## Sum up to 6

Posted by Chris on September 29, 2011 – 12:46 pm

Roll a ~~dice~~ die until you get a 6. On average, what will the sum of the rolls be before you get the 6?

e.g. if you rolled 1,2,5,6 the sum would be 8.

Posted by Chris on September 29, 2011 – 12:46 pm

Roll a ~~dice~~ die until you get a 6. On average, what will the sum of the rolls be before you get the 6?

e.g. if you rolled 1,2,5,6 the sum would be 8.

September 29th, 2011 at 2:54 pm

Your chances of NOT getting a 6 after n throws is (5/6)^n.

This becomes less than 50% after 4 throws, i.e. on average you will get 3 throws without a 6 before you get a 6.

These 3 throws can each range from 1 to 5, averaging 3.

So the average sum of rolls before getting a 6 is 3×3 = 9.

September 29th, 2011 at 3:42 pm

Hi Wiz. You are saying that on average you only need 4 rolls to get any given number.

Take a look at http://trickofmind.com/?p=437#comment-2355 to see what another Wizard of Oz said

September 30th, 2011 at 12:20 am

I’m going to say 15.

September 30th, 2011 at 3:16 am

Hi Jan. Why 15?

September 30th, 2011 at 5:38 am

infinity?

September 30th, 2011 at 6:09 am

I got 21. Did I miss something?

r-the roll#

t=sub total

a=count of sixes

b=grand total

Sub test()

Randomize

Dim a, b, t, r As Long

For i = 1 To 10000000

r = Int(Rnd() * 6 + 1)

t = t + r

If r = 6 Then

a = a + 1

b = b + t

t = 0

End If

Next i

Debug.Print b / a

End Sub

Result of b/a=20.9999999999

September 30th, 2011 at 7:53 am

Basically i deduced the answer from the two posts above me.

On average 6 rolls to get a 6, while not rolling a 6 the average roll is 3. 3×5=15.

ragknot, i think your program adds the final 6 to the sum: 15+6=21.

September 30th, 2011 at 9:16 am

Hi Jan. You’re right, 15 is the correct answer

The average number of rolls to get a 6 is 6. So the average number for the previous 5 rolls is (1+2+3+4+5)/5 = 3, and of course the average total is 3*5 = 15.

You’re also right, ragknot included the 6. Including the 6 makes the average roll be (1+2+3+4+5+6)/6 = 3.5, and 3.5*6 = 21 for the average total.

September 30th, 2011 at 1:20 pm

Hi all. I don’t mean to be picayune, but the term for a single cube is ‘die’. I started looking at this problem in terms of a pair of dice and was coming up with completely different results. But my misunderstanding is a good basis for the question, what is the average sum when rolling a pair of dice until a six is rolled?

September 30th, 2011 at 1:41 pm

Hi JB. I goofed. I normally say “die”.

September 30th, 2011 at 4:06 pm

yeah, someone was right, I should not have counted that last six, so 21 -6 would be 15.

September 30th, 2011 at 5:20 pm

Hi ragknot. I must have goofed again, I had written a reply along the lines of you had done the opposite if missing something. I must have forgotten to click submit.

September 30th, 2011 at 5:38 pm

Just to be clear on the problem, I am talking about the sum of the 2 dice equaling 6.

September 30th, 2011 at 5:47 pm

Hi JB. I’ve put your problem (and more) up on a new page.