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aceside
aceside
aceside
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December 10th, 2025 at 5:40:55 PM permalink
I looked up an online calculator. For a streak of 7 dice rolling in a row, the average number of rolls is 335922.
Ace2
Ace2 
Ace2
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December 10th, 2025 at 6:30:07 PM permalink
You can get that answer via linear equations, or you can calculate it directly as:

6^1 + 6^2 + 6^3 + 6^4 + 6^5 + 6^6 + 6^7 = (6^(7+1) - 1) / (6 - 1) - 1 = 335,922
It’s all about making that GTA
aceside
aceside
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December 11th, 2025 at 6:57:04 AM permalink
This is easy. When the number of states is large, we can just neglect all terms except the last one. The above as an example,

6^1 + 6^2 + 6^3 + 6^4 + 6^5 + 6^6=55,986;
6^7=279,936.

6^7/(6^1 + 6^2 + 6^3 + 6^4 + 6^5 + 6^6 +6^7)=0.833.

So, this approximation is already very good.
ThatDonGuy
ThatDonGuy
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December 11th, 2025 at 7:08:30 AM permalink
Quote: aceside

I looked up an online calculator. For a streak of 7 dice rolling in a row, the average number of rolls is 335922.
link to original post


Here is the general solution for rolling t 6s in a row:

Let E(n) be the expected number needed to reach t in a row when you currently have n in a row
Note E(n) = 1/6 E(n+1) + 5/6 E(0)
E(0) is the solution, as you start with zero 6s

Proof by induction that E(n) = 1 + 1/6 + (1/6)^2 + ... + (1/6)^(t-n-1) + 5/6 E(0) (1 + 1/6 + (1/6)^2 + ... + (1/6)^(t-n-1))
For n = t-1: E(t-1) = 1 + 1/6 E(t) + 5/6 E(0) = 1 + 5/6 E(0)
Assume E(n) = 1 + 1/6 + (1/6)^2 + ... + (1/6)^(t-n-1) + 5/6 E(0) (1 + 1/6 + (1/6)^2 + ... + (1/6)^(t-n-1))
E(n-1) = 1 + 1/6 E(n) + 5/6 E(0)
= 1 + 5/6 E(0) + 1/6 (1 + 1/6 + (1/6)^2 + ... + (1/6)^(t-n-1) + 5/6 E(0) (1 + 1/6 + (1/6)^2 + ... + (1/6)^(t-n-1)))
= 1 + 1/6 + (1/6)^2 + ... + (1/6)^(t-(n-1)+1) + 5/6 E(0) (1 + 1/6 + (1/6)^2 + ... + (1/6)^(t-(n-1)+1)))
= 1 + 1/6 + (1/6)^2 + ... + (1/6)^(t-(n+1)-1) + 5/6 E(0) (1 + 1/6 + (1/6)^2 + ... + (1/6)^(t-(n+1)-1)))

E(0) = 1 + 1/6 + (1/6)^2 + ... + (1/6)^(t-0-1) + 5/6 E(0) (1 + 1/6 + (1/6)^2 + ... + (1/6)^(t-0-1))
= (1 - (1/6)^t) / (1 - 1/6) + 5/6 E(0) (1 - (1/6)^t) / (1 - 1/6)
= 6/5 (1 - (1/6)^t) + E(0) (1 - (1/6)^t)
E(0) (1 - (1 - (1/6)^t) = 6/5 (1 - (1/6)^t)
(1/6)^t E(0) = 6/5 (1 - (1/6)^t)
E(0) = 6/5 (1 - (1/6)^t) / (1/6)^t
= 6^t (6/5) (1 - (1/6)^t)
= 6^(t + 1) (1 - (1/6)^t) / 5
= 6^(t + 1) (6^t - 1) / (6^t x 5)
= 6 (6^t - 1) / 5

For t = 7, this is 6 (6^7 - 1) / 5 = 335,922

Ace2
Ace2 
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December 11th, 2025 at 8:29:29 AM permalink
Quote: aceside

This is easy. When the number of states is large, we can just neglect all terms except the last one. The above as an example,

6^1 + 6^2 + 6^3 + 6^4 + 6^5 + 6^6=55,986;
6^7=279,936.

6^7/(6^1 + 6^2 + 6^3 + 6^4 + 6^5 + 6^6 +6^7)=0.833.

So, this approximation is already very good.
link to original post

17% off is not a “very good” approximation of something that can be easily calculated exactly. Would you say 2 + 2 = 3.3 is a very good approximation?
It’s all about making that GTA
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