Easy math puzzles
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51 members have voted
January 19th, 2026 at 11:07:33 AM
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DisagreeQuote: ThatDonGuy
I'll go with the obvious answer: I expect him to be on the sidewalk.
Since the sidewalk runs north-south, the distance from the sidewalk will be the difference between the number of steps east and the number of steps west. However, since these are equally likely with each step, the expected number of steps east equals the expected number of steps west.
For that matter, since the expected number of steps north equals the number of steps south, I expect him to be at his starting location.
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It’s all about making that GTA
January 20th, 2026 at 1:47:08 PM
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The answer is (10,000/π)^.5 =~ 56.419 steps from the sidewalk. That’s technically an approximation though a very accurate one. The formula comes from integrating the normal curve times x from zero to infinity then multiplying times two. This gives the expected distance from the mean.Quote: Ace2Here’s another easy one:
A drunkard starts walking on a straight sidewalk that runs north-south. His goal is to walk northward, but for each step he’s equally likely to go one foot directly north, south, east or west.
After 10,000 steps, how far off the sidewalk would you expect him to be?
Closed form solutions only.
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The exact answer (shown to ten digits) of 56.41825312 can be obtained via Markov chain, though that’s not a closed form solution.
If you were to start searching for the drunkard, the best place to start would be at the sidewalk since you don’t know which side of it he’s on, but you’d be an average of ~56 steps away from it when you found him. The chance of him being on the sidewalk is <1%
It’s all about making that GTA
January 20th, 2026 at 2:41:16 PM
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The issue of where to start to search can be an issue in Orienteering. The concept of the sport is you're given a map of, typically, a forest and you have to find a series of controls. Let's say you're trying to find a point on a footpath that runs East-West but is due North of where you are. You're good at following a compass heading but have an in-built error, so the point you reach on the footpath follows a normal curve. Assuming you're trying to pick the fastest option, I suspect your optimal choice is to aim towards one side, say slightly West of North, and then turn East when you reach the footpath. At some stage you recognise your likely error and need to turn back on yourself. etc. (In the original puzzle you could only be whole degrees out and never greater than 4SDs.)
January 21st, 2026 at 6:32:44 AM
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Quote: Ace2The answer is (10,000/π)^.5 =~ 56.419 steps from the sidewalk. That’s technically an approximation though a very accurate one. The formula comes from integrating the normal curve times x from zero to infinity then multiplying times two. This gives the expected distance from the mean.
The exact answer (shown to ten digits) of 56.41825312 can be obtained via Markov chain, though that’s not a closed form solution.
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Technically, an approximation isn't one either.
The closest to a "closed form solution" that I got was this:
Each step has probability 1/4 of adding 1 to the number of steps to the left of the sidewalk, 1/4 of subtracting 1, and 1/2 of keeping it the same.
Let (m, n, 10,000 - (m + n)) represent the number of steps to the left, the number to the right, and the number forward/backward, with each one being >= 0 and <= 10,000.
The number of steps away from the sidewalk is |m - n|, and the probability of that particular triple is C(10,000, m) C(10,000 - m, n) (1/4)^10,000, or C(10,000, m) C(10,000 - m, n) / 2^20,000.
The solution is the sum over all m from 0 to 10,000 of (the sum over all n from 0 to (10,000 - n) of ( |m - n| C(10,000, m) C(10,000 - m, n) ) ), divided by 2^20,000.
However, I could not find a way to simplify that.
January 24th, 2026 at 12:31:06 AM
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I get a lot of simple math problems on my Facebook feed. Usually there is a strong consensus on the answer, but this one is split about 50/50.
-62-36
-62-36
-72
"My life is spent in one long effort to escape from the commonplace of existence. These little problems help me to do so." -- Sherlock Holmes
January 24th, 2026 at 1:10:35 AM
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Quote: WizardI get a lot of simple math problems on my Facebook feed. Usually there is a strong consensus on the answer, but this one is split about 50/50.
-62-36-72
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The placement of brackets can alter the answer, however, since the question did not include these, they can still be used but should not alter the context of the question.
My answer would be:
- 6² - 36
= - (6 * 6)-36
= - 36 - 36
= - 72
An alternate solution using brackets is:
(-6 * -6) - 36
= 0
My answer would be:
- 6² - 36
= - (6 * 6)-36
= - 36 - 36
= - 72
An alternate solution using brackets is:
(-6 * -6) - 36
= 0
Casino Enemy No.1
January 24th, 2026 at 1:15:39 AM
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Quote: WizardI get a lot of simple math problems on my Facebook feed. Usually there is a strong consensus on the answer, but this one is split about 50/50.
-62-36-72
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It's ambiguous enough, but a minus sign in front of a numeral with nothing else in front of the minus sign I have never seen used to indicate other than just that it's a negative number, while a minus sign between two numbers indicates subtraction. On the other hand if it was -x2-36, where x=6 or -6, I'd probably say -72 too.
January 24th, 2026 at 6:51:22 AM
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Quote: WizardI get a lot of simple math problems on my Facebook feed. Usually there is a strong consensus on the answer, but this one is split about 50/50.
-62-36-72
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Wizard is correct
Exponent comes before unary minus; this makes it -(62) - 36 = -36 - 36 = -72
January 24th, 2026 at 1:45:43 PM
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From the bastion of all human knowledge, Wikipedia: https://en.wikipedia.org/wiki/Order_of_operations
Thus, state the convention you are using to decide whether exponentiation or unary minus has precedence.
Dog Hand
Thus, state the convention you are using to decide whether exponentiation or unary minus has precedence.
Dog Hand
January 24th, 2026 at 5:18:17 PM
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Excel says it’s 0. I say it’s -72
It’s all about making that GTA
January 24th, 2026 at 6:09:38 PM
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I was thinking a negative times a negative is a positive so I can see where they came up with zero. Been a while since I have been in high school math class though
January 25th, 2026 at 5:13:49 AM
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It's an order of operations problem and ThatDonGuy gave the correct answer.
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
January 25th, 2026 at 1:01:46 PM
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If parentheses are implied around the -6 to get (-6)^2 - 36 =0, then we should be consistent and assume parentheses are implied around the -36. Therefore:
(-6)^2 (-36) = -1,296
I don’t agree with that answer but one could make the argument if parentheses can be implied
(-6)^2 (-36) = -1,296
I don’t agree with that answer but one could make the argument if parentheses can be implied
It’s all about making that GTA
January 25th, 2026 at 1:16:40 PM
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Quote: Ace2If parentheses are implied around the -6 to get (-6)^2 - 36 =0, then we should be consistent and assume parentheses are implied around the -36. Therefore:
(-6)^2 (-36) = -1,296
I don’t agree with that answer but one could make the argument if parentheses can be implied
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Except that, since that minus sign is between two numbers, it is not unary, but binary.
January 25th, 2026 at 2:39:41 PM
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Are you saying “unary” -6 is not the same as “binary” 0 - 6 ?
It’s all about making that GTA
January 25th, 2026 at 10:25:23 PM
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Thank you all for the feedback. I too thought to do the exponent first.
"My life is spent in one long effort to escape from the commonplace of existence. These little problems help me to do so." -- Sherlock Holmes
January 26th, 2026 at 6:59:56 AM
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Quote: Ace2Are you saying “unary” -6 is not the same as “binary” 0 - 6 ?
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Not at all.
However, I am saying that 0 - 6 = -6, and not 0 - 6 = (0)(-6) = 0.
February 3rd, 2026 at 4:01:05 PM
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Using a random number generator, you pick a number (x) between zero and one. You then apply the formula x^(x^2) = y. What x value will give the lowest value of y?
It’s all about making that GTA
February 3rd, 2026 at 4:55:23 PM
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Quote: Ace2Using a random number generator, you pick a number (x) between zero and one. You then apply the formula x^(x^2) = y. What x value will give the lowest value of y?
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We are asked to find the minimum of y = x^(x^2) where 0 <= x <= 1.
dy = d(x^(x^2))
= d(e^(x^2 ln x))
= e^(x^2 ln x) d(x^2 ln x)
= x^(x^2) (x + 2x ln x) dx
This = 0 when 2x ln x = -x, so ln x = -1/2, and x = e^(-1/2) = 1 / the square root of e, or about 0.60653
February 3rd, 2026 at 5:27:54 PM
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Agree.
For some reason I was surprised to find an e in that one
For some reason I was surprised to find an e in that one
It’s all about making that GTA
February 3rd, 2026 at 5:45:00 PM
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Quote: Ace2Agree.
For some reason I was surprised to find an e in that one
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e shows up a lot with exponentiation, since the first derivative of a^x is (a^x ln a) dx.
Bonus problem:
Prove that x^(x^2)) approaches 1 as x approaches 0 from the positive side
February 4th, 2026 at 9:53:46 AM
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Any positive number raised to the exponent of zero equals one. So as (x^2) approaches zero, x^(x^2) approaches one
It’s all about making that GTA
February 5th, 2026 at 3:15:38 PM
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Quote: Ace2Any positive number raised to the exponent of zero equals one. So as (x^2) approaches zero, x^(x^2) approaches one
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Not passing the smell test. As x^2 approaches 0, x^(x^2) approaches 0^0.
x^(x^2)) = e^(ln (x^(x^2))
= e^(x^2 ln x)
Since e^x is strictly increasing, the limit of e^(x^2 ln x) = e^(the limit of x^2 ln x)
= e^(the limit of ln x / x^(-2))
Using L'Hopital's rule, this is e^(the limit of x^(-1) / ((-2) x^(-3)))
= e^(the limit of -x^2 / 2)
Since x is approaching 0, this approaches e^(-0/2) = 1.
February 5th, 2026 at 4:06:13 PM
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The method I was told at college of showing that something tended towards a limit was, I may have got the letters the wrong way round, given a delta>0 there exists an epsilon such that the value is within delta of the limit.
As a simple example consider 1+1/2+1/4... which tends to 2. If you give me a delta of say 1/500, then I can show the total of 1+1/2+1/4...+1/512 is within 1/500 of 2. If you give me a smaller delta, then I just keep on going until I reach a small enough fraction.
As a simple example consider 1+1/2+1/4... which tends to 2. If you give me a delta of say 1/500, then I can show the total of 1+1/2+1/4...+1/512 is within 1/500 of 2. If you give me a smaller delta, then I just keep on going until I reach a small enough fraction.
February 5th, 2026 at 4:20:58 PM
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Exactly. Since 0^0 = 1, it confirms the aforementioned proof. Additionally, the exponent of x^2 will always be less than the base of x for any value of 0<x<1, so the function gets closer to 1 as x approaches zeroQuote: ThatDonGuyAs x^2 approaches 0, x^(x^2) approaches 0^0.
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Last edited by: Ace2 on Feb 5, 2026
It’s all about making that GTA
February 6th, 2026 at 6:34:03 AM
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Quote: Ace2Exactly. Since 0^0 = 1, it confirms the aforementioned proof. Additionally, the exponent of x^2 will always be less than the base of x for any value of 0<x<1, so the function gets closer to 1 as x approaches zeroQuote: ThatDonGuyAs x^2 approaches 0, x^(x^2) approaches 0^0.
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Excuse me? 0^0 = 1?
February 6th, 2026 at 8:48:43 AM
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In certain areas of mathematics, such as combinatorics and algebra, 0^0 is conventionally defined as 1 because this assignment simplifies many formulas and ensures consistency in operations involving exponents. For instance, in combinatorics, defining 0^0 = 1 aligns with the interpretation of choosing 0 elements from a set and simplifies polynomial and binomial expansions.
Would you agree that, for example, combin(5,0) = 1?
Regardless, we are evaluating the function as x approaches zero, not at x equals zero
Would you agree that, for example, combin(5,0) = 1?
Regardless, we are evaluating the function as x approaches zero, not at x equals zero
Last edited by: Ace2 on Feb 6, 2026
It’s all about making that GTA
February 7th, 2026 at 11:08:34 AM
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You go to Vegas over a long weekend to play a single-deck blackjack game with a 0% edge ($500 minimum). You decide to play 1,200 hands but you will quit if you bust or double your initial bankroll before reaching 1200 hands.
What size bankroll should you bring to give yourself a 1/3 chance of busting, 1/3 chance of doubling and 1/3 chance of finishing the 1200 hands without busting or doubling?
Assume a standard deviation of 1.1547, flat betting $500 one hand at a time and perfect basic strategy to realize the 0% edge
What size bankroll should you bring to give yourself a 1/3 chance of busting, 1/3 chance of doubling and 1/3 chance of finishing the 1200 hands without busting or doubling?
Assume a standard deviation of 1.1547, flat betting $500 one hand at a time and perfect basic strategy to realize the 0% edge
Last edited by: Ace2 on Feb 7, 2026
It’s all about making that GTA
February 7th, 2026 at 4:50:08 PM
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Quote: Ace2You go to Vegas over a long weekend to play a single-deck blackjack game with a 0% edge ($500 minimum). You decide to play 1,200 hands but you will quit if you bust or double your initial bankroll before reaching 1200 hands.
What size bankroll should you bring to give yourself a 1/3 chance of busting, 1/3 chance of doubling and 1/3 chance of finishing the 1200 hands without busting or doubling?
Assume a standard deviation of 1.1547, flat betting $500 one hand at a time and perfect basic strategy to realize the 0% edge
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What does the SD of 1.1547 refer to?
February 7th, 2026 at 5:21:44 PM
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The blackjack game being played.Quote: ThatDonGuy[
What does the SD of 1.1547 refer to?
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Out of curiosity, what else could it refer to ?
It’s all about making that GTA
February 8th, 2026 at 6:27:55 AM
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Quote: Ace2The blackjack game being played.Quote: ThatDonGuy[
What does the SD of 1.1547 refer to?
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Out of curiosity, what else could it refer to ?
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I understand it has to do with the blackjack game - but in what way? 1.1547 of what, exactly, is within 1 SD of the mean? Is it the final win/loss result after 1200 hands?
February 8th, 2026 at 9:06:28 AM
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If you play n hands of a game with an edge of x, your expectation is nx +/- n^.5 * SD. N, x and SD are defined in the problem.
I’m quite surprised I have to explain what standard deviation means. Look up the SD for just about any blackjack game and it’s in the 1.13 - 1.16 range
I’m quite surprised I have to explain what standard deviation means. Look up the SD for just about any blackjack game and it’s in the 1.13 - 1.16 range
It’s all about making that GTA
February 8th, 2026 at 6:01:33 PM
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I'm buying Big Game boxes. Each pool has 100 boxes and has four prizes: $10, $30, $10, $50. $1 buys one box. $1 = 1/100 chance of winning $50, but what are the chances of winning a prize?
I have $50 to buy tickets. If I buy $50 worth of boxes in one pool, I have a very good chance of winning something, but am limited to one final prize. If I buy $10 worth of boxes in five pools, I have less chance in each pool, but my odds of winning multiple top prizes increase.
How do I determine the odds in each scenario?
I have $50 to buy tickets. If I buy $50 worth of boxes in one pool, I have a very good chance of winning something, but am limited to one final prize. If I buy $10 worth of boxes in five pools, I have less chance in each pool, but my odds of winning multiple top prizes increase.
How do I determine the odds in each scenario?
The older I get, the better I recall things that never happened
February 9th, 2026 at 5:23:11 AM
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Quote: Ace2If you play n hands of a game with an edge of x, your expectation is nx +/- n^.5 * SD. N, x and SD are defined in the problem.
I’m quite surprised I have to explain what standard deviation means. Look up the SD for just about any blackjack game and it’s in the 1.13 - 1.16 range
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I counted cards at bj for several years
but I admit that standard deviation as indicated in the question is something I did not fully understand
but I knew enough to know there was a large difference possible in the outcomes of a session or sessions even though I had a smallish advantage
the following does not answer the question posed but it helped me understand calculations behind standard deviation
it's not based on the game being equal but in a game where the player is using basic strategy and the house has about a .5% edge
it indicates that because of the rules of bj - splitting, doubling, and drawing bj the standard deviation of a game of bj is greater than that of a 50/50 coin flip
the standard deviation of a coin flip (assuming equal chance of coming heads or tails) is 0.5
the question posed is interesting and hopefully Ace or someone else will answer it and explain the calculations
"AI Overview
In blackjack, a standard deviation (SD) of approximately 1.15 units per round is the standard measure of volatility for a player using basic strategy. It quantifies how much your bankroll will fluctuate around the expected outcome (the average) due to the luck of the draw. Here is a detailed breakdown of what 1.1547 (often rounded to 1.14–1.15) means and how to use it: 1. What "1.15 Units" Means The Unit: The unit refers to your base bet. If you bet $10 per hand, your SD is $11.50 ($10 \(\times \) 1.15).Volatily Measure: Because blackjack involves doubling down and splitting pairs, the variance is higher than a simple 50/50 coin flip. The 1.15 value accounts for these extra payouts, indicating that a session of blackjack is quite volatile.Context: While the house edge might be a tiny fraction of a percent (e.g., -0.5%), the standard deviation is massive in comparison (115% of your bet).
2. How to Use 1.15 for Session Planning Standard deviation helps you determine how much money (bankroll) you need to survive, as actual results are likely to fall within 1 SD of your expected value (EV) about 68% of the time. Example (1 Hour/100 Hands):Bet: $10/hand.Expected Value (EV): Roughly -$5 (assuming -0.5% edge).SD per round: $11.50.SD for 100 hands: $11.50 \(\times \) \(\sqrt{100}\) (10) = $115.Result: In one hour, you are likely to be somewhere between winning $110 ($115 - $5) and losing $120 (-$115 - $5).
3. Key Takeaways Short-Term vs. Long-Term: In the short term (one night), standard deviation dwarfs the house edge. You can win with a negative expectation or lose with a positive one.Bankroll Requirement: The 1.15 figure helps players calculate the necessary bankroll to avoid going broke (risk of ruin).Accumulation: Standard deviation does not grow linearly. 400 hands do not have 4 times the SD of 100 hands; they have roughly 2 times (square root of the number of hands)."
.
the foolish sayings of a rich man often pass for words of wisdom by the fools around him
February 9th, 2026 at 9:43:39 AM
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The proper comparison would be a coin flip game in which you lose on tails and get paid 2 for 1 every time a head comes up. The standard deviation of the game is 1.Quote: lilredrooster
the standard deviation of a coin flip (assuming equal chance of coming heads or tails) is 0.5
Blackjack’s SD is a bit higher than 1 since you have <50% chance of winning a hand (before cards are dealt) and since some payouts are higher than 2 for 1.
It’s all about making that GTA
