Easy math puzzles
Poll
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| | 6 votes (11.76%) | ||
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| | 12 votes (23.52%) | ||
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51 members have voted
December 10th, 2025 at 5:40:55 PM
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I looked up an online calculator. For a streak of 7 dice rolling in a row, the average number of rolls is 335922.
December 10th, 2025 at 6:30:07 PM
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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
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
December 11th, 2025 at 6:57:04 AM
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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.
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.
December 11th, 2025 at 7:08:30 AM
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Quote: acesideI looked up an online calculator. For a streak of 7 dice rolling in a row, the average number of rolls is 335922.
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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
December 11th, 2025 at 8:29:29 AM
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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?Quote: acesideThis 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.
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It’s all about making that GTA
December 16th, 2025 at 11:35:59 AM
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Quote: acesideI looked up an online calculator. For a streak of 7 dice rolling in a row, the average number of rolls is 335922.
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I agree. The general answer for the expected number of rolls to get a particular face n times in a row on a d sided die is d^1 + d^2 + d^3 + ... + d^n. In this case 6^1 + 6^2 + ... + 6^7 = 335,922.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
December 16th, 2025 at 11:39:54 AM
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Fair warning I found the following puzzle quite hard.
You are among three gods, which are labeled A, B and C. One always speaks the truth, one always lies and one answers yes/no randomly without even listening to the question. The gods know the identity of each other. You may ask three yes/no questions directed to any particular god one at a time. Questions must have clear yes/no answers, so no paradoxical questions. Your goal is to determine which god is which.
What should be your line of questioning, which you may adapt according to previous responses.
You are among three gods, which are labeled A, B and C. One always speaks the truth, one always lies and one answers yes/no randomly without even listening to the question. The gods know the identity of each other. You may ask three yes/no questions directed to any particular god one at a time. Questions must have clear yes/no answers, so no paradoxical questions. Your goal is to determine which god is which.
What should be your line of questioning, which you may adapt according to previous responses.
Last edited by: Wizard on Dec 16, 2025
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
December 16th, 2025 at 1:49:28 PM
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Quote: WizardFair warning I found the following puzzle quite hard.
You are among three gods. One always speaks the truth, one always lies and one answers yes/no randomly without even listening to the question. You may ask three yes/no questions directed to a particular god one at a time. Questions must have clear yes/no answers, so no paradoxical questions. What should be your line of questioning, which you may adapt according to previous responses.
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Question: does each god know how the other two gods will answer? For example, if god A tells the truth, does A know which of B or C is the liar?
December 16th, 2025 at 2:03:38 PM
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Quote: ThatDonGuyQuestion: does each god know how the other two gods will answer? For example, if god A tells the truth, does A know which of B or C is the liar?
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Yes, they know the identity of each other. I also forgot to say the goal is to determine who is whom (is that the correct grammar?).
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
December 16th, 2025 at 2:19:00 PM
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Quote: WizardQuote: ThatDonGuyQuestion: does each god know how the other two gods will answer? For example, if god A tells the truth, does A know which of B or C is the liar?
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Yes, they know the identity of each other. I also forgot to say the goal is to determine who is whom (is that the correct grammar?).
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No, it is not correct grammar; "whom" is used with objects, so "who is who" is correct.
Also, defining the goal like that is important, as otherwise you could get an answer in two questions.
Label the gods A, B, and C
Question 1, to A: "Is B the truthteller or is C the liar?"
Of the six possible sets of gods (let T be the truthletter, F be the liar, and R be random):
T, F, R: the answer is No, so the (truthful) response will be No
T, R, F: the answer is Yes (C is the liar), so the (truthful) response will be Yes
F, T, R: the answer is Yes (B is the truthteller), so the (false) response will be No
F, R, T: the answer is No, so the (false) response will be Yes
R, T, F and R, F, T: the response will be random
If the response is Yes, then C is not random; if the response is No, then B is not random
Question 2, to whichever of B or C is not random: "Does 1 + 1 = 2?"
If the answer is Yes, that god is the truthteller; if it is No, that god is the liar
Question 3, to the same god: "Is A random?"
If the question is asked to the truthteller, Yes means that A is random, and No means that the other god is random; whichever one is not random is the liar
If the question is asked to the liar, No means that A is random, and Yes means that the other god is random; whichever one is not random is the truthteller
Extra credit (at least, I think I have this right): obtain the answer to any yes-or-no question in two questions to the gods.
