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The probability is 1/36

Q: When you roll three number cubes what is the probability that all three numbers will match?

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The answer depends on the lotto. The relevant variables are:How many numbers you chose from,How many numbers you have to choose,How many numbers you need to match to win - something - not necessarily the jackpot.The answer depends on the lotto. The relevant variables are: How many numbers you chose from,How many numbers you have to choose,How many numbers you need to match to win - something - not necessarily the jackpot.The answer depends on the lotto. The relevant variables are: How many numbers you chose from,How many numbers you have to choose,How many numbers you need to match to win - something - not necessarily the jackpot.The answer depends on the lotto. The relevant variables are: How many numbers you chose from,How many numbers you have to choose,How many numbers you need to match to win - something - not necessarily the jackpot.

The probability of at least 1 match is equivalent to 1 minus the probability of there being no matches. The first person's birthday can fall on any day without a match, so the probability of no matches in a group of 1 is 365/365 = 1. The second person's birthday must also fall on a free day, the probability of which is 364/365 The probability of the third person also falling on a free day is 363/365, which we must multiply by the probability of the second person's birthday being free as this must also happen. So for a group of 3 the probability of no clashes is (363*364)/(365*365). Continuing this way, the probability of no matches in a group of 41 is (365*364*363*...326*325)/36541 This can also be written 365!/(324!*36541) Which comes to 0.09685... Therefore the probability of at least one match is 1 - 0.09685 = 0.9032 So the probability of at least one match is roughly 90%

The more samples you use, the closer your results will match probability.

None of the experimental probabilities need match the corresponding theoretical probabilities exactly.

Think of watching the lottery draw. You have 6 numbers and there are 40 to draw from. On the first pick, you have a 6/40 chance of getting a match. If you're successful, then on the second pick you have a 5/39 chance on the next pick (because one number is gone from your card and the draw), and so on. Because all of the events are required to happen for you to get your 6 numbers, we multiply the individual probabilities together to get the overall probability. 6/40 x 5/39 x 4/38 x 3/37 x 2/36 x 1/35 = 1/3838380 So the chance of getting all six numbers is a little better than 1 in 4 million.

Related questions

The first thing to do is find how many possible combinations of those numbers match the criteria. All numbers that start with 7 match it. And there are 5x4x3=60 possible numbers that start with 7. All numbers that start with 5 also match it, and there are 5x4x3=60 numbers that start with 5. All numbers that start with 35, 37 or 39 match it. And there are 4x3x3=36 numbers that do this. And finally, all numbers that start with 325, 327 and 329 match it, and there are 3x3=9 numbers that do this. In total this is 60+60+36+9=165 possible 4 digit numbers that match the criteria. The number of possible four digit numbers in total is 6x5x4x3 = 360. This means the probability of a 4 digit number using only 1,2,3,5,7 and 9 being greater than 3251 but less than 8825 is 165/360 which can be simplified to 11/24.

Assuming that you cannot repeat any of the numbers that you are using to guess (e.g., would not guess 1, 4 and 4) AND that a match only counts once (e.g., in the serial number A23345678D, the matching 3 only counts as a single match despite appearing more than one time), the probability is between 15.4% and 15.5% (via 5MM simulations). If we loosen the rules and say that there can be double matches with the numbers that you are guessing (e.g., in the serial number above, if 3 was a guess number, there would be two matches, not just one), the probability is between 25.4% and 25.5% (via 5MM simulations) The spirit of the question suggests that we would never repeat a guessed number.

For a four digit pin number: You receive the first PIN number, let's say WXYZ. The probability that the next pin number you receive would match (assuming they are randomly provided), is: For each digit, they are 10 possibilities [0 1 2 3 4 5 6 7 8 8]. The probability that one specific number is chosen is thus of 1/10. For the are four digits, hence four independent selection of one digit, each with a probability of 1/10. The probability of an event, combination of independent events, is the product of the the probability of the independent event. Thus, the probability that the next pin number you receive would match (assuming they are randomly provided), is: 1/10*1/10*1/10*1/10 or 1/10^4 or 0.0001 or 1 out 10000

Out of the six faces, only 3 and 5 match the requirements, so the probability is 2/6 or 1/3.

no because serial numbers all have a unique number

The answer depends on the lotto. The relevant variables are:How many numbers you chose from,How many numbers you have to choose,How many numbers you need to match to win - something - not necessarily the jackpot.The answer depends on the lotto. The relevant variables are: How many numbers you chose from,How many numbers you have to choose,How many numbers you need to match to win - something - not necessarily the jackpot.The answer depends on the lotto. The relevant variables are: How many numbers you chose from,How many numbers you have to choose,How many numbers you need to match to win - something - not necessarily the jackpot.The answer depends on the lotto. The relevant variables are: How many numbers you chose from,How many numbers you have to choose,How many numbers you need to match to win - something - not necessarily the jackpot.

The probability that the second coin matches the first is 0.5 .The probability that the third coin matches the first is 0.5 .The probability that the second and third coins both match the first is (0.5 x 0.5) = 0.25 = 25%

Nothing, unless all the others match up in the proper order.

All numbers in your Pokemon's ID number count towards the lottery. The lottery does not omit any number. The days number has to match all numbers, including the zeros. No numbers are left out in the Pokemon's ID number.

The probability of at least 1 match is equivalent to 1 minus the probability of there being no matches. The first person's birthday can fall on any day without a match, so the probability of no matches in a group of 1 is 365/365 = 1. The second person's birthday must also fall on a free day, the probability of which is 364/365 The probability of the third person also falling on a free day is 363/365, which we must multiply by the probability of the second person's birthday being free as this must also happen. So for a group of 3 the probability of no clashes is (363*364)/(365*365). Continuing this way, the probability of no matches in a group of 41 is (365*364*363*...326*325)/36541 This can also be written 365!/(324!*36541) Which comes to 0.09685... Therefore the probability of at least one match is 1 - 0.09685 = 0.9032 So the probability of at least one match is roughly 90%

For complex events, it is possible to calculate the probability of events, but often extremely difficult. In the given example, for an "average" person (that would need some definition to start with) you would need to know the probability of them scoring a basket without the blindfold - this can be found by observing a number of "average" people attempting a number of baskets and seeing how many are successful (the greater the number of observations, the better the accuracy of the [estimation of the] probability. Also, the effect of blindfolding them needs to be found - this is not so easy, but some measure could possibly be made - and then combining this effect and the probability found some estimation of the probability of the required event can be calculated. Someone has analysed tennis scoring and given the probability of one of the players winning a point (which can be estimated fairly accurately through past observation) has calculated the probability of them winning the match; however, each match (and even a game within a match) can be affected by further factors (eg one player suffering a small injury) which modify the probability of winning a point, but a calculated probability can still be made.

In the multiplicative context, possibly -32329 and -41.