Probability Q: Given one toss of 7 (fair) dice, how easy is it roll at least a 4-straight? A 5-straight? A 6-straight?

There are 6^7 = 279936 possible rolls

A 6 straight means you have 123456x, not necessarily in that order.

The x must be one of the other numbers. You need to know how many permutations there are of123456x/ That's 7! / 2! = 5040 / 2 = 2520.

Then x can be any of 6 numbers so the final total is 2520x6 = 15120.

So the probability of a 6 straight is 15120 / 279936 = 0.054012346

I had this wrong (forgot to multiply by 6) so this has been corrected.

For a 5-straight there are several cases to consider

Case 1. 12345xy

Neither x nor y can be a 6 or else you have a 6-straight.

There are two subcases:

(a) x = y= 1,2,3,4 or 5.

Then you have a triple, and the number of permutations is 7!/3! = 5040/6 = 840. You multiply that by 5 for the number of possible triples, giving you 840 x 5 = 4200.

(b) x <> y

Now you have two doubles, so you have 7! / 2!2! =1260

You multiply that by 5C2 = 10 for the number of ways to choose x and y, giving you 1260 x 10 = 12600 ways.

The total for Case 1 is 4200 +12600 = 16800.

Case 2. 23456xy

This also has 16800 cases.

The total is 33600 cases

The probability of a 5-straight is 33600 / 279936 = 0.120027435

The original answer for this part was also wrong and has been corrected (after being checked in Excel).

There are even more cases to consider for a 4-straight.

There are cases with a quad:

1234 xxx with x = 1,2,3 or 4 (7!/4! = 210, times 4 choices for x = 840)

2345 xxx (840)

3456 xxx (840)

Total with a quad = 2520 (checked with Excel)

There are cases with a triple like

1234 666 (7!/3! = 840 permutations)

3456 111 (840)

1234 6xx with x = 1,2,3 or 4 (840 times 4 choices for x = 3360)

3456 1xx (3360)

1234 xxy with x and y = 1,2,3,or 4 (7! / 3!2! = 420 times 4 choices for x and 3 for y = 5040)

2345 xxy (same 5040)

3456 xxy (same 5040)

Total with a triple = 840+840+3360+3360+5040+5040+5040 = 23520 (checked with Excel).

The cases with a double but no longer multiple are:

Let's look at 2345 first, because there are no 1's or 6's.

The pattern is 2345 xyz with x, y and z all different (7! / 2!2!2! = 630)

There are 4C3 = 4 choices for x, y and z.

630x4 = 2520

Now look at 1234 with doubles

1234 66x gives you 7!/2!2! = 1260, with 4 choice of x, for 5040

1234 6xy gives you 7!/2!2! = 1260 with 4 choices 4C3 = 6 choices for x and y, for a total of 7560

1234 xyz with x,y and z = 1,2,3,or 4 gives you 7!/2!2!2! = 630 times 4C3 = 4 choices for xyz, for a total of 2520.

The total for 1234 is 5040+7560+2520 = 15120

With 3456 there are another 15120

Total number with no more than a double is

2520 + 15120+15120 = 32760 (checked with Excel)

So the grand total is 2520 + 23520 + 32760 = 58800

58800 / 279936 = 0.210048011

The complete breakout into runs of length k is:

k Ways(k) Probability

6 15120 5.40%

5 33600 12.00%

4 58800 21.00%

3 91224 32.59%

2 72702 25.97%

1 8490 3.03%

Tot 279936 100.00%

Tot =6^7

This was all done by generating all the permutations in Excel, and using formulas to find how may had runs of 1 to 6. This was invaluable for checking the 3 cases you asked for, since this is so error prone when you try to do it by formulas.

For the OCD people (like myself) out there:

91224 = 10x7!/3! + 12x7!/3!3! + 72x7!/3!2! + 12x7!3!2! + 20x7!/2!2! + 24x7!/2!2!2! if you want to work out all the cases with triples and doubles.

Did you mean all 4 ? If so:

P( all 4) = 1 / 6^7

P(all 5) = 1/ 6^7

P( all 6) = 1/ 6^7