Thread Rating:

Poll

21 votes (45.65%)
14 votes (30.43%)
6 votes (13.04%)
3 votes (6.52%)
12 votes (26.08%)
3 votes (6.52%)
6 votes (13.04%)
5 votes (10.86%)
12 votes (26.08%)
10 votes (21.73%)

46 members have voted

Gialmere
Gialmere
  • Threads: 45
  • Posts: 3044
Joined: Nov 26, 2018
Thanked by
aceside
December 29th, 2021 at 4:29:06 PM permalink
Consider a deck with only 2 cards (2, 3). How 'bout a 3 card (2, 3, 4) deck?
Have you tried 22 tonight? I said 22.
ThatDonGuy
ThatDonGuy
  • Threads: 122
  • Posts: 6696
Joined: Jun 22, 2011
Thanked by
acesideGialmere
December 29th, 2021 at 4:34:17 PM permalink
I assume the strategy is, for each card, count the number of remaining cards higher than that card, and the number lower than that card, and making the decision for the next card based on that (as opposed to just going higher or lower based on a full deck).


2-10: 775 / 4,536 (1 / 5.8529)
2-11: 115,991 / 907,200 (1 / 7.8213)
2-12: 950,171 / 9,979,200 (1 / 10.5025)

Anything larger than that took too long to calculate with brute force, but I do have some simulation-based results:
2-20: 1 / 119
2-30: around 1 / 2750
Pretty much anything 50 or higher had too few "wins" to simulate with any degree of accuracy without running for hours



As for starting with 2 and going forward...

I couldn't find a way to generate the pattern 1, 5/6, 16/24, 62/120,...; it is listed in the Online Encyclopedia of Integer Sequences, but the only listed generating formula is recursive

Last edited by: ThatDonGuy on Dec 29, 2021
aceside
aceside
  • Threads: 2
  • Posts: 582
Joined: May 14, 2021
December 29th, 2021 at 5:29:07 PM permalink
Quote: ThatDonGuy

I assume the strategy is, for each card, count the number of remaining cards higher than that card, and the number lower than that card, and making the decision for the next card based on that (as opposed to just going higher or lower based on a full deck).


2-10: 775 / 4,536 (1 / 5.8529)
2-11: 115,991 / 907,200 (1 / 7.8213)
2-12: 950,171 / 9,979,200 (1 / 10.5025)

Anything larger than that took too long to calculate with brute force, but I do have some simulation-based results:
2-20: 1 / 119
2-30: around 1 / 2750
Pretty much anything 50 or higher had too few "wins" to simulate with any degree of accuracy without running for hours



As for starting with 2 and going forward...

I couldn't find a way to generate the pattern 1, 5/6, 16/24, 62/120,...; it is listed in the Online Encyclopedia of Integer Sequences, but the only listed generating formula is recursive


link to original post


Wonderful! Can you also compare this sequence and the 0.5^(n-delta) sequence to see if these two converge to each other?
ThatDonGuy
ThatDonGuy
  • Threads: 122
  • Posts: 6696
Joined: Jun 22, 2011
December 29th, 2021 at 5:44:17 PM permalink
Quote: aceside


Wonderful! Can you also compare this sequence and the 0.5^(n-delta) sequence to see if these two converge to each other?
link to original post


I only see the values through 23 cards, but it doesn't look like they do; the 0.5^(n-delta) sequence is reduced by 1/2 for each increase of 2 cards, whereas the actual values don't reduce quite that fast.
For example, for 23 cards, the actual value is about 1 / 410.769, whereas the 0.5-based value is 1 / 2048.
Gialmere
Gialmere
  • Threads: 45
  • Posts: 3044
Joined: Nov 26, 2018
December 29th, 2021 at 5:50:03 PM permalink
Quote: ThatDonGuy

I assume the strategy is, for each card, count the number of remaining cards higher than that card, and the number lower than that card, and making the decision for the next card based on that (as opposed to just going higher or lower based on a full deck).


2-10: 775 / 4,536 (1 / 5.8529)
2-11: 115,991 / 907,200 (1 / 7.8213)
2-12: 950,171 / 9,979,200 (1 / 10.5025)

Anything larger than that took too long to calculate with brute force, but I do have some simulation-based results:
2-20: 1 / 119
2-30: around 1 / 2750
Pretty much anything 50 or higher had too few "wins" to simulate with any degree of accuracy without running for hours



As for starting with 2 and going forward...

I couldn't find a way to generate the pattern 1, 5/6, 16/24, 62/120,...; it is listed in the Online Encyclopedia of Integer Sequences, but the only listed generating formula is recursive


link to original post


Correct!!


It’s about 17 percent.

The optimal strategy is fairly simple: You should say “bigger” if there are more bigger cards remaining and “smaller” if there are more smaller cards remaining. (If there is an equal number of bigger and smaller cards, you can just guess randomly — for simplicity, let’s say you always guess “bigger” in this case.)

Getting to the probability of winning is a little trickier. Because we are dealing with cards 2 through 10, we are dealing with nine cards. Therefore, there are 9! — or 362,880 — ways that these cards might be arranged. What we need to do next is figure out how many of those arrangements will lead us to a win. If we divide the number of winning arrangements by the total number of arrangements, we’ll get the probability that we win the game.

To ease into these calculations, let’s first imagine a smaller game — just two cards, numbered 2 and 3. In this case, you’re guaranteed to win because you’ll know for sure after seeing the first card whether the second and final card is bigger or smaller. Now, let’s move to a slightly larger game — three cards, numbered 2, 3 and 4. Things get a little more interesting here. There are six (3!) decks you might face. Five of these — every arrangement except 324 — leads to a win. In a game with four cards, there are 24 (4!) possible decks, 16 of which lead to victory.

Solver Keith Hudson developed a clever way to find and visualize the underlying pattern here, which we can extend to solve our nine-card game. Imagine numbers in a pyramid, where each row going down represents a game of a certain size: one card, two cards, etc. The numbers in each row of the pyramid add up to the number of ways you can win that game. From left to right, the individual numbers in each row are the number of ways you can win that game if you happen to start with the lowest card, then the second-lowest card, and so on to the highest card. For example, in a one-card game, there is only one card you can start with and one way to win. In a two-card game, there are two cards you can start with, and one way to win in each of those cases. In a three-card game, there are three cards you can start with, and two, one and two ways, respectively, to win in those scenarios. The first six rows look like this:



We can build this pyramid with some pretty simple rules. If, for example, we start by drawing either the smallest or biggest card in our deck, we are essentially then playing the game as if we had started with one fewer card, so the outer numbers (for example, 62 and 62 in the bottom row above) in the rows are the sums of the whole row that comes above them.

You’ll notice that you want to draw the biggest or smallest cards — the closer you get to the middle, the fewer ways you have to win the game. For example, say you’re playing with six cards (the last row in our pyramid above). If you draw the biggest or smallest card (the outer numbers of the row), you have 62 ways to win. But if you draw the third-biggest or third-smallest, you have only 35 ways to win.

You’ll also notice that the numbers have a relationship with each other. If a number is on the interior of the pyramid and on the left side, it is the sum of everything above it and to the right. Same goes for a number on the interior of the pyramid and to the right — it’s the sum of everything above it to the left.

Why is this? Because if we draw the biggest or smallest cards (the outer numbers), it’s like we haven’t added a new card at all — we have the same number of ways to win as we did in total in the smaller game (the row above). But as you move closer to the middle of a row, you’re adding more and more complications, so your options to win are fewer. Remarkably, though, the past games help us calculate the number of winning paths in the new ones.

Continuing in this way, the sum of the numbers in the ninth row of this pyramid is exactly 62,000. So our probability of winning is 62,000/362,880, or about 17 percent.


Puzzle by Freddie Simmons (at Riddler forum)

-------------------------------------------------------

Have you tried 22 tonight? I said 22.
charliepatrick
charliepatrick
  • Threads: 39
  • Posts: 3014
Joined: Jun 17, 2011
December 30th, 2021 at 2:01:25 AM permalink
I only have had a quick look at this and saw some recursion as a valid approach. So to get from N cards left to N-1 you have to get the next one correct.
Consider N cards in a row, if you pick the smallest one then there's (N-1) / (N-1) chances of winning (as all the others are larger). If you pick the next one then the chances are (N-2)/(N-1). Continue this and you get (N-1) (N-2) (N-3) etc and either get one or two occurrences in the middle.
With 2 cards left your chances are 1/1 1/1 (as you'll always be correct) which add up to 2/1 so average out to 2/2. (Note 2 = 2*1).
With 3 cards left your chances are 2/2 1/2 2/2 which add up to 5/2 so average out to 5/6. (Note 6 = 3*2, 5 is 3 more than 2)
With 4 cards 3/3 2/3 2/3 3/3 which add up to 10/3 so average out to 10/12 or 5/6. (Note 12 = 4*3, 10 is 5 more than 5)
With 5 cards 4/4 3/4 2/4 3/4 4/4: 16/4, 16/20 or 4/5. (Note 20 = 5*4, 16 is 6 more than 10)
With 6 cards 5/5 4/5 3/5 3/5 4/5 5/5, 24/5, 24/30 or 5/6. (Note 30 = 6*5, 24 is 8 more than 16)
With 7 cards 6/6 5/6 4/6 3/6 4/6 5/6 6/6, 33/6, 33/42 or 11/14. (note 42 = 7*6, 33 is 9 more than 24)
Thus you can see the pattern emerging.
Wizard
Administrator
Wizard
  • Threads: 1520
  • Posts: 27079
Joined: Oct 14, 2009
January 10th, 2022 at 4:19:41 PM permalink
We're overdue for our first puzzle of 2022. This one comes to us courtesy of The Riddler Express.

Let's assume NFL rules. Consider the following situation:

  • The Red team is down 14 points late in the game
  • The Red team will have two more possessions
  • The Blue team will have zero more possessions
  • Let's ignore field goals and safeties, as the Red team must score two touchdowns to have a chance at winning
  • If the game goes into overtime, each team will have a 50% chance to win. The game cannot end in a tie
  • The probability of making a one-point kick after a touchdown is 100%.
  • The probability of making a two-point conversion is p.


At what value of p should the red team be indifferent to kicking and going for a two-point conversion after a first touchdown (now down by 8)?



Fun fact: The probability of making a two-point conversion, based on the 2018 and 2019 seasons, was 49.4% . Source: Sporting News.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
ChesterDog
ChesterDog
  • Threads: 9
  • Posts: 1720
Joined: Jul 26, 2010
January 10th, 2022 at 5:03:32 PM permalink
Quote: Wizard

We're overdue for our first puzzle of 2022. This one comes to us courtesy of The Riddler Express.

Let's assume NFL rules. Consider the following situation:

  • The Red team is down 14 points late in the game
  • The Red team will have two more possessions
  • The Blue team will have zero more possessions
  • Let's ignore field goals and safeties, as the Red team must score two touchdowns to have a chance at winning
  • If the game goes into overtime, each team will have a 50% chance to win. The game cannot end in a tie
  • The probability of making a one-point kick after a touchdown is 100%.
  • The probability of making a two-point conversion is p.


At what value of p should the red team be indifferent to kicking and going for a two-point conversion after a first touchdown (now down by 8)?



Fun fact: The probability of making a two-point conversion, based on the 2018 and 2019 seasons, was 49.4% . Source: Sporting News.
link to original post




I get p = ( 3 - sqrt(5) ) / 2, which is about 38.2%.

Let x = probability of making a touchdown with a possession.

Going for a 2-point conversion, Red can win by either making the 2 points and then another touchdown, or missing the 2-point conversion, making another touchdown with 2-point conversion and then winning in overtime.

The probability of Red's winning with the above strategy is: p * x + (1 - p) * x * p * (1 / 2)

Forgoing 2-point conversion attempts, Red can win by making one more touchdown and then winning in overtime.

The probability of Red's winning with this strategy is: x * (1 / 2).

Equating the two probabilities and simplifying yields: p2 - 3p + 1 = 0

Solving for p yields: p = (3 - sqrt(5)) / 2.
ksdjdj
ksdjdj
  • Threads: 94
  • Posts: 1707
Joined: Oct 20, 2013
January 10th, 2022 at 5:15:24 PM permalink
~38.2% for p, when you go for it on the first touch down.
Ace2
Ace2
  • Threads: 32
  • Posts: 2706
Joined: Oct 2, 2017
January 10th, 2022 at 5:35:56 PM permalink
Agree with above. Incidentally, the answer of ~38.2% is 2 minus the golden ratio

It’s all about making that GTA
ThatDonGuy
ThatDonGuy
  • Threads: 122
  • Posts: 6696
Joined: Jun 22, 2011
January 10th, 2022 at 6:30:07 PM permalink

There are two answers: (3 - sqrt(5)) / 2, and 1

If p >= 1/2, the probability of winning by going for 1 the first time and 2 the second time is p; the probability of successfully going for 2 the first time, which is an automatic win since you can go for 1 the second time, is p, and even if it is missed, there is a p/2 chance of being successful the second time and winning in overtime, so the overall winning probability by going for 2 the first time > p -- unless p = 1, in which case, going for 2 the second time is always successful, so it doesn't matter what is done the first time.

If p < 1/2, the probability of winning by going for 1 the first time and 2 the second time = p, but the probability of winning by going for 1 both times = 1/2, so going for 1 the first time has a win probability of 1/2. The probability of winning by going for 2 the first time is p + (1 - p) p / 2 = p + p/2 - p^2/2 = p (3 - p) / 2. The two are equal when 3p - p^2 = 1; the only solution between 0 and 1/2 is 3/2 +/- sqrt(5)/2, or about 38.1966%.

Wizard
Administrator
Wizard
  • Threads: 1520
  • Posts: 27079
Joined: Oct 14, 2009
January 11th, 2022 at 6:54:04 AM permalink
Congratulations to ChesterDog, ksdjdj, Ace2, and ThatDonGuy -- I agree!
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
Gialmere
Gialmere
  • Threads: 45
  • Posts: 3044
Joined: Nov 26, 2018
Thanked by
charliepatrick
January 19th, 2022 at 8:17:44 AM permalink


Find a nine digit number, using the numbers 1 to 9, and using each number once without repeats, such that; the first digit is a number divisible by 1. The first two digits form a number divisible by 2; the first three digits form a number divisible by 3 and so on until we get a nine digit number divisible by 9.

You might try, for example, the number 923,156,784. But this number doesn't work — the first three digit number, 923, is not divisible by 3.

Can you find a nine digit number that works?
Have you tried 22 tonight? I said 22.
charliepatrick
charliepatrick
  • Threads: 39
  • Posts: 3014
Joined: Jun 17, 2011
Thanked by
Gialmere
January 19th, 2022 at 10:39:05 AM permalink
Here is a logical way to get to the answer.

(2,4,6,8) Since the first two digits form an even number it follows the second digit must be even. Similar logic applies for positions 4,6 and 8. Hence the final nine digit number is formed from odd,even,odd...odd,even,odd digits.
(3,6,9) The first three digits for a number divisible by three, so the sum of the first three digits is divisible by three. Similarly digits 4-6 and 7-9 have the same property.
(5) The first five digits are divisible by 5, so the middle digit has to be a "5"; this means the three digits must be 258 456 654 or 852 as their total is divisible by 3.

Looking at the four-digit number, the third digit will always be odd. Since the hundreds are divisible by 4, it is necessary for the tens and digits to be divisible by 4. So the number can be 12,32,72,92 or 16,36,76,96; thus the middle three have to be either 654 or 258.
Looking at the eight-digit number, the first six digits form an even number, so the hundreds part is divisible by 8. Thus the tens and units have to be divisible by 8, this means 16,96,32 or 72.

This means the number must be xxx65432y,xxx65472y,xxx25816x or xxx25896y.
Using the last three digits rule gives: xxx654321, xxx654327, xxx654723, xxx654729, xxx25816# (none), xxx258963
The remaining front three digits can be either way round, the even digit in position 2.

This gives ten numbers to check - the only factor left is whether the number is divisible by 7!
This gives only one solution 381 654 729.
ThatDonGuy
ThatDonGuy
  • Threads: 122
  • Posts: 6696
Joined: Jun 22, 2011
Thanked by
Gialmere
January 19th, 2022 at 12:19:59 PM permalink

The second, fourth, sixth, and eighth digits must be even; since these are the only even digits, the others must be odd
The fifth digit must be 5
The first digit is odd and not 5
The second digit is even
The third digit is 1, 3, 7, or 9, and the sum of the first three digits is a multiple of 3
Since the third digit is odd, the fourth must be 2 or 6
The fourth digit is even, and since the sum of the first three digits is a multiple of 3, the sum of the second three digits is also a multiple of 3
Since the fourth and fifth digits must be 25 or 65, the sixth is either 8 (with 25) or 4 (with 65)
Skip 7 for now...
Since three of the four even digits are in place, place the remaining one in position 8
Digits 6-8 are a multiple of 8, with the middle digit odd and the first digit 4 or 8
The possibilities are 416, 432, 456, 472, 496, 816, 832, 856, 872, and 896
This leaves:
147258963
183654729
189654327
189654723
381654729
741258963
789654321
981654327
981654723
987654321
Of these, only 381654729 has its first 7 digits as a multiple of 7

Gialmere
Gialmere
  • Threads: 45
  • Posts: 3044
Joined: Nov 26, 2018
January 19th, 2022 at 4:51:41 PM permalink
Quote: charliepatrick

Here is a logical way to get to the answer.

(2,4,6,8) Since the first two digits form an even number it follows the second digit must be even. Similar logic applies for positions 4,6 and 8. Hence the final nine digit number is formed from odd,even,odd...odd,even,odd digits.
(3,6,9) The first three digits for a number divisible by three, so the sum of the first three digits is divisible by three. Similarly digits 4-6 and 7-9 have the same property.
(5) The first five digits are divisible by 5, so the middle digit has to be a "5"; this means the three digits must be 258 456 654 or 852 as their total is divisible by 3.

Looking at the four-digit number, the third digit will always be odd. Since the hundreds are divisible by 4, it is necessary for the tens and digits to be divisible by 4. So the number can be 12,32,72,92 or 16,36,76,96; thus the middle three have to be either 654 or 258.
Looking at the eight-digit number, the first six digits form an even number, so the hundreds part is divisible by 8. Thus the tens and units have to be divisible by 8, this means 16,96,32 or 72.

This means the number must be xxx65432y,xxx65472y,xxx25816x or xxx25896y.
Using the last three digits rule gives: xxx654321, xxx654327, xxx654723, xxx654729, xxx25816# (none), xxx258963
The remaining front three digits can be either way round, the even digit in position 2.

This gives ten numbers to check - the only factor left is whether the number is divisible by 7!
This gives only one solution 381 654 729.

link to original post


Quote: ThatDonGuy


The second, fourth, sixth, and eighth digits must be even; since these are the only even digits, the others must be odd
The fifth digit must be 5
The first digit is odd and not 5
The second digit is even
The third digit is 1, 3, 7, or 9, and the sum of the first three digits is a multiple of 3
Since the third digit is odd, the fourth must be 2 or 6
The fourth digit is even, and since the sum of the first three digits is a multiple of 3, the sum of the second three digits is also a multiple of 3
Since the fourth and fifth digits must be 25 or 65, the sixth is either 8 (with 25) or 4 (with 65)
Skip 7 for now...
Since three of the four even digits are in place, place the remaining one in position 8
Digits 6-8 are a multiple of 8, with the middle digit odd and the first digit 4 or 8
The possibilities are 416, 432, 456, 472, 496, 816, 832, 856, 872, and 896
This leaves:
147258963
183654729
189654327
189654723
381654729
741258963
789654321
981654327
981654723
987654321
Of these, only 381654729 has its first 7 digits as a multiple of 7


link to original post


Correct!!

Well done.


-----------------------------------------------------

Have you tried 22 tonight? I said 22.
charliepatrick
charliepatrick
  • Threads: 39
  • Posts: 3014
Joined: Jun 17, 2011
January 22nd, 2022 at 3:20:30 AM permalink
^ Seeing whether a (long) number is divisible by 7.
I gave this puzzle to a friend who managed to get to the penultimate stage, so has the ten numbers and now wants to know which are divisible by 7. Personally I just put them into a calculator (as I have a 12-digit calculator) but have since noticed you could use a simple trick even if you had an 8-digiti calculator.

When checking for a number being divisible by 9 you add up all the digits of the number, repeat the process if you get more than 9, and hey presto.
I was wondering, since 999999 is divisible by 7, whether you could stack the digits in groups of sixes for long numbers and so do the same. For instance 1000006 is divisible by 7 as is 1+6, similarly 2000005.
Interestingly it would also work for 11 using 99, 37 and 27 using 999, 13 using 999999.
Dieter
Administrator
Dieter
  • Threads: 16
  • Posts: 6040
Joined: Jul 23, 2014
January 22nd, 2022 at 4:24:11 AM permalink
Quote: charliepatrick


When checking for a number being divisible by 9 you add up all the digits of the number, repeat the process if you get more than 9, and hey presto.
link to original post



The repeated digit summing seems to work to test divisibility of a number by base-1. If you're comfortable working in octal and converting, this may be useful.

Converting from decimal to octal may be nontrivial for you, and most people I know only think somewhere between base 2 (binary) and base 16 (hexadecimal), with a few excursions up to base 36 and base 62 (generally for non-math reasons).

I do not have a proof that this always works, but it seems to, at least for octal, decimal, and hexadecimal. It wouldn't make much sense to try binary (everything is divisible by 1).

I do not currently have a number theorist in residence to bounce your grouping idea off of, but it is intriguing.
May the cards fall in your favor.
ThatDonGuy
ThatDonGuy
  • Threads: 122
  • Posts: 6696
Joined: Jun 22, 2011
January 22nd, 2022 at 9:05:04 AM permalink
Quote: Dieter

Quote: charliepatrick


When checking for a number being divisible by 9 you add up all the digits of the number, repeat the process if you get more than 9, and hey presto.
link to original post



The repeated digit summing seems to work to test divisibility of a number by base-1. If you're comfortable working in octal and converting, this may be useful.

Converting from decimal to octal may be nontrivial for you, and most people I know only think somewhere between base 2 (binary) and base 16 (hexadecimal), with a few excursions up to base 36 and base 62 (generally for non-math reasons).

I do not have a proof that this always works, but it seems to, at least for octal, decimal, and hexadecimal. It wouldn't make much sense to try binary (everything is divisible by 1).

I do not currently have a number theorist in residence to bounce your grouping idea off of, but it is intriguing.
link to original post



I am pretty sure this is true for all bases >= 2. In fact, I am also pretty sure the sum of the digits of a base B number mod (B - 1) = the number mod (B - 1).
Start with 1; the number = 1 mod (B - 1), and the sum of its digits = 1 mod (B - 1).
When you add 1 to a number, if the last digit is not B - 1, you add 1 to the number and 1 to the sum of its digits; this adds 1 to both the number and the sum of its digits, so the number and the sum of its digits are still the same mod (B - 1).
If the last digit is B - 1, you subtract B - 1 from the last digit and all digits immediately to its left that are also B - 1, and add 1 to the rightmost digit that is not B - 1 (assume the leftmost digit is 0 if necessary). You subtract a multiple of B - 1, which does not affect the sum of the digits mod (B - 1), and add 1, which adds 1 to the sum, but you also added 1 to the number itself.
GM
GM
  • Threads: 0
  • Posts: 49
Joined: Jun 16, 2021
Thanked by
charliepatrick
January 22nd, 2022 at 1:52:40 PM permalink
Quote: charliepatrick

^ Seeing whether a (long) number is divisible by 7.

I gave this puzzle to a friend who managed to get to the penultimate stage, so has the ten numbers and now wants to know which are divisible by 7. Personally I just put them into a calculator (as I have a 12-digit calculator) but have since noticed you could use a simple trick even if you had an 8-digiti calculator.

When checking for a number being divisible by 9 you add up all the digits of the number, repeat the process if you get more than 9, and hey presto.
I was wondering, since 999999 is divisible by 7, whether you could stack the digits in groups of sixes for long numbers and so do the same. For instance 1000006 is divisible by 7 as is 1+6, similarly 2000005.
Interestingly it would also work for 11 using 99, 37 and 27 using 999, 13 using 999999.
link to original post


Yes, this idea works. If the number consisting of d 9s is divisible by k, then you can test for divisibility by k by splitting the decimal representation of your number into groups of d digits starting from the end, adding up these d digit numbers and repeating the process if the sum has more than d digits. The number you obtain is divisible by k if and only if the original number is divisible by k.

Fermat's Little Theorem implies that 10p-1-1, i. e., the number consisting of (p-1) 9s is divisible by p if p is a prime other than 2 or 5, so you can use groups of length p-1 to test for divisibility by p. For a lot of primes some divisor of p-1 will also work, but there are also primes like 7, 17, 19, 23 and 29, for which you have to use groups of length p-1.
Ace2
Ace2
  • Threads: 32
  • Posts: 2706
Joined: Oct 2, 2017
January 25th, 2022 at 9:50:11 AM permalink
Henry and Tom decide to bet on a coin flip. Henry wins on heads, Tom wins on tails.

It’s $1 per flip and they are really bored, so they decide to do one million flips. At the end of the session, the loser will write a check to the winner for the final balance.

What is the expected value and standard deviation of the check amount ?
It’s all about making that GTA
ThatDonGuy
ThatDonGuy
  • Threads: 122
  • Posts: 6696
Joined: Jun 22, 2011
January 25th, 2022 at 10:51:19 AM permalink
Quote: Ace2

Henry and Tom decide to bet on a coin flip. Henry wins on heads, Tom wins on tails.

It’s $1 per flip and they are really bored, so they decide to do one million flips. At the end of the session, the loser will write a check to the winner for the final balance.

What is the expected value and standard deviation of the check amount ?
link to original post




The expected value is zero
Variance =
2 x C(1,000,000, 0) x 1 / 2^1,000,000 x 1,000,000^2
+ 2 x C(1,000,000, 1) x 1 / 2^1,000,000 x 999,998^2
+ 2 x C(1,000,000, 2) x 1 / 2^1,000,000 x 999,996^2
+ ...
+ 2 x C(1,000,000, 499,999) x 1 / 2^1,000,000 x 2^2
= 8 / 2^1,000,000 x (
C(1,000,000, 0) x 500,000^2
+ C(1,000,000, 1) x 499,999^2
+ ...
+ C(1,000,000, 499,999) x 1^2
)
= 1 / 2^999,997 x (
C(1,000,000, 0) x 500,000^2
+ C(1,000,000, 1) x 499,999^2
+ ...
+ C(1,000,000, 499,999) x 1^2
)

My computer claims the result is exactly 1,000,000; I'll have to see if I can compute that manually
Anyway, the SD = sqrt(variance) = 1000

Expected Value = 0; SD = $1000
Ace2
Ace2
  • Threads: 32
  • Posts: 2706
Joined: Oct 2, 2017
January 25th, 2022 at 11:18:47 AM permalink
Quote: ThatDonGuy

Quote: Ace2

Henry and Tom decide to bet on a coin flip. Henry wins on heads, Tom wins on tails.

It’s $1 per flip and they are really bored, so they decide to do one million flips. At the end of the session, the loser will write a check to the winner for the final balance.

What is the expected value and standard deviation of the check amount ?
link to original post




The expected value is zero
Variance =
2 x C(1,000,000, 0) x 1 / 2^1,000,000 x 1,000,000^2
+ 2 x C(1,000,000, 1) x 1 / 2^1,000,000 x 999,998^2
+ 2 x C(1,000,000, 2) x 1 / 2^1,000,000 x 999,996^2
+ ...
+ 2 x C(1,000,000, 499,999) x 1 / 2^1,000,000 x 2^2
= 8 / 2^1,000,000 x (
C(1,000,000, 0) x 500,000^2
+ C(1,000,000, 1) x 499,999^2
+ ...
+ C(1,000,000, 499,999) x 1^2
)
= 1 / 2^999,997 x (
C(1,000,000, 0) x 500,000^2
+ C(1,000,000, 1) x 499,999^2
+ ...
+ C(1,000,000, 499,999) x 1^2
)

My computer claims the result is exactly 1,000,000; I'll have to see if I can compute that manually
Anyway, the SD = sqrt(variance) = 1000

Expected Value = 0; SD = $1000

link to original post

I disagree.

If, for instance, they flipped three times, then there is a 2/8 chance of a $3 check and a 6/8 chance of a $1 check going either way, for an expected check amount of $1.50.

Yes it’s true than both players have an expected gain/loss of zero (1/8 * 3 + 3/8 * 1 - 3/8 * 1 + 1/8 * 3), but that’s not what’s being asked. Looking for the expected settlement amount.

Also disagree on standard deviation
It’s all about making that GTA
ThatDonGuy
ThatDonGuy
  • Threads: 122
  • Posts: 6696
Joined: Jun 22, 2011
January 25th, 2022 at 11:26:56 AM permalink
Quote: Ace2

Quote: ThatDonGuy

Quote: Ace2

Henry and Tom decide to bet on a coin flip. Henry wins on heads, Tom wins on tails.

It’s $1 per flip and they are really bored, so they decide to do one million flips. At the end of the session, the loser will write a check to the winner for the final balance.

What is the expected value and standard deviation of the check amount ?
link to original post




The expected value is zero
Variance =
2 x C(1,000,000, 0) x 1 / 2^1,000,000 x 1,000,000^2
+ 2 x C(1,000,000, 1) x 1 / 2^1,000,000 x 999,998^2
+ 2 x C(1,000,000, 2) x 1 / 2^1,000,000 x 999,996^2
+ ...
+ 2 x C(1,000,000, 499,999) x 1 / 2^1,000,000 x 2^2
= 8 / 2^1,000,000 x (
C(1,000,000, 0) x 500,000^2
+ C(1,000,000, 1) x 499,999^2
+ ...
+ C(1,000,000, 499,999) x 1^2
)
= 1 / 2^999,997 x (
C(1,000,000, 0) x 500,000^2
+ C(1,000,000, 1) x 499,999^2
+ ...
+ C(1,000,000, 499,999) x 1^2
)

My computer claims the result is exactly 1,000,000; I'll have to see if I can compute that manually
Anyway, the SD = sqrt(variance) = 1000

Expected Value = 0; SD = $1000

link to original post

I disagree.

If, for instance, they flipped three times, then there is a 2/8 chance of a $3 check and a 6/8 chance of a $1 check going either way, for an expected check amount of $1.50.

Yes it’s true than both players have an expected gain/loss of zero (1/8 * 3 + 3/8 * 1 - 3/8 * 1 + 1/8 * 3), but that’s not what’s being asked. Looking for the expected settlement amount.

Also disagree on standard deviation
link to original post



Understood. You want the mean and the SD of the value of the check, regardless of which player pays it to the other.


For the mean, I get about 797.884361. I do have an exact answer, but it is a fraction with a numerator having 301,029 digits.

Because of this, calculating the SD may take a while...

Last edited by: ThatDonGuy on Jan 25, 2022
Wizard
Administrator
Wizard
  • Threads: 1520
  • Posts: 27079
Joined: Oct 14, 2009
January 25th, 2022 at 12:59:40 PM permalink
797.88



I ask a related question in Ask the Wizard #358.

Based on the method in that solution, for n flips my estimate is 2^2*sqrt(n/4) * sqrt(2/pi).

The reason for multiplying by 2^2 is with every flip there is a change in money of $2. We square that $2 to get the variance.


Note: Answer updated 7:15 PST 1/25/22, because I didn't correctly account for $2 difference between winning and losing for each flip.
Last edited by: Wizard on Jan 25, 2022
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
Ace2
Ace2
  • Threads: 32
  • Posts: 2706
Joined: Oct 2, 2017
January 25th, 2022 at 2:39:56 PM permalink
Quote: ThatDonGuy



For the mean, I get about 797.884361. I do have an exact answer, but it is a fraction with a numerator having 301,029 digits.

Because of this, calculating the SD may take a while...


link to original post

I agree with that answer ($797.88) for the mean, but I’d like to see a formulaic solution for full credit. The solution can be an estimate, albeit a very accurate one (to at least five digits).

Also looking for a formulaic solution for the standard deviation of the check amount, which can also be an accurate estimate


It’s all about making that GTA
ThatDonGuy
ThatDonGuy
  • Threads: 122
  • Posts: 6696
Joined: Jun 22, 2011
January 25th, 2022 at 2:55:17 PM permalink
Quote: Ace2

Quote: ThatDonGuy



For the mean, I get about 797.884361. I do have an exact answer, but it is a fraction with a numerator having 301,029 digits.

Because of this, calculating the SD may take a while...


link to original post

I agree with that answer ($797.88) for the mean, but I’d like to see a formulaic solution for full credit. The solution can be an estimate, albeit a very accurate one (to at least five digits).

Also looking for a formulaic solution for the standard deviation of the check amount, which can also be an accurate estimate



link to original post


I assume by "formulatic solution," you mean something other than the sum of:
C(1,000,000, 0) x (1/2)^1,000,000 x (1/2)^0 x 1,000,000
C(1,000,000, 1) x (1/2)^999,999 x (1/2)^1 x 999,998
C(1,000,000, 2) x (1/2)^999,998 x (1/2)^2 x 999,996
...
C(1,000,000, 499,999) x (1/2)^500,001 x (1/2)^499,999 x 2
C(1,000,000, 500,000) x (1/2)^500,000 x (1/2)^500,000 x 0
C(1,000,000, 500,001) x (1/2)^499,999 x (1/2)^500,001 x 2
...
C(1,000,000, 999,999) x (1/2)^1 x (1/2)^999,999 x 999,998
C(1,000,000, 1,000,000) x (1/2)^0 x (1/2)^1,000,000 x 1,000,000
Ace2
Ace2
  • Threads: 32
  • Posts: 2706
Joined: Oct 2, 2017
January 25th, 2022 at 3:27:20 PM permalink
Quote: ThatDonGuy

Quote: Ace2

Quote: ThatDonGuy



For the mean, I get about 797.884361. I do have an exact answer, but it is a fraction with a numerator having 301,029 digits.

Because of this, calculating the SD may take a while...


link to original post

I agree with that answer ($797.88) for the mean, but I’d like to see a formulaic solution for full credit. The solution can be an estimate, albeit a very accurate one (to at least five digits).

Also looking for a formulaic solution for the standard deviation of the check amount, which can also be an accurate estimate



link to original post


I assume by "formulatic solution," you mean something other than the sum of:
C(1,000,000, 0) x (1/2)^1,000,000 x (1/2)^0 x 1,000,000
C(1,000,000, 1) x (1/2)^999,999 x (1/2)^1 x 999,998
C(1,000,000, 2) x (1/2)^999,998 x (1/2)^2 x 999,996
...
C(1,000,000, 499,999) x (1/2)^500,001 x (1/2)^499,999 x 2
C(1,000,000, 500,000) x (1/2)^500,000 x (1/2)^500,000 x 0
C(1,000,000, 500,001) x (1/2)^499,999 x (1/2)^500,001 x 2
...
C(1,000,000, 999,999) x (1/2)^1 x (1/2)^999,999 x 999,998
C(1,000,000, 1,000,000) x (1/2)^0 x (1/2)^1,000,000 x 1,000,000
link to original post

Yes. I guess I forgot that some people have software to “brute force” things of this magnitude. When I ran some tests in excel, it would not let me calculate combinations too far above 1000.
It’s all about making that GTA
ThatDonGuy
ThatDonGuy
  • Threads: 122
  • Posts: 6696
Joined: Jun 22, 2011
January 26th, 2022 at 6:38:19 AM permalink

Note that the mean is very close to sqrt(1,000,000) / sqrt(2 PI)
Change the problem so that the mean is 1 / sqrt(2 PI); the SD for 1,000,000 samples is sqrt(M (1 - M) / 1,000,000) = sqrt(M (1 - M)) / 1000
For M = 1 / sqrt(2 PI), SD = sqrt((sqrt(2 PI) - 1) / sqrt(2 PI)) / 1000 = 0.77528 / 1000
If you multiply this by 1,000,000, which is how many tosses were made, you get SD = 775.28

tuttigym
tuttigym
  • Threads: 10
  • Posts: 2052
Joined: Feb 12, 2010
January 26th, 2022 at 8:21:27 AM permalink
Quote: Ace2

Henry and Tom decide to bet on a coin flip. Henry wins on heads, Tom wins on tails.

It’s $1 per flip and they are really bored, so they decide to do one million flips. At the end of the session, the loser will write a check to the winner for the final balance.

What is the expected value and standard deviation of the check amount ?
link to original post


Expected value= death by sleep deprivation and kidney failure.

Standard deviation= arthritis and thumb amputation from flipping
GM
GM
  • Threads: 0
  • Posts: 49
Joined: Jun 16, 2021
January 26th, 2022 at 8:23:37 AM permalink
Quote: Ace2

Henry and Tom decide to bet on a coin flip. Henry wins on heads, Tom wins on tails.

It’s $1 per flip and they are really bored, so they decide to do one million flips. At the end of the session, the loser will write a check to the winner for the final balance.

What is the expected value and standard deviation of the check amount ?
link to original post


If we use the normal approximation to the binomial distribution, we get that the expected value of the check is about sqrt(2n/π)≈0.7978845*sqrt(n) and the standard deviation is about sqrt((π-2)n/π)≈0.6028103*sqrt(n). For n=1000000, this gives expected value $797.8845 and standard deviation $602.8103. The values I calculated without approximation are $797.8843 and $602.8105.

A related question (it is easy if you look at it at the right way): show that the expected value of the check is the same for n flips and for n-1 flips, if n is even.
Ace2
Ace2
  • Threads: 32
  • Posts: 2706
Joined: Oct 2, 2017
January 26th, 2022 at 8:44:57 AM permalink
Thanks GM, that is the correct answer for the check amount. Still reviewing your SD answer, which is very close to mine

The average check amount is (1,000,000 * 2 / π)^.5 = $797.88

I described this method back in April (link below). At that time the Wizard’s reply was that I was “overthinking this”, but since then he published a pdf and Ask the Wizard problem using this method

https://wizardofvegas.com/forum/questions-and-answers/gambling/35821-sports-bet-size/#post802206

The average settlement amount is a very useful number. For instance, let’s say you’re going to Vegas for the weekend and expect to play 600 blackjack hands at $100 per hand. Your expectation is to lose $300 +/- 2,800, but that -$300 is basically meaningless…it’s a very long term average of results. But take $2,800 * (2/π)^.5 to get an average settlement amount of about $2,200 and that is much more useful information…it tells you that your result for the weekend will probably resemble something like -$2,500 or +$1,900. As mentioned, I originally posted about the average settlement amount earlier to get a practical figure like this
It’s all about making that GTA
ThatDonGuy
ThatDonGuy
  • Threads: 122
  • Posts: 6696
Joined: Jun 22, 2011
January 26th, 2022 at 10:51:18 AM permalink
Quote: GM

Quote: Ace2

Henry and Tom decide to bet on a coin flip. Henry wins on heads, Tom wins on tails.

It’s $1 per flip and they are really bored, so they decide to do one million flips. At the end of the session, the loser will write a check to the winner for the final balance.

What is the expected value and standard deviation of the check amount ?
link to original post


If we use the normal approximation to the binomial distribution, we get that the expected value of the check is about sqrt(2n/π)≈0.7978845*sqrt(n) and the standard deviation is about sqrt((π-2)n/π)≈0.6028103*sqrt(n). For n=1000000, this gives expected value $797.8845 and standard deviation $602.8103. The values I calculated without approximation are $797.8843 and $602.8105.

A related question (it is easy if you look at it at the right way): show that the expected value of the check is the same for n flips and for n-1 flips, if n is even.
link to original post


Shows you how much I know, at least when it comes to normal distributions.

Is there an online source that shows how you came up with those mean and SD values?
Ace2
Ace2
  • Threads: 32
  • Posts: 2706
Joined: Oct 2, 2017
January 27th, 2022 at 4:32:26 PM permalink
Quote: ThatDonGuy

Quote: GM

Quote: Ace2

Henry and Tom decide to bet on a coin flip. Henry wins on heads, Tom wins on tails.

It’s $1 per flip and they are really bored, so they decide to do one million flips. At the end of the session, the loser will write a check to the winner for the final balance.

What is the expected value and standard deviation of the check amount ?
link to original post


If we use the normal approximation to the binomial distribution, we get that the expected value of the check is about sqrt(2n/π)≈0.7978845*sqrt(n) and the standard deviation is about sqrt((π-2)n/π)≈0.6028103*sqrt(n). For n=1000000, this gives expected value $797.8845 and standard deviation $602.8103. The values I calculated without approximation are $797.8843 and $602.8105.

A related question (it is easy if you look at it at the right way): show that the expected value of the check is the same for n flips and for n-1 flips, if n is even.
link to original post


Shows you how much I know, at least when it comes to normal distributions.

Is there an online source that shows how you came up with those mean and SD values?
link to original post

The normal distribution has a height of 1/(2π)^.5 at x=0, so that’s it’s mean value. For the coin flip problem, we are only looking at one side of the distribution (since we don’t care if it’s + or -) but we are now saying that one side is the entire distribution. It still has to total 1, so the mean becomes 2 * 1/(2π)^.5 = (2/π)^.5

(2π)^.5 also appears in Stirling’s formula for factorials :
(2πn)^.5 * (n/e)^n. Probably appears in many places
Last edited by: Ace2 on Jan 27, 2022
It’s all about making that GTA
ThatDonGuy
ThatDonGuy
  • Threads: 122
  • Posts: 6696
Joined: Jun 22, 2011
January 27th, 2022 at 5:06:56 PM permalink
Quote: Ace2

Quote: ThatDonGuy


Shows you how much I know, at least when it comes to normal distributions.

Is there an online source that shows how you came up with those mean and SD values?


The normal distribution has a height of 1/(2π)^.5 at x=0, so that’s it’s mean value.


Er...huh? Isn't the mean of a normal distribution zero?

I can get sqrt(2/π) as the mean by dividing the integral over 0 to positive infinity of x f(x) dx by the integral over 0 to positive infinity of f(x) dx, where f(x) = 1 / sqrt(2π) * e^(-x%2/2), but I am looking for an easier way to get that number.

Oh, and while I'm in pedantic mode: "it's mean value" means "it is mean value." The word for "something belonging to it" - "its" - does not have an apostrophe.
Ace2
Ace2
  • Threads: 32
  • Posts: 2706
Joined: Oct 2, 2017
January 28th, 2022 at 6:13:39 PM permalink
Quote: ThatDonGuy


Oh, and while I'm in pedantic mode: "it's mean value" means "it is mean value." The word for "something belonging to it" - "its" - does not have an apostrophe.
link to original post

Are you kidding me?

Though I’ll admit that language skills are not my strong point, I do know the difference between its and it’s. My iPhone’s “autocorrect” probably did that.

I’m quite surprised you nitpicked that so allow me to make a constructive criticism.

Before I posted this basic coin flip problem, I triple-checked the wording, ensuring there was absolutely no ambiguity to what was being asked. The problem ends with:

“What is the expected value and standard deviation of the check amount ?”

I thought there was no possible way someone could misunderstand that question. But, I was wrong…you did not comprehend it and you answered that the expected value is zero (which is the answer to a different question). You didn’t actually “get it” until after I made a second post with a detailed example.

So your reading comprehension skills have lots of room for improvement
Last edited by: Ace2 on Jan 28, 2022
It’s all about making that GTA
aceside
aceside
  • Threads: 2
  • Posts: 582
Joined: May 14, 2021
January 29th, 2022 at 2:52:53 PM permalink
Quote: Ace2

Henry and Tom decide to bet on a coin flip. Henry wins on heads, Tom wins on tails.

It’s $1 per flip and they are really bored, so they decide to do one million flips. At the end of the session, the loser will write a check to the winner for the final balance.

What is the expected value and standard deviation of the check amount ?
link to original post


It seems this is exactly the same problem as the one-dimensional random walk problem on Wikipedia. To find the exact solution, you need the Pascal's triangle to solve it, but usually the number N is large enough to guarantee the use of its limit. Is this right?
Standard deviation hasn't been considered very much. what is the S.D. limit when N approaches infinity? It seems a previous post says it is sqrt((π-2)n/π).
Wizard
Administrator
Wizard
  • Threads: 1520
  • Posts: 27079
Joined: Oct 14, 2009
January 29th, 2022 at 9:10:31 PM permalink
Quote: Ace2

So your reading comprehension skills have lots of room for improvement
link to original post



I have tremendous respect for both of you, especially your math skills. There was obviously some miscommunication on this problem.

I'd like to suggest we pass the peace pipe.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
TwelveOr21
TwelveOr21
  • Threads: 3
  • Posts: 72
Joined: Nov 18, 2018
January 29th, 2022 at 9:17:07 PM permalink
Quote: Wizard


I'd like to suggest we pass the peace pipe.
link to original post



Is that a Futurama reference there ? :)

https://youtu.be/SCFkr7Wj9TQ?t=66
Ace2
Ace2
  • Threads: 32
  • Posts: 2706
Joined: Oct 2, 2017
February 1st, 2022 at 2:28:41 PM permalink
Quote: Ace2

Henry and Tom decide to bet on a coin flip. Henry wins on heads, Tom wins on tails.

It’s $1 per flip and they are really bored, so they decide to do one million flips. At the end of the session, the loser will write a check to the winner for the final balance.

What is the expected value and standard deviation of the check amount ?


In case anyone is interested, here is how the EV and SD are found via integration

The normal distribution is represented by the function (2π)^(-.5) * e^(-.5x^2).  To find the mean value of x on one side of the distribution, we must take the product of all probabilities (of one side) and their corresponding x values and then sum them. This is given by the integral from zero to infinity of: (2π)^(-.5) * e^(-.5x^2) * x  dx

There is an antiderivative for this integral as follows:

Let u = (-.5x^2) and substitute into above integral
du/dx = -x
-du = x dx

Now we can rewrite (2π)^(-.5) * e^u * x  dx as: 
-(2π)^(-.5) * e^u du

For which the antiderivative is simply: -(2π)^(-.5) * e^u

Undoing the substitution gives us the antiderivative of: 
-(2π)^(-.5) * e^(-.5x^2)

We are evaluating this from zero to infinity, which is:
0 - (-(2π)^(-.5)) = (2π)^(-.5)

Since this is for half of the distribution, multiply by 2 to get the mean value of: (2/π)^(.5) =~ 0.79788. Multiply by 1,000,000^.5 to get the expected check value of $797.88

Now that we know the mean value of x (on one side) is (2/π)^(.5), we can calculate the variance by taking the integral from zero to infinity of:

 (2π)^(-.5) * e^(-.5x^2) * (x - (2/π)^(.5))^2 dx

I don't believe there's a way to get an exact antiderivative for this function, so I used an integral calculator to get the answer of: (π - 2) / (2π)

That is the variance for half the distribution, so multiply by 2 to get the full variance of: (π - 2) / π

Therefore the standard deviation is ((π - 2) / π)^.5 =~ 0.60281. Multiply by 1,000,000^.5 to get the check’s SD of $602.81
Last edited by: Ace2 on Feb 1, 2022
It’s all about making that GTA
GM
GM
  • Threads: 0
  • Posts: 49
Joined: Jun 16, 2021
February 1st, 2022 at 4:38:23 PM permalink
Quote: Ace2


Now that we know the mean value of x (on one side) is (2/π)^(.5), we can calculate the variance by taking the integral from zero to infinity of:

 (2π)^(-.5) * e^(-.5x^2) * (x - (2/π)^(.5))^2 dx

I don't believe there's a way to get an exact antiderivative for this function, so I used an integral calculator to get the answer of: (π - 2) / (2π)


We can expand (x - (2/π)^(.5))^2, (x - (2/π)^(.5))^2=x^2-2*(2/π)^(.5)*x+2/π and we can calculate the integral of each term multiplied by (2π)^(-.5) * e^(-.5x^2) separately.

The integral of (2π)^(-.5)*e^(-.5x^2)*x^2 dx from x=0 to x=∞ is 1/2 of the integral of the same function from x=-∞ to x=∞, which is the integral expressing the variance of the normal distribution, so the integral from x=-∞ to x=∞ is 1 and the integral from x=0 to x=∞ is 1/2.

We have already calculated that the integral of (2π)^(-.5)*e^(-.5x^2)*x dx from x=0 to x=∞ is 1/(2π)^(.5), so the integral of (2π)^(-.5)*e^(-.5x^2)*2*(2/π)^(.5)*x dx from x=0 to x=∞ is 1/(2π)^(.5)*2*(2/π)^(.5)=2/π.

The integral of (2π)^(-.5)*e^(-.5x^2) dx from x=0 to x=∞ is 1/2, it is just the probability that a standard normal random variable is positive.

Putting everything together, the value of the integral is 1/2-2/π+1/2=1-2/π=(π-2)/π.
Ace2
Ace2
  • Threads: 32
  • Posts: 2706
Joined: Oct 2, 2017
February 1st, 2022 at 6:24:59 PM permalink
Quote: GM


The integral of (2π)^(-.5)*e^(-.5x^2)*x^2 dx from x=0 to x=∞ is 1/2 of the integral of the same function from x=-∞ to x=∞, which is the integral expressing the variance of the normal distribution, so the integral from x=-∞ to x=∞ is 1 and the integral from x=0 to x=∞ is 1/2.

Thanks again, GM. It took me a minute to grasp that step, but once I did, getting the antiderivative of the function was easy. I suspected there was a simple way to get there since (pi - 2) / pi is such a “clean” answer…using only 2 constants from the original function

My calculus skills are quite basic, especially when it comes to rearranging/simplifying functions. I rarely use calculus so I don’t get much practice
Last edited by: Ace2 on Feb 1, 2022
It’s all about making that GTA
Gialmere
Gialmere
  • Threads: 45
  • Posts: 3044
Joined: Nov 26, 2018
Thanked by
charliepatrick
February 2nd, 2022 at 8:17:15 AM permalink
This puzzle is rather wordy so read carefully...



The number N represents the first 6 digits of a special number.

N consists of three prime numbers put side by side. These three prime numbers come xth, yth and zth on the list of primes, where x, y and z are themselves three consecutive primes (for example, x, y and z could be 3, 5 and 7, in which case we'd be looking at the third, fifth and seventh prime numbers).

In addition, if N is split in the middle into two separate numbers, the prime factors of the left part of N add up to its right part.

What is N? And what is the special number whose first 6 digits it forms?
Have you tried 22 tonight? I said 22.
charliepatrick
charliepatrick
  • Threads: 39
  • Posts: 3014
Joined: Jun 17, 2011
Thanked by
Gialmere
February 2nd, 2022 at 8:54:15 AM permalink
Quote: Gialmere

This puzzle...What is N?...link to original post

First one lists the first few primes being 2,3,5,7,11,13,17,19,23 and then list the 2nd, 3rd, 5th etc primes. The latter gives 3, 5, 11, 17, 31, 41, 59, 67, 83, 109. Then the six-digits numbers can be 11-17-31 thru 59-67-83. What is interesting is 314159 has a lower second half (159) than first half (314) and 314 = 157*2. Thus N is nearly Pi*100000! (I'm guessing this is what the last part refers to.)
Gialmere
Gialmere
  • Threads: 45
  • Posts: 3044
Joined: Nov 26, 2018
February 2nd, 2022 at 8:10:51 PM permalink
Quote: charliepatrick

First one lists the first few primes being 2,3,5,7,11,13,17,19,23 and then list the 2nd, 3rd, 5th etc primes. The latter gives 3, 5, 11, 17, 31, 41, 59, 67, 83, 109. Then the six-digits numbers can be 11-17-31 thru 59-67-83. What is interesting is 314159 has a lower second half (159) than first half (314) and 314 = 157*2. Thus N is nearly Pi*100000! (I'm guessing this is what the last part refers to.)

link to original post


Correct!!

The numbers 31, 41, and 59 are the 11th, 13th, and 17th primes where 11, 13 and 17 form three consecutive primes. So the number we get by putting them together, 314159, fulfills the first part of our requirement for the number N.

Now split 314159 into the two numbers 314 and 159. The prime factors of 314 are 2 and 157, which add up to 159. So 314159 also fulfills the second requirement — hence N=314159 is the number we are looking for. It also represents the first six digits of π.
----------------------------------------------

This puzzle is by Aziz Inan, Professor of Electrical Engineering at the University of Portland. His own name, when written in capital letters, is a geometric word puzzle.

----------------------------------------------

Thank you for your undivided attention.
Have you tried 22 tonight? I said 22.
Ace2
Ace2
  • Threads: 32
  • Posts: 2706
Joined: Oct 2, 2017
Thanked by
Gialmere
February 2nd, 2022 at 9:45:13 PM permalink
Quote: Gialmere



Thank you for your undivided attention.

link to original post

Your jokes are the best
It’s all about making that GTA
Wizard
Administrator
Wizard
  • Threads: 1520
  • Posts: 27079
Joined: Oct 14, 2009
February 3rd, 2022 at 10:49:38 AM permalink
Two dice are rolled until either event happens:

A) A total of 2 is rolled.
B) A total of 7 is rolled twice consecutively.

What is the probability the total of 2 occurs first?
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
ThatDonGuy
ThatDonGuy
  • Threads: 122
  • Posts: 6696
Joined: Jun 22, 2011
February 3rd, 2022 at 12:13:17 PM permalink

From the starting state, one of four things will happen:
Only three things need to be considered:
(a) A 2;
(b) A 7 followed by a 2;
(c) A 7 followed by a 7;
(d) anything else, which returns to the starting state, so only (a), (b), and (c) need to be considered.

(a) has probability 1/36
(b) has probability 1/216
(c) has probability 1/36
P(a + b) / P(a + b + c) = 7/13

Wizard
Administrator
Wizard
  • Threads: 1520
  • Posts: 27079
Joined: Oct 14, 2009
February 3rd, 2022 at 2:34:15 PM permalink
Quote: ThatDonGuy


From the starting state, one of four things will happen:
Only three things need to be considered:
(a) A 2;
(b) A 7 followed by a 2;
(c) A 7 followed by a 7;
(d) anything else, which returns to the starting state, so only (a), (b), and (c) need to be considered.

(a) has probability 1/36
(b) has probability 1/216
(c) has probability 1/36
P(a + b) / P(a + b + c) = 7/13


link to original post



I agree. I solved it with a simple Markov chain, but your way looks easier to explain.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
Ace2
Ace2
  • Threads: 32
  • Posts: 2706
Joined: Oct 2, 2017
Thanked by
rsactuary
February 3rd, 2022 at 3:26:46 PM permalink
Avg rolls to get two consecutive 7s is: 6^1 + 6^2 = 42

Avg rolls to get a 2 is 36.

42 / (42 + 36) = 7/13
It’s all about making that GTA
  • Jump to: