Easy math puzzles
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51 members have voted
November 4th, 2025 at 9:13:06 AM
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Sorry to step on the craps puzzle, but I think we can handle two at the same time.
On Jan 1 a hospital nursery has 3 boys and an unknown number of girls. On Jan 2 a baby is born and added to the nursery. On Jan 3 a random baby is selected from the nursery and it is a boy.
What is the probability the baby born on Jan 2 was a boy?
On Jan 1 a hospital nursery has 3 boys and an unknown number of girls. On Jan 2 a baby is born and added to the nursery. On Jan 3 a random baby is selected from the nursery and it is a boy.
What is the probability the baby born on Jan 2 was a boy?
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
November 4th, 2025 at 10:15:02 AM
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This comes from old racing odds used in the UK (US tracks tend to use 6/5 7/5 3/2 8/5 9/5 etc). Also we used to have Pounds, Shillings and Pence which is why some early prices were things like 100/6 100/7 100/8 and showed how many shillings you needed to put on to win £5 = 100/-. I'm guessing 1/8th (e.g. 13/8) were used as there was a half-crown (2/6) coin, and 6/4 used as it fitted into the Ev 11/10 5/4 6/4 7/4 2/1 9/4 5/2 11/4 3/1 100/30 7/2 4/1 simple pattern. (Some small greyhound tracks don't use 6/5 11/8 13/8 15/8 85/40 17/2.)Quote: Ace2Out of curiosity, why did you use 6:4 instead of 3:2 ? I’ve also seen UK sportsbooks use 6:4.Quote: charliepatrick
5/9 pays 6 to 4]
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In the old days Starting Prices were from an "average" of "leading" on-course bookmakers. This was done via a group of people going round checking the prices and, as the race started, having a "huddle". It wasn't a mathematical average but a fair price that was available on a number of good bookmakers. They, nearly always, used the nearest odds from the approved list, although in some big race tracks I've known prices such as 95/40, 35/1, 7/5 to be used.
Nowadays, at UK racetracks, Starting Price is calculated using a computer which can directly read the bookmakers prices. I'm not sure whether they still use on course bookies or have moved to include off course ones, but it uses a sample of (say) 12 and takes the median (i.e. a price available on half of the sample). This means it's a price from a specific bookie, so you now do get intermediate prices such as 7/5 8/5 11/5 16/5.
November 4th, 2025 at 10:36:14 AM
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Interesting. Incidentally, 6:4 is the true inverse of winning a 5 or 9 as in 6 ways to lose and 4 ways to win
It’s all about making that GTA
November 5th, 2025 at 8:24:51 AM
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My zero-edge craps version:
Point win on hard six pays 3:1.
Point win on five (2-3) pays 1:2
So 3.5% of resolutions are affected with two changes for a score of 0.07. The only caveat is bets should be made in even amounts due to the 1:2 payout on a 2-3 five point win, but the only adjustment would probably be for a $25 bettor to increase to $30.
Zero-edge can also be accomplished as:
Point win on hard six/eight pays 2:1.
Point win on five (2-3) pays 1:2
Though this has a higher score of 0.14 since a higher percentage of resolutions are affected and there are three changes instead of two
There are probably more ways to get to zero edge with minimal changes and practical payout ratios, though I don’t know of any method to find them besides trial and error. I believe there’s no way to get there with less than two changes since none of the individual outcomes’ probabilities are divisible by the 7 of 495
Point win on hard six pays 3:1.
Point win on five (2-3) pays 1:2
So 3.5% of resolutions are affected with two changes for a score of 0.07. The only caveat is bets should be made in even amounts due to the 1:2 payout on a 2-3 five point win, but the only adjustment would probably be for a $25 bettor to increase to $30.
Zero-edge can also be accomplished as:
Point win on hard six/eight pays 2:1.
Point win on five (2-3) pays 1:2
Though this has a higher score of 0.14 since a higher percentage of resolutions are affected and there are three changes instead of two
There are probably more ways to get to zero edge with minimal changes and practical payout ratios, though I don’t know of any method to find them besides trial and error. I believe there’s no way to get there with less than two changes since none of the individual outcomes’ probabilities are divisible by the 7 of 495
Last edited by: Ace2 on Nov 5, 2025
It’s all about making that GTA
November 5th, 2025 at 11:15:10 AM
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Wiz Idea #3
Point win on a 4 pays 3-2.
If point is a 3-3 and made with another 3-3, pays 11-10.
All other rules stay the same.
I know that second rule seems nearly worthless, but I had to find 0.0002525253 somewhere.
Wiz Idea #4
My best yet!
If point is established and made with a hard 6 or hard 8, then win is 19 to 5.
Point win on a 4 pays 3-2.
If point is a 3-3 and made with another 3-3, pays 11-10.
All other rules stay the same.
I know that second rule seems nearly worthless, but I had to find 0.0002525253 somewhere.
Wiz Idea #4
My best yet!
If point is established and made with a hard 6 or hard 8, then win is 19 to 5.
Last edited by: Wizard on Nov 5, 2025
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
November 5th, 2025 at 3:07:33 PM
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Great answers!Quote: WizardWiz Idea #3
Point win on a 4 pays 3-2.
If point is a 3-3 and made with another 3-3, pays 11-10.
All other rules stay the same.
I know that second rule seems nearly worthless, but I had to find 0.0002525253 somewhere.
Wiz Idea #4
My best yet!
If point is established and made with a hard 6 or hard 8, then win is 19 to 5.
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They both beat my score of 0.07. Your second version only affects about one half of one percent (1/198) of all resolutions
It’s all about making that GTA
November 5th, 2025 at 4:56:09 PM
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Good spot since it creates 1/11, so can be factored against 495=5*99.
Chance of coming out using hard 6 = 1/36
Chance of then winning with hard 6 = 1/11
Hence chance of either 6 or 8 = 1/198 = .5/99
House edge required = 7/495 = 1.4/99 (= 2.8/198)
Hence additional payout needs to be 1.4/.5 = 2.8
This means the actualy payout required = 3.8 = 19/5.
Chance of coming out using hard 6 = 1/36
Chance of then winning with hard 6 = 1/11
Hence chance of either 6 or 8 = 1/198 = .5/99
House edge required = 7/495 = 1.4/99 (= 2.8/198)
Hence additional payout needs to be 1.4/.5 = 2.8
This means the actualy payout required = 3.8 = 19/5.
November 5th, 2025 at 5:22:06 PM
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Deleted
Last edited by: Ace2 on Nov 5, 2025
It’s all about making that GTA
November 6th, 2025 at 3:19:05 PM
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Quote: WizardSorry to step on the craps puzzle, but I think we can handle two at the same time.
On Jan 1 a hospital nursery has 3 boys and an unknown number of girls. On Jan 2 a baby is born and added to the nursery. On Jan 3 a random baby is selected from the nursery and it is a boy.
What is the probability the baby born on Jan 2 was a boy?
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Since no one else seems to have bothered trying to answer this one, I'll give it a shot:
1/2, assuming the laws of genetics still apply.
Other than that, I don't know where to start working on this one. Care to clue us in?
November 6th, 2025 at 4:58:31 PM
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Quote: WizardSorry to step on the craps puzzle, but I think we can handle two at the same time.
On Jan 1 a hospital nursery has 3 boys and an unknown number of girls. On Jan 2 a baby is born and added to the nursery. On Jan 3 a random baby is selected from the nursery and it is a boy.
What is the probability the baby born on Jan 2 was a boy?
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I get 4/7.
Let BB represent the probability a boy is born and a boy is chosen at random. Irrespective of the starting number of girls, the probability of BB is 4/3 of GB. We can eliminate BG and GG since we know a boy was chosen. So 4 / (4+3) =~ 57 %
Let BB represent the probability a boy is born and a boy is chosen at random. Irrespective of the starting number of girls, the probability of BB is 4/3 of GB. We can eliminate BG and GG since we know a boy was chosen. So 4 / (4+3) =~ 57 %
Last edited by: Ace2 on Nov 6, 2025
It’s all about making that GTA
November 6th, 2025 at 5:24:07 PM
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Quote: ThatDonGuy
1/2, assuming the laws of genetics still apply.
Other than that, I don't know where to start working on this one. Care to clue us in?
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Incorrect. I'll give a little more time to see if you wake up the problem.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
November 6th, 2025 at 5:25:32 PM
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Quote: Ace2I get 4/7.
Let BB represent the probability a boy is born and a boy is chosen at random. Irrespective of the starting number of girls, the probability of BB is 4/3 of GB. We can eliminate BG and GG since we know a boy was chosen. So 4 / (4+3) =0.57 %
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I agree!
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
November 6th, 2025 at 6:13:29 PM
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Addendum to hospital nursery answer:
4/7 assumes boys/girls are born at a 50/50 ratio. The actual ratio is more like 51/49. Applying that ratio, the answer is (51 * 4) / (51 *4 + 49 * 3) = 68/117 =~58.1%
It’s all about making that GTA
November 8th, 2025 at 9:04:20 AM
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You can solve the Monty Hall problem with the same method used for the hospital nursery problem
If there are three doors A,B,C and you choose A, there are four possibilities:
AB 1/6
AC 1/6
BC 1/3
CB 1/3
The first letter is the location of the prize and the second letter is the door opened by the host, followed by their probabilities
If the host opens door C, that leaves AC and BC as possibilities, with BC having a probability of 1/3 / (1/3 + 1/6) = 2/3. So you should change to door B.
If there are three doors A,B,C and you choose A, there are four possibilities:
AB 1/6
AC 1/6
BC 1/3
CB 1/3
The first letter is the location of the prize and the second letter is the door opened by the host, followed by their probabilities
If the host opens door C, that leaves AC and BC as possibilities, with BC having a probability of 1/3 / (1/3 + 1/6) = 2/3. So you should change to door B.
It’s all about making that GTA
November 10th, 2025 at 4:11:11 AM
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This conditional probability problem can be seen in the 2008 movie "21" where student Ben Campbell explains the solution to his onscreen professor Mickey Rosa (played by Kevin Spacey).
Casino Enemy No.1
November 10th, 2025 at 10:30:48 AM
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This is well relevant. We have mathematicians, entertainment and gamblers. People learn while they are entertained.
November 10th, 2025 at 2:54:34 PM
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Here's one I found on the internet, and I'm "pretty sure" I have the answer...
A large number of students take identical tests, each consisting of the same 10 true-or-false questions in the same order.
What is the maximum number of students that can be taking the test such that, if you choose any two of the completed tests, at least two of the questions will have different answers on them?
A large number of students take identical tests, each consisting of the same 10 true-or-false questions in the same order.
What is the maximum number of students that can be taking the test such that, if you choose any two of the completed tests, at least two of the questions will have different answers on them?
November 10th, 2025 at 3:17:49 PM
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C(10,2)x2^8=45x256=11,520
November 10th, 2025 at 3:41:05 PM
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Quote: acesideC(10,2)x2^8=45x256=11,520
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I have a feeling you read the problem wrong
According to the Pigeonhole Principle, if there are 1025 tests, then at least two of them will have all ten questions with the same answers
November 10th, 2025 at 3:45:50 PM
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This question is confusing. I hesitated between these two answers. “at least two of the questions will have different answers on them?” Should be “ at least one of the questions will have different answers on them?”
November 10th, 2025 at 4:00:30 PM
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Quote: acesideThis question is confusing. I hesitated between these two answers. “at least two of the questions will have different answers on them?” Should be “ at least one of the questions will have different answers on them?”
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No. The question is, what is the maximum number such that, if you pull two of them out at random, at least two of the ten questions will have the answer "true" on one of them and "false" on the other? Note that the questions with different answers do not all have to be true on one of the tests and all false on the other; for example, on question 1, test A can have "true" and test B can have "false," while on question 8, test A can have "false" and test B can have "true."
November 10th, 2025 at 4:30:00 PM
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So, if there were only four questions on the test, there could be at least three people, as follows:
Ace: TTTT
Don: TTFF
Charlie: FFFF
Note you can pull any two of them and they will disagree on at least two questions.
I'm still working on the ten-question case.
Ace: TTTT
Don: TTFF
Charlie: FFFF
Note you can pull any two of them and they will disagree on at least two questions.
I'm still working on the ten-question case.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
November 10th, 2025 at 8:33:23 PM
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Quote: WizardSo, if there were only four questions on the test, there could be at least three people, as follows:
Ace: TTTT
Don: TTFF
Charlie: FFFF
Note you can pull any two of them and they will disagree on at least two questions.
I'm still working on the ten-question case.
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Why couldn’t there also be:
Betsy: TFTF
Ernie: FTFT
Freddie: FFTT
Ernie: FTFT
Freddie: FFTT
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
November 10th, 2025 at 8:36:01 PM
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512.
I started with someone getting 0 out of 10 questions correct and then saw that no-one could only get 1 correct and satisfy the solution. However anyone with 2 correct would be OK. From this it follows that if everyone got an even number of tests correct, then those with different numbers of tests correct would be OK. So now see what happens inside the groups.
Suppose someone has N correct and someone else also has (a different set of) N correct. Then there might be N-1 where they both got it correct but that leaves 2 where they disagreed. c.f. 0XXXX and XXXX0 where there are two 0-X. Hence each group is allowed every combination.
Hence it's the total ways to have 0-questions correct, 2-questions correct (10*9/2), 4, 6, 8 and 10.
It transpires that the total is 1+45+210+210+45+1=512.
This sort of makes sense as the same logic could apply to everyone having answered an odd number of questions correct.
I started with someone getting 0 out of 10 questions correct and then saw that no-one could only get 1 correct and satisfy the solution. However anyone with 2 correct would be OK. From this it follows that if everyone got an even number of tests correct, then those with different numbers of tests correct would be OK. So now see what happens inside the groups.
Suppose someone has N correct and someone else also has (a different set of) N correct. Then there might be N-1 where they both got it correct but that leaves 2 where they disagreed. c.f. 0XXXX and XXXX0 where there are two 0-X. Hence each group is allowed every combination.
Hence it's the total ways to have 0-questions correct, 2-questions correct (10*9/2), 4, 6, 8 and 10.
It transpires that the total is 1+45+210+210+45+1=512.
This sort of makes sense as the same logic could apply to everyone having answered an odd number of questions correct.
November 11th, 2025 at 4:04:15 AM
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Deleted
Last edited by: aceside on Nov 11, 2025
November 11th, 2025 at 6:09:49 AM
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Quote: unJonWhy couldn’t there also be:
Betsy: TFTF
Ernie: FTFT
Freddie: FFTT
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There can - and if you look closer, you're getting close to a solution
November 11th, 2025 at 12:25:19 PM
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Quote: ThatDonGuyQuote: unJonWhy couldn’t there also be:
Betsy: TFTF
Ernie: FTFT
Freddie: FFTT
link to original postThere can - and if you look closer, you're getting close to a solution
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Yes I see it. Also Gretchen and Heather.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
November 12th, 2025 at 6:33:52 AM
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511 = combin(10,8)+combin(10,6)+combin(10,4)+combin(10,2)+combin(10,0)
The way I think of it is there is shall we say a key test. Call it one with all the right answers, but it can be any test.
combin(10,8) = ways 8 questions can match on another test, leaving 2 that don't.
combin(10,6) = ways 6 questions can match on another test, leaving 4 that don't.
Etc.
I think you can compare any two of the other tests and there will be at least two answers that are different.
The way I think of it is there is shall we say a key test. Call it one with all the right answers, but it can be any test.
combin(10,8) = ways 8 questions can match on another test, leaving 2 that don't.
combin(10,6) = ways 6 questions can match on another test, leaving 4 that don't.
Etc.
I think you can compare any two of the other tests and there will be at least two answers that are different.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
November 12th, 2025 at 7:25:21 AM
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Let me make a guess, 172. This is based on the fact that there are 172 prime numbers less than 1024. These two questions are very similar.
November 12th, 2025 at 7:26:26 AM
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You were close...Quote: Wizard511 = combin(10,8)+combin(10,6)+combin(10,4)+combin(10,2)+combin(10,0)
The way I think of it is there is shall we say a key test. Call it one with all the right answers, but it can be any test.
combin(10,8) = ways 8 questions can match on another test, leaving 2 that don't.
combin(10,6) = ways 6 questions can match on another test, leaving 4 that don't.
Etc.
I think you can compare any two of the other tests and there will be at least two answers that are different.
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This is correct.Quote: charliepatrick512.
I started with someone getting 0 out of 10 questions correct and then saw that no-one could only get 1 correct and satisfy the solution. However anyone with 2 correct would be OK. From this it follows that if everyone got an even number of tests correct, then those with different numbers of tests correct would be OK. So now see what happens inside the groups.
Suppose someone has N correct and someone else also has (a different set of) N correct. Then there might be N-1 where they both got it correct but that leaves 2 where they disagreed. c.f. 0XXXX and XXXX0 where there are two 0-X. Hence each group is allowed every combination.
Hence it's the total ways to have 0-questions correct, 2-questions correct (10*9/2), 4, 6, 8 and 10.
It transpires that the total is 1+45+210+210+45+1=512.
This sort of makes sense as the same logic could apply to everyone having answered an odd number of questions correct.
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I will provide a general solution for N questions on the test.
For the first N-1 questions there are 2 possible answers for the first question, 2 for the second, ..., and 2 for the (N-1)th, so there are a total of 2N-1 possible different sets of answers for the first N-1 questions.
If there are 2N-1 + 1 or more test takers, then, by the Pigeonhole Principle, at least two of them have identical answers for the first N-1 questions. Whether or not their answers to the Nth question are the same, they cannot have more than one questions with different answers.
This means that it is always possible to choose two tests where the number of different answers < 2, so the solution must be < 2N-1 + 1.
Now, suppose there are 2N-1 test takers. Each one can have a different set of answers for the first N-1 questions; one can have all of them false, one can have only the (N-1)th question true, one can have only the (N-2)th question true, one can have only the (N-1)th and (N-2)th questions true, one can have only the (N-3)th question true, and so on.
Also, each test has its Nth question answered as follows: if the number of true answers for the first N-1 questions is odd, then the Nth answer is true, and if the number is even, then the Nth answer is false.
Choose any two tests. One of two conditions must be true: either they have two or more of the first N-1 questions with different answers, or they have exactly one of the first N-1 questions with different answers. In the latter case, since N-2 of the first N-1 questions have the same answer, and the remaining question is answered true on one test and false on the other, one test has one more true answer in the first N-1 questions, so one has an odd number of answers and the other has an even number, which means the answers to question N must also be different, so there are two questions with different answers on the tests. This means that every pair of tests have at least two questions with different answers.
Therefore, the solution to the problem is 2N-1.
In the problem as stated, N = 1, so the maximum number is 512.
November 12th, 2025 at 7:32:09 AM
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Can you also generalize this to “at least 3 of the questions will have different answers” ?
November 12th, 2025 at 8:37:01 AM
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Quote: acesideCan you also generalize this to “at least 3 of the questions will have different answers” ?
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I will give it a try.
November 12th, 2025 at 9:32:27 AM
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I'm guessing 26 but I don't know an upper bound for the answer to the latest puzzle.
I have heard of Hamming Codes. They are used in transmitting data or storing data and add extra non-data bits for error detection and correction. The aim is either to detect there has been an error or actually know which bit is in error. For example with a simple parity bit, you know one of the bits has been flipped but you don't know which one. With (better) Hamming Codes if there is only one bit wrong, you also have the ability to know which bit was flipped or if there were two bits flipped the ability to detect the error but not know which bits are wrong.
There is a formula for working out how many bits work and I'm guessing the solution to this is based on the best Hamming Code using 10 bits. Therefore the answer is likely to be 2n where n is probably 6. c.f. https://en.wikipedia.org/wiki/Hamming_code and SECDED.
The layout of the data is P1 P2 D1 P3 D2 D3 D4 P4 D5 D6. So you choose the D (one of 64 possibilities) and that sets P according to the following.
P1=D1\D2\D4\D5
P2=D1\D3\D4\D6
P3=D2\D3\D4
P4=D5\D6
(where \ means Xor)
So, consider the starting position and see what happens if you change some of the Ds.
Changing D1 flips D1,P1 and P2. so a change of three bits
Changing D2 flips D2,P1 and P3.
etc.
Changing D1 and D2 flips P2 and P4.
Changing D1 and D3 flips P1 and P4.
etc.
Changing more then two D's means three (or more) are already different.
So for any two ten-bit entries that have different Ds, they will have at least three bits different.
Hence there are at least 64 (26) different possibilities.
I have heard of Hamming Codes. They are used in transmitting data or storing data and add extra non-data bits for error detection and correction. The aim is either to detect there has been an error or actually know which bit is in error. For example with a simple parity bit, you know one of the bits has been flipped but you don't know which one. With (better) Hamming Codes if there is only one bit wrong, you also have the ability to know which bit was flipped or if there were two bits flipped the ability to detect the error but not know which bits are wrong.
There is a formula for working out how many bits work and I'm guessing the solution to this is based on the best Hamming Code using 10 bits. Therefore the answer is likely to be 2n where n is probably 6. c.f. https://en.wikipedia.org/wiki/Hamming_code and SECDED.
The layout of the data is P1 P2 D1 P3 D2 D3 D4 P4 D5 D6. So you choose the D (one of 64 possibilities) and that sets P according to the following.
P1=D1\D2\D4\D5
P2=D1\D3\D4\D6
P3=D2\D3\D4
P4=D5\D6
(where \ means Xor)
So, consider the starting position and see what happens if you change some of the Ds.
Changing D1 flips D1,P1 and P2. so a change of three bits
Changing D2 flips D2,P1 and P3.
etc.
Changing D1 and D2 flips P2 and P4.
Changing D1 and D3 flips P1 and P4.
etc.
Changing more then two D's means three (or more) are already different.
So for any two ten-bit entries that have different Ds, they will have at least three bits different.
Hence there are at least 64 (26) different possibilities.
November 12th, 2025 at 4:38:51 PM
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Quote: acesideCan you also generalize this to “at least 3 of the questions will have different answers” ?
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The best answer I have found so far is 2^(N-2).
If there are 2^(N-2) test takers, then each one can have a different set of answers for the first (N-2) questions.
Use the "parity formula" from the 2-answer solution, but this time, anyone who answers true for the (N-1)th question answers true for the Nth question as well, and anyone who answers false for the (N-1)th question answers false for the Nth question as well, resulting in at least 3 different answers.
This can be generalized for 4, 5, 6, ... questions by dividing the result by 2, 4, 8, ...
November 12th, 2025 at 5:23:08 PM
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Let’s say there are 300 students and I randomly select two tests and the answers are all “TRUE” on both.
Doesn’t that disprove the theory that there must be at least two different answers?
Doesn’t that disprove the theory that there must be at least two different answers?
It’s all about making that GTA
November 12th, 2025 at 6:31:16 PM
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Where there are 6 digits, I can see a way of having eight different test results (I'm using binary digits as it's easier for the spreadsheet to check).
It seems obvious that once you've picked the front 3 digits, then there can only be two entries using those 3. This is where you flip the last three digits. e.g. 101101 and 101010 could work, however the problem with 101010 is it's too similar to 111000. Also, given that 000000 has been used you can't have numbers with 1 or 2 ones.
What you notice is in the table below the last three digits are either all 0's or two 1's, and the front half is either something or its inverse. (I suspect trying to find front three digits for xxx111, xxx100, xxx010 or xxx001 won't find any other entries.)
If this is true then for larger numbers, e.g. 10-digit ones, once you've picked positions 5,6 and 7, then this sets 8, 9 and 10. Hence the maximum it could be is 27.
It seems obvious that once you've picked the front 3 digits, then there can only be two entries using those 3. This is where you flip the last three digits. e.g. 101101 and 101010 could work, however the problem with 101010 is it's too similar to 111000. Also, given that 000000 has been used you can't have numbers with 1 or 2 ones.
What you notice is in the table below the last three digits are either all 0's or two 1's, and the front half is either something or its inverse. (I suspect trying to find front three digits for xxx111, xxx100, xxx010 or xxx001 won't find any other entries.)
N1 | N2 | N3 | BC | BD | CD |
0 | 0 | 0 | 0 | 0 | 0 |
0 | 0 | 1 | 0 | 1 | 1 |
0 | 1 | 0 | 1 | 0 | 1 |
0 | 1 | 1 | 1 | 1 | 0 |
1 | 0 | 0 | 1 | 1 | 0 |
1 | 0 | 1 | 1 | 0 | 1 |
1 | 1 | 0 | 0 | 1 | 1 |
1 | 1 | 1 | 0 | 0 | 0 |
If this is true then for larger numbers, e.g. 10-digit ones, once you've picked positions 5,6 and 7, then this sets 8, 9 and 10. Hence the maximum it could be is 27.
