a die is rolled 6 times find probability for getting three different numbers?

Answer 1

(6x5x4)/(6x6x6)

Answer 2

So far I can tell you that the answer should be in the ball park of 0.0463 ± error. This is based on a program that I wrote which simulates the Yahtzee random and decision making process. I doubt there is an analytical approach other than brute force. Then again maybe there is (see Zo Maar's answer below). Let's consider each case one by one. Case 1: yahtzee in one roll Case 2: 4 of kind in first roll Case 3: 3 of kind in first roll (including full house) Case 4: 2 of kind (but not three of kind, includes 2 pairs) Case 5: 5 distinct numbers on first roll We can work out the probability of getting a Yahtzee given each of these cases. P(case 1) = 6/6^5 P(case 2) = 6*5*(1/6)^4 * (5/6) P(case 3) = 6 * 5C3 * (1/6)^3 * (5/6)^2 P(case 5) = 6! / 6^5 = 720 / 6^5 P(case 4) = 1 - ∑ P(other cases) = 25/36 Clearly case 4 is tough to work out otherwise. Y - event that you eventually get a yahtzee P(Y | case 1) = 1 Case 2 At this point you hold the number, X, that occurs 4 times, and roll the last die until you get X. P(Y | case 2) = 1/6 + 5/6 * 1/6 = 11/6^2 Case 3 Again you hold the number, X, occurring 3 times P(Y | case 3) = 121 / 6^4 Case 4 This is where it gets nasty. Let's assume that if you get more than one pair, then you hold the larger one (it doesn't really matter). The only way you'll change that hold is if you get 3 of something else in the 2nd roll. You can use a probability tree to get hte prob of a yahtzee once you have 2 of a kind in the first round. P(Y | case 4) = 167/5832 *** see edit below Case 5 At this point, it makes no sense holding anything. You might as well roll all the dice again. You can make use of the above results. Now it’s like trying to get a yahtzee in 2 rolls rather than three. P(Y | case 5) = 5279 / (54*6^5) *** see edit below Finally P(Yahtzee) = ∑ P(Y | case i) * P(case i) = 0.0457 *** see edit below ***EDIT*** Zo Maar is right. I've checked over my work and indeed P(Y | case 4) = 113/3888 P(Y | case 5) = 221/17496 hence P(Y) = 0.04603 And this result is more consistent with my earlier computational result. Hmmm, Markov chains. The good stuff I never learned as an engineer.

Answer 3

When you throw a die once, the probability of getting any number is 6/6 = 1, because a die has 6 faces.

When you throws die a second time, the probability that you will get a different number is 5/6, because one number (from the first throw) is excluded and here are 5 possible numbers out of 5.

Therefore the probability that you will get two different numbers from two throws of a die is (6/6) * (5/6)

When you throw the die a third time, if you are to get yet another number, the numbers from the first two throws are excluded. Hence there are 4 possibilities out of 6.

Therefore the probability that you will get three different numbers from three throws of a die is (6/6) * (5/6)* (4/6), i.e. 20/36 or 5/9.

Answer 4

I think this is an example of permuting question.

First we calculate the total number of all the possible answers which is 6^6= 46656

the values can be 666666 or 666665 or 123456 or 654321......
but we want at least 3 different numbers.

so we calculate the permutation for that 3 numbers first.

the first number has 6 choices, the second one has 5 choices and the third one has 4 choices as first and second one have taken the 2 choices out of 6.
So the possible permutation of the 3 numbers is:

6P3(find function P on calculator) or 6x5x4 = 120

all the other 3 numbers have 6 choices so 6^3= 216 possible ways to permute.

I'll just ignore the fact above and permute the two permutation of the same three and different three together:

216x120= 25920 possible outcomes with 3 different numbers.

P(getting 3 different numbers)= 25920/46656
= 5/9

this could also be obtained this way:
P(getting 3 different numbers)= 1(first no. can be anything)x 5/6(one number taken)x 4/6(2 numbers taken)x 1x1x1
= 5/9

This is my idea ( yeah really long one I know) but there're still room for mistakes. For example, the three different numbers above, can't they be further more placed in different combinations among the three same numbers. ( I couldn't calculate that as it's getting too complicated.)

Maybe I just thought too hard than necessary

Answer 5

There are 6 x 6 x 6 different ways of rolling 3 dice. That's 6 ways of rolling the first, 6 ways of rolling the second and 6 ways of rolling the third.

There are 6 x 5 x 4 different ways of rolling 3 dice so that they're all different. That's 6 ways of rolling the first, 5 ways of rolling the second so that it is different from the first, and 4 ways of rolling the third so that it is different from the other two dice.

So the probability is the number of favourable outcomes divided by the total number of (equally likely) outcomes.

p = (6x5x4)/ (6x6x6) = 5/9

Answer 6

I would assume the same as getting three times the same number. The probability is 1:6 in each roll of the dice, no matter what, so the probability here would be 1:6x6x6x6x6x6.

Answer 7

1/12

Answer 8

p(3 different numbers) = (6/6)(5/6)(4/6) = 20/36 = 5/9

Answer 9

6x5x4
b/c you have a 1/6*1/5*1/4 probability

Answer 10

It should be 20/36.

Answer 11

are you krishia go? anyway...
it would be 1 i guess.
no wait it would be 1/6