Easy math puzzles

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).

Solution for 2-10 through 2-12

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...

General solution

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

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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?
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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).

Solution for 2-10 through 2-12

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...

General solution

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

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Official Solve

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)

Joke

• 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.

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.
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Quote: charliepatrick

an approach to the answer

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.

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Quote: ThatDonGuy

My Answer

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

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Old Number Joke Punchline...

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.
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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.
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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.
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Quote: charliepatrick

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

Show Spoiler

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.
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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 ?
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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 ?
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My Answer

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

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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 ?
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My Answer

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

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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
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Quote: ThatDonGuy

Answer for Mean

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...

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Quote: Ace2

Quote: ThatDonGuy

Answer for Mean

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...

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Show Spoiler

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

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Quote: ThatDonGuy

Quote: Ace2

Quote: ThatDonGuy

Answer for Mean

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...

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Show Spoiler

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

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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
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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 ?
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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 ?
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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 ?
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Show Spoiler

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.
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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 ?
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Show Spoiler

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.
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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?
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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.

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.
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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 ?
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Quote: Ace2

So your reading comprehension skills have lots of room for improvement
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Quote: Wizard

I'd like to suggest we pass the peace pipe.
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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 ?

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)

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.

Quote: Gialmere

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

Quote: charliepatrick

Show Spoiler

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.)

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Quote: Gialmere

Just take a minute and think about prime numbers...

Thank you for your undivided attention.

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Quote: ThatDonGuy

My Answer

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

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