60,466,176[/B] or [B]1.65381717e-8[/B] [/LIST]

Write a sample space for rolling a dice twice Each die roll has 6 possible outcomes. So 2 die-rolls has 6^2 = 36 possible outcomes: [LIST=1] [*]1,1 [*]1,2 [*]1,3 [*]1,4 [*]1,5 [*]1,6 [*]2,1 [*]2,2 [*]2,3 [*]2,4 [*]2,5 [*]2,6 [*]3,1 [*]3,2 [*]3,3 [*]3,4 [*]3,5 [*]3,6 [*]4,1 [*]4,2 [*]4,3 [*]4,4 [*]4,5 [*]4,6 [*]5,1 [*]5,2 [*]5,3 [*]5,4 [*]5,5 [*]5,6 [*]6,1 [*]6,2 [*]6,3 [*]6,4 [*]6,5 [*]6,6 [/LIST]

Free Yahtzee-1st Roll Calculator - Calculates the probability of various scoring hands in the game of Yahtzee on the 1st roll of the dice.

You roll a red die and a green die. What is the size of the sample space of all possible outcomes of rolling these two dice, given that the red die shows an even number and the green die shows an odd number greater than 1? [LIST] [*]Red Die Sample Space {2, 4, 6} [*]Green Die Sample Space {3, 5} [*]Total Sample Space {(2, 3), (2, 5), (4, 3), (4, 5), (6, 3), (6, 5)} [*]The sie of this is 6 elements. [/LIST]

[URL=' our 2 dice calculator:[/URL] [MEDIA=youtube]MmLjl5TRsBM[/MEDIA]

You roll two six-sided dice. What is the probability that the sum is less than 13? The probability is [B]1, or 100%[/B], since the maximum sum of two six-sided dice is 12.

You throw two dice. The red dice is fair but on the blue dice the probability of a 1=15%, probability of a 2 is 25%, and the probability of any other number is 15%. What is the probability of getting 4? Possible Rolls with a sum of 4: [LIST] [*]R = 1, B = 3 [*]R = 2, B = 2 [*]R = 3, B = 1 [/LIST] Probabilities: [LIST] [*]R = 1, B = 3 = 1/6 * 15/100 = 15/600 = 1/40 = 0.025 [*]R = 2, B = 2 = 1/6 * 25/100 = 25/600 = 1/24 = 0.041667 [*]R = 3, B = 1 = 1/6 * 15/100= 15/600 = 1/40 = 0.025 [/LIST] We add all three probabilities up to get: 0.025 + 0.025 + 0.014667 = [B]0.09166667[/B]