Rolling dice statistics, probability of rolling a straight 

How do I correctly set up an equation to solve for the probability of rolling a straight of X amount. First with 5, then 10 then 15 dice and so on. See image.

So for example: (6 sided die) I would roll 10 dice all at once, order does not matter. What is the probability I would get at least one straight of 3. Straight of 3 meaning a set of one of these numbers 123 or 234 or 345 or 456. 

Edit (rephrasing example question):
 What is the probability of rolling 1,2,3 or 2,3,4 or 3,4,5 or 4,5,6 out of 10 dice? 

I am making a statistics chart that I can pull data from when making cards for a board game. If this can be done in excel that would be helpful too. I am guessing this is a one and done scenario, so I am not too focus on retaining the knowledge.
  • The bounty is low!

  • What do you believe a fair amount for this would be?

  • Your example is not clear

  • Well, it would take at least 30 minutes to write a good solution. Offer whatever you think is a fair bounty for a skilled professional to work for half hour on your problem.

  • So you roll n die at the same time and you want to compute the probability of say rolling at least one 1, one 2, and one 3 so you have a 123, and you don't care what happens for the remaining (n-3) die?

  • I will just leave it at $10 for the 4 days and see if its worth anyone's time. Any higher and Chegg is a better route.

  • Philip, Yes but it can also can be one 2, one 3, one 4 ( 234) or one 3, one 4, one 5 (345) and finally (456). Say I roll 10 dice: 1,1,1,1,2,2,4,4,5,5 = this is a fail, because there is no sequence of 3 (straight of 3). Say I roll again: 1,1,2,2,3,3,4,5,6,6= success, because of multiple sequences of 3. So what is the probability of a success if I roll ten dice. Hope that helps

Answer

Answers can be viewed only if
  1. The questioner was satisfied and accepted the answer, or
  2. The answer was disputed, but the judge evaluated it as 100% correct.
View the answer
  • I don't believe it is correct. As 'n' increases the probability of success should too. It is actually the reverse. If you roll 10 dice we get a probably of 0.43 success rate. If you roll 20 dice the success rate is .077. Wouldn't you have better chances of seeing at least 1 straight of 3 as 'n' increases? Or did I interpret it wrong?

  • you are doing something wrong. P_{Failure} is clearly a decreasing function. So 1-P_{Failure} is increasing in n.

The answer is accepted.