Dice Probability?

hi

Firstly consider just 3 dice. The probability of throwing 123 in that order is (1/6)^3. Now the number of arrangements of 123 is 3!. So the probability of throwing 123 in any order with 3 dice is (1/6)^3*3!, which is 1/36. If we have 6 dice we multiply this by 6C3 (the number of ways of choosing 3 from 6 = 20). So the answer is 20/36 or 5/9.

Sorry that can't be right. Otherwise with a large number of dice the probability would be >1. That might, however be the expectation value.

I'm going to try this one before sleeping:

1 - (35/36)^20 = 0.4307

Reasoning:

35/36 is the probability of not throwing 123 in any order with 3 dice. 6C3 = 20 which means that in throwing 6 dice you are in effect making 20 throws of 3 dice. So (35/36)^20 is the probability of not throwing 123 in any order with 6 dice.

So 1 - (35/36)^20 is the probability of obtaining 1 or more 123 by rolling 6 dice, And this is 0.4307.

Not 100% sure of this because these 20 throws are not really independent. However the answer seems realistic.

Your sample roll of 1,3,4,4,2,2 does not include 1,2,3 "in order" as you say. I'm going to assume they don't have to be in order.

It doesn't matter when the 1,2,3 occur, so let's do it this way.

Oops. I misinterpreted my own interpretation of your question, and now I have to go. Correct the error in your question and someone may be better able to help you.

********EDIT********

Okay, ready to solve this correctly. Best way, in my opinion, is to do it in reverse.

The probability that no die will show a 1, 2 or 3 is (3/6)^6, or 0.015625.

The probability that just one die will show a 1, 2 or 3 is also (3/6)^6, or 0.015625.

The probability that exactly two dice will show two of the numbers in 1, 2 or 3 is (3/6)*(2/6)*((5/6)^4), or 0.0803754. Explanation for this probability:

First die must be 1, 2 or 3 (3/6).

Second die must be one of the other two (2/6)

Last four dice can be anything but the remaining number (5/6).

The probability of rolling a 1, 2 or 3 on any three dice is 1 minus the probability that any of the above three will occur, or 1 minus the sum of those three probabilities.

1 - (0.015625 + 0.015625 + 0.0803754) = 0.8883746

A high probability makes sense as the chances of throwing 123456 in any order is, in theory, very high given the law of averages. So, that's my answer and I'm stickin' with it.

0.8883746

The odds that one (or more) of the 6 dice will be a "1" is 1/6 multiplied by 6 = 6/6.

The odds that one (or more) of the remaining 5 is a "2" is 5/6.

The odds that one (or more) of the remaining 4 is a "3" is 4/6.

That gives 1*5/6*4/6 = 20/36 = 5/9.

First link is an interactive model to try for yourself :-)

OR

See 2nd link for :-

What is the probability of rolling 1,2,3,4,5,6 with six dice, six times in a row?

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.

You mean the expectation not the probability

E = 15 * ( 3/6 * 2/6 ) * 4 * 1/6 = 5/3 = 1.6666

# of pairs * prob of good pair * # of 3rd dice * prob of good 3rd dice

I hate stats, but I think it is 5/6 X 4/5 X 3/4 = 60/120

50% chance

1- (1/2*1/2*1/2*1/2*2/3*5/6)