## Poll

 I love math! 21 votes (45.65%) Math is great. 14 votes (30.43%) My religion is mathology. 6 votes (13.04%) Women didn't speak to me until I was 30. 3 votes (6.52%) Total eclipse reminder -- 04/08/2024 12 votes (26.08%) I steal cutlery from restaurants. 3 votes (6.52%) I should just say what's on my mind. 6 votes (13.04%) Who makes up these awful names for pandas? 5 votes (10.86%) I like to touch my face. 12 votes (26.08%) Pork chops and apple sauce. 10 votes (21.73%)

46 members have voted

Ace2
• Posts: 2706
Joined: Oct 2, 2017
Thanked by
June 8th, 2021 at 9:01:00 AM permalink
You should win a million dollars just for understanding the rules of the game
It’s all about making that GTA
ThatDonGuy
• Posts: 6471
Joined: Jun 22, 2011
June 8th, 2021 at 2:31:05 PM permalink
Quote: Ace2

You should win a million dollars just for understanding the rules of the game

I think I can rephrase the problem:

There are n persons standing around a circle, numbered in order 1, 2, ..., n.
Starting with person 1, then going to 2, 3, and so on, count q persons, then remove that person from the circle.
Starting with the person after the one removed, count another q persons (in the same direction), then remove that person as well.
Stop when k people remain.
For all n < 9, for every value of k, every "ending condition" of k people still around the circle is possible - i.e. there is at least one value of q where you will end up with that set of people.
For n = 9 and k = 5, you cannot end with (1, 2, 5, 8, 9), (2, 3, 4, 5, 8), or (2, 5, 6, 7, 8), but you can end with every one of the other 123 sets of 5 people out of 9.
The problem: find a value of n such that at least one impossible ending condition exists for k = n - 7.
ThatDonGuy
• Posts: 6471
Joined: Jun 22, 2011
June 9th, 2021 at 6:21:52 AM permalink
Quote: Gialmere

It's toughie Tuesday. Here's a puzzle from the brainiacs at IBM...

Your goal: Find an n such that there is a set of unwinnable numbers for seven steps (i.e., the set is of size n-7). In your answer, supply the number n and the elements of the unwinnable set.

I think I am on the right track, but I want to make sure I'm not missing something...is there a solution with n < 19? I am trying brute force, but my code claims that every set of (n - 7) with n < 19 can be formed after seven steps, and it's going to take quite a bit of time to check n = 19.
Gialmere
• Posts: 3006
Joined: Nov 26, 2018
June 9th, 2021 at 8:04:09 AM permalink
n > 19

There's also a bonus question for finding an unwinnable set which takes more than seven steps.
Have you tried 22 tonight? I said 22.
ThatDonGuy
• Posts: 6471
Joined: Jun 22, 2011
Thanked by
June 9th, 2021 at 9:37:55 AM permalink
Quote: Gialmere

n > 19

There's also a bonus question for finding an unwinnable set which takes more than seven steps.

I found a problem with my code that was causing it to run much slower than it should.

n = 20 has 2 solutions:
1,2,3,4,5,6,7,8,11,14,15,16,17
4,5,6,7,10,13,14,15,16,17,18,19,20

As for the bonus question:

n = 12, k = 8:
1,2,3,4,5,7,9,11
1,2,3,4,6,8,10,11
1,2,5,6,7,8,11,12
1,3,4,5,6,8,9,11
1,3,5,7,8,9,10,11
2,3,4,5,6,8,10,12
2,3,5,7,9,10,11,12
2,4,5,7,8,9,10,12
2,4,6,8,9,10,11,12

n = 15, k = 11:
1,2,3,4,5,6,7,8,11,14,15
1,2,3,4,5,6,8,9,10,11,14
1,2,3,4,5,6,8,11,12,13,14
1,2,3,4,5,8,9,10,11,14,15
1,2,3,4,5,8,11,12,13,14,15
1,2,3,5,6,7,8,10,11,12,14
1,2,3,5,6,7,8,11,13,14,15
1,2,3,5,7,8,9,11,12,13,14
1,2,3,5,8,9,10,11,13,14,15
1,2,4,5,6,7,8,9,11,14,15
1,2,4,5,6,8,10,11,12,14,15
1,2,5,6,7,8,9,10,11,14,15
1,2,5,6,7,8,11,12,13,14,15
1,2,5,7,8,9,10,11,12,14,15
1,2,5,8,9,10,11,12,13,14,15
2,3,4,5,6,7,8,9,10,11,14
2,3,4,5,6,7,8,11,12,13,14
2,3,4,5,7,8,9,10,11,12,14
2,3,4,5,7,8,9,11,13,14,15
2,3,4,5,8,9,10,11,12,13,14
2,3,4,5,8,10,11,12,13,14,15
2,4,5,6,7,8,9,11,12,13,14
2,4,5,6,8,9,10,11,13,14,15
2,5,6,7,8,9,10,11,12,13,14
2,5,6,7,8,10,11,12,13,14,15

n = 15, k = 9:
3,4,5,6,8,10,11,12,13

n = 18, k = 14 has 54 solutions
n = 20, k = 14 has 236 solutions

Last edited by: ThatDonGuy on Jun 9, 2021
ThatDonGuy
• Posts: 6471
Joined: Jun 22, 2011
June 9th, 2021 at 10:28:04 AM permalink
Okay, not exactly a math puzzle...

In a game of chess, after White's fifth move, h1 is empty. (h1 is where White's king-side rook begins.)
Black's fifth move is rook to h1, checkmate.
What were the first 9 moves of the game?
Last edited by: ThatDonGuy on Jun 9, 2021
gordonm888
• Posts: 5196
Joined: Feb 18, 2015
June 9th, 2021 at 3:43:18 PM permalink
I have been really enjoying the posts in which ThatDonGuy argues with himself. Very entertaining.
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
ThatDonGuy
• Posts: 6471
Joined: Jun 22, 2011
June 9th, 2021 at 5:11:09 PM permalink
Quote: gordonm888

I have been really enjoying the posts in which ThatDonGuy argues with himself. Very entertaining.

You're lucky - you don't have to live with him 24/7
Gialmere
• Posts: 3006
Joined: Nov 26, 2018
June 9th, 2021 at 6:16:05 PM permalink
Quote: ThatDonGuy

I found a problem with my code that was causing it to run much slower than it should.

n = 20 has 2 solutions:
1,2,3,4,5,6,7,8,11,14,15,16,17
4,5,6,7,10,13,14,15,16,17,18,19,20

As for the bonus question:

n = 12, k = 8:
1,2,3,4,5,7,9,11
1,2,3,4,6,8,10,11
1,2,5,6,7,8,11,12
1,3,4,5,6,8,9,11
1,3,5,7,8,9,10,11
2,3,4,5,6,8,10,12
2,3,5,7,9,10,11,12
2,4,5,7,8,9,10,12
2,4,6,8,9,10,11,12

n = 15, k = 11:
1,2,3,4,5,6,7,8,11,14,15
1,2,3,4,5,6,8,9,10,11,14
1,2,3,4,5,6,8,11,12,13,14
1,2,3,4,5,8,9,10,11,14,15
1,2,3,4,5,8,11,12,13,14,15
1,2,3,5,6,7,8,10,11,12,14
1,2,3,5,6,7,8,11,13,14,15
1,2,3,5,7,8,9,11,12,13,14
1,2,3,5,8,9,10,11,13,14,15
1,2,4,5,6,7,8,9,11,14,15
1,2,4,5,6,8,10,11,12,14,15
1,2,5,6,7,8,9,10,11,14,15
1,2,5,6,7,8,11,12,13,14,15
1,2,5,7,8,9,10,11,12,14,15
1,2,5,8,9,10,11,12,13,14,15
2,3,4,5,6,7,8,9,10,11,14
2,3,4,5,6,7,8,11,12,13,14
2,3,4,5,7,8,9,10,11,12,14
2,3,4,5,7,8,9,11,13,14,15
2,3,4,5,8,9,10,11,12,13,14
2,3,4,5,8,10,11,12,13,14,15
2,4,5,6,7,8,9,11,12,13,14
2,4,5,6,8,9,10,11,13,14,15
2,5,6,7,8,9,10,11,12,13,14
2,5,6,7,8,10,11,12,13,14,15

n = 15, k = 9:
3,4,5,6,8,10,11,12,13

n = 18, k = 14 has 54 solutions
n = 20, k = 14 has 236 solutions

Correct!

Very good. (I liked the "bubble gum, bubble gum in a dish" analogy.)
---------------------------------------------------

Have you tried 22 tonight? I said 22.
Ace2
• Posts: 2706
Joined: Oct 2, 2017
June 9th, 2021 at 7:20:32 PM permalink
Assuming an infinite standard deck of cards, you draw until you have all thirteen cards of any suit (diamonds, hearts, spades or clubs).

On average, how many draws will it take?
It’s all about making that GTA
aceside
• Posts: 544
Joined: May 14, 2021
June 11th, 2021 at 6:04:54 AM permalink
Quote: ThatDonGuy

Quote: aceside

Quote: Gialmere

It's easy Monday. Let's change things up...

g]

without being able to change a \$20 bill?

I’ve read this part several times but still haven’t got its exact meaning. Can you rephrase this part into precise mathematical terms?

You cannot make exactly \$20 with any combination of bills and coins. You can have more than \$20, but not exactly \$20.

I have thought about this again. I would state the question like this “without being able to change any partial amount to a \$20 bill.”
unJon
• Posts: 4662
Joined: Jul 1, 2018
June 11th, 2021 at 6:07:37 AM permalink
Quote: aceside

Quote: ThatDonGuy

Quote: aceside

Quote: Gialmere

It's easy Monday. Let's change things up...

g]

without being able to change a \$20 bill?

I’ve read this part several times but still haven’t got its exact meaning. Can you rephrase this part into precise mathematical terms?

You cannot make exactly \$20 with any combination of bills and coins. You can have more than \$20, but not exactly \$20.

I have thought about this again. I would state the question like this “without being able to change any partial amount to a \$20 bill.”

That phrasing is confusing to me. Don’t understand “partial amount.” I would say either “without being able to break a 20” or “without having exact change for a 20.”
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
aceside
• Posts: 544
Joined: May 14, 2021
June 11th, 2021 at 6:15:38 AM permalink
Quote: unJon

Quote: aceside

Quote: ThatDonGuy

Quote: aceside

Quote: Gialmere

It's easy Monday. Let's change things up...

g]

without being able to change a \$20 bill?

I’ve read this part several times but still haven’t got its exact meaning. Can you rephrase this part into precise mathematical terms?

You cannot make exactly \$20 with any combination of bills and coins. You can have more than \$20, but not exactly \$20.

I have thought about this again. I would state the question like this “without being able to change any partial amount to a \$20 bill.”

That phrasing is confusing to me. Don’t understand “partial amount.” I would say either “without being able to break a 20” or “without having exact change for a 20.”

You are right. “Without having exact change for a \$20 bill” is good.
aceside
• Posts: 544
Joined: May 14, 2021
June 11th, 2021 at 6:29:54 AM permalink
Quote: aceside

Quote: unJon

Quote: aceside

Quote: ThatDonGuy

Quote: aceside

Quote: Gialmere

It's easy Monday. Let's change things up...

g]

without being able to change a \$20 bill?

I’ve read this part several times but still haven’t got its exact meaning. Can you rephrase this part into precise mathematical terms?

You cannot make exactly \$20 with any combination of bills and coins. You can have more than \$20, but not exactly \$20.

I have thought about this again. I would state the question like this “without being able to change any partial amount to a \$20 bill.”

That phrasing is confusing to me. Don’t understand “partial amount.” I would say either “without being able to break a 20” or “without having exact change for a 20.”

You are right. “Without having exact change for a \$20 bill” is good.

Also, I want to simply this question a little as below.
Using only pennies, nickels, dimes, quarters, \$1 bills, and \$5 bills, what is the most money that you can have without having exact change for a \$10 bill?
Joeman
• Posts: 2428
Joined: Feb 21, 2014
Thanked by
June 11th, 2021 at 7:48:49 AM permalink
Quote: ThatDonGuy

Okay, not exactly a math puzzle...

In a game of chess, after White's fifth move, h1 is empty. (h1 is where White's king-side rook begins.)
Black's fifth move is rook to h1, checkmate.
What were the first 9 moves of the game?

This was a toughie. I kept coming up with 6 move solutions until I realized that white's knight had to wait until black's rook passed to move.

1. g4 h5
2. Bg2 h5xg4
3. Bxb7 Rxh2
4. Nh3 Bxb7
5. O-O Rh1++

A very odd set of moves for both sides!
"Dealer has 'rock'... Pay 'paper!'"
aceside
• Posts: 544
Joined: May 14, 2021
June 11th, 2021 at 8:35:43 AM permalink
Quote: Joeman

This was a toughie. I kept coming up with 6 move solutions until I realized that white's knight had to wait until black's rook passed to move.

1. g4 h5
2. Bg2 h5xg4
3. Bxb7 Rxh2
4. Nh3 Bxb7
5. O-O Rh1++

A very odd set of moves for both sides!

Can anybody make a 30 second video to demonstrate these moves? That would be a lot more straight.
unJon
• Posts: 4662
Joined: Jul 1, 2018
June 11th, 2021 at 8:54:29 AM permalink
Quote: aceside

Quote: aceside

Quote: unJon

Quote: aceside

Quote: ThatDonGuy

Quote: aceside

Quote: Gialmere

It's easy Monday. Let's change things up...

g]

without being able to change a \$20 bill?

I’ve read this part several times but still haven’t got its exact meaning. Can you rephrase this part into precise mathematical terms?

You cannot make exactly \$20 with any combination of bills and coins. You can have more than \$20, but not exactly \$20.

I have thought about this again. I would state the question like this “without being able to change any partial amount to a \$20 bill.”

That phrasing is confusing to me. Don’t understand “partial amount.” I would say either “without being able to break a 20” or “without having exact change for a 20.”

You are right. “Without having exact change for a \$20 bill” is good.

Also, I want to simply this question a little as below.
Using only pennies, nickels, dimes, quarters, \$1 bills, and \$5 bills, what is the most money that you can have without having exact change for a \$10 bill?

Same concept I think. \$10.19. Simplist is to ask most money you can have in pennies, nickels, dimes and quarters without having exact change for a dollar.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
Gialmere
• Posts: 3006
Joined: Nov 26, 2018
June 11th, 2021 at 9:15:40 AM permalink
Quote: aceside

Quote: Joeman

This was a toughie. I kept coming up with 6 move solutions until I realized that white's knight had to wait until black's rook passed to move.

1. g4 h5
2. Bg2 h5xg4
3. Bxb7 Rxh2
4. Nh3 Bxb7
5. O-O Rh1++

A very odd set of moves for both sides!

Can anybody make a 30 second video to demonstrate these moves? That would be a lot more straight.

Have you tried 22 tonight? I said 22.
aceside
• Posts: 544
Joined: May 14, 2021
June 11th, 2021 at 9:21:10 AM permalink
Quote: unJon

Quote: aceside

Quote: aceside

Quote: unJon

Quote: aceside

Quote: ThatDonGuy

Quote: aceside

Quote: Gialmere

It's easy Monday. Let's change things up...

g]

without being able to change a \$20 bill?

I’ve read this part several times but still haven’t got its exact meaning. Can you rephrase this part into precise mathematical terms?

You cannot make exactly \$20 with any combination of bills and coins. You can have more than \$20, but not exactly \$20.

I have thought about this again. I would state the question like this “without being able to change any partial amount to a \$20 bill.”

That phrasing is confusing to me. Don’t understand “partial amount.” I would say either “without being able to break a 20” or “without having exact change for a 20.”

You are right. “Without having exact change for a \$20 bill” is good.

Also, I want to simply this question a little as below.
Using only pennies, nickels, dimes, quarters, \$1 bills, and \$5 bills, what is the most money that you can have without having exact change for a \$10 bill?

Same concept I think. \$10.19. Simplist is to ask most money you can have in pennies, nickels, dimes and quarters without having exact change for a dollar.

Very good. Following Wizard’s solution, we have a similar:
1 quarter = \$0.25
9 dimes = \$0.90
4 pennies = \$0.04
Total = \$1.19
No nickels are required in this case.
aceside
• Posts: 544
Joined: May 14, 2021
June 11th, 2021 at 9:37:24 AM permalink
Quote: Gialmere

Quote: aceside

Quote: Joeman

This was a toughie. I kept coming up with 6 move solutions until I realized that white's knight had to wait until black's rook passed to move.

1. g4 h5
2. Bg2 h5xg4
3. Bxb7 Rxh2
4. Nh3 Bxb7
5. O-O Rh1++

A very odd set of moves for both sides!

Can anybody make a 30 second video to demonstrate these moves? That would be a lot more straight.

This is cool. Well done.
ThatDonGuy
• Posts: 6471
Joined: Jun 22, 2011
June 11th, 2021 at 12:20:20 PM permalink
Quote: Joeman

This was a toughie. I kept coming up with 6 move solutions until I realized that white's knight had to wait until black's rook passed to move.

1. g4 h5
2. Bg2 h5xg4
3. Bxb7 Rxh2
4. Nh3 Bxb7
5. O-O Rh1++

A very odd set of moves for both sides!

When I first saw it, I didn't think it was possible, until I discovered...

...that you are allowed to castle even when your rook is being attacked - it's just the king (or the spaces the king traverses) that can't be attacked.

ThatDonGuy
• Posts: 6471
Joined: Jun 22, 2011
June 11th, 2021 at 1:18:09 PM permalink
Quote: Ace2

Assuming an infinite standard deck of cards, you draw until you have all thirteen cards of any suit (diamonds, hearts, spades or clubs).

On average, how many draws will it take?

47,070,163,786,203,595,393,879,102,723 / 1,237,940,039,285,380,274,899,124,224, which is about 38.

EdCollins
• Posts: 1739
Joined: Oct 21, 2011
June 11th, 2021 at 2:40:42 PM permalink
If the original question is changed from this:

Using only pennies, nickels, dimes, quarters, half dollars, \$1 bills, \$5 bills, and \$10 bills, what is the most money that you can have without being able to change a \$20 bill?

to this:

Using only current U.S. currency \$10 and under, and current U.S. coin denominations, what is the most money you can have without being able to change a \$20 bill?

charliepatrick
• Posts: 2976
Joined: Jun 17, 2011
June 11th, 2021 at 2:46:17 PM permalink
Quote: EdCollins

... Using only current U.S. currency \$10 and under, and current U.S. coin denominations, what is the most money you can have without being able to change a \$20 bill?...what is the answer?

\$23 being \$5 \$5 \$5 \$2 \$2 \$2 \$2 would be a different answer, and that's ignoring the pennies part!

btw I think you might need to have some reference to exactly change \$20.
Ace2
• Posts: 2706
Joined: Oct 2, 2017
June 11th, 2021 at 3:02:19 PM permalink
Quote: ThatDonGuy

47,070,163,786,203,595,393,879,102,723 / 1,237,940,039,285,380,274,899,124,224, which is about 38.

I disagree. My answer is significantly higher than yours.

I did a basic simulation in excel and there is only about a 1 in 3,000 chance of success with that many draws
It’s all about making that GTA
gordonm888
• Posts: 5196
Joined: Feb 18, 2015
June 11th, 2021 at 6:40:11 PM permalink
Quote: Ace2

Quote: ThatDonGuy

47,070,163,786,203,595,393,879,102,723 / 1,237,940,039,285,380,274,899,124,224, which is about 38.

I disagree. My answer is significantly higher than yours.

I did a basic simulation in excel and there is only about a 1 in 3,000 chance of success with that many draws

I agree that ThatDonGuy's answer seems to be too low - and only by doing calculations in my head, not on my computer. I wonder if ThatDonGuy mis-read the problem in some way.

I'm away from my number-grinding CPU system, so no math: But with full replacement, if you have drawn 38 cards, how many distinct cards do you expect to have drawn? I don't know, maybe 30-32? And if you have 30-32 cards out of a possible 52, what is the liklihood that you have all 13 cards from one of the 4 suits? Not very good.
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
ThatDonGuy
• Posts: 6471
Joined: Jun 22, 2011
June 11th, 2021 at 7:01:23 PM permalink
Quote: Ace2

Quote: ThatDonGuy

47,070,163,786,203,595,393,879,102,723 / 1,237,940,039,285,380,274,899,124,224, which is about 38.

I disagree. My answer is significantly higher than yours.

I did a basic simulation in excel and there is only about a 1 in 3,000 chance of success with that many draws

I did read the problem wrong. The number I got is how many it takes to get 13 cards of any suit, regardless of ranks.

And here is the correct answer:

712,830,140,335,392,780,521 / 6,621,889,966,337,599,800, or about 107.64753627

If you limit it to a single 52-card deck, it takes 28,561 / 630 = 45.335 cards.

Last edited by: ThatDonGuy on Jun 12, 2021
Ace2
• Posts: 2706
Joined: Oct 2, 2017
June 12th, 2021 at 10:18:11 AM permalink
Quote: ThatDonGuy

Quote: Ace2

Quote: ThatDonGuy

47,070,163,786,203,595,393,879,102,723 / 1,237,940,039,285,380,274,899,124,224, which is about 38.

I disagree. My answer is significantly higher than yours.

I did a basic simulation in excel and there is only about a 1 in 3,000 chance of success with that many draws

I did read the problem wrong. The number I got is how many it takes to get 13 cards of any suit, regardless of ranks.

And here is the correct answer:

712,830,140,335,392,780,521 / 6,621,889,966,337,599,800, or about 107.64753627

If you limit it to a single 52-card deck, it takes 28,561 / 630 = 45.335 cards.

It’s all about making that GTA
ThatDonGuy
• Posts: 6471
Joined: Jun 22, 2011
June 12th, 2021 at 10:39:33 AM permalink
Quote: Ace2

Brute force / Markov chain.
Let E(s,h,c,d) be the expected number of cards needed when you already have s different spades, h different hearts, c different clubs, and d different diamonds.
Note that this is 0 if any of the numbers are 13.
Of the 52 cards:
(13 - s) will be a spade you didn't draw yet, so you will move to state (s+1, h, c, d)
(13 - h) will be a heart you didn't draw yet, so you will move to state (s, h+1, c, d)
(13 - c) will be a club you didn't draw yet, so you will move to state (s, h, c+1, d)
(13 - d) will be a diamond you didn't draw yet, so you will move to state (s, h, c, d+1)
The remaining (s + h + c + d) cards will be cards you have already drawn, so you will remain at state (s, h, c, d)
Since it is an infinite deck, each card always has a 1/52 chance of being drawn.
E(s, h, c, d) = 1 + (13 - s) / 52 x E(s+1, h, c, d) + (13 - h) / 52 x E(s, h+1, c, d) + (13 - c) / 52 x E(s, h, c+1, d) + (13 - d) / 52 x E(s, h, c, d+1) + (s + h + c + d)/52 x E(s, h, c, d)
(52 - (s + h + c + d)) x E(s, h, c, d) = 52 + (13 - s) x E(s+1, h, c, d) + (13 - h) x E(s, h+1, c, d) + (13 - c) x E(s, h, c+1, d) + (13 - d) x E(s, h, c, d+1)
E(s, h, c, d) = (52 + (13 - s) x E(s+1, h, c, d) + (13 - h) x E(s, h+1, c, d) + (13 - c) x E(s, h, c+1, d) + (13 - d) x E(s, h, c, d+1)) / (52 - (s + h + c + d))
Start with calculating E(12, 12, 12, 12), then E(12, 12, 12, 11), and so on to E(12, 12, 12, 0), then E(12, 12, 11, 12), E(12, 12, 11, 11), and so on, until you finally get E(0, 0, 0, 0), which is the solution.

In the single-deck version, there are 52 - (s + h + d + c) cards left at any point, and drawing a duplicate is impossible:
E(s, h, c, d) = 1 + (13 - s) / (52 - (s + h + c + d)) x E(s+1, h, c, d) + (13 - h) / (52 - (s + h + c + d)) x E(s, h+1, c, d) + (13 - c) / (52 - (s + h + c + d)) x E(s, h, c+1, d) + (13 - d) / (52 - (s + h + c + d)) x E(s, h, c, d+1)
Ace2
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June 12th, 2021 at 11:29:52 AM permalink
You can also get the answer by integrating the following formula over all t:

(1-(1-1/e^(t/52))^13)^4 dt

Which gives us the answer of 712830140335392780521 / 6621889966337599800 =~ 108 draws

Starting from the middle, 1/e^(t/52) is the probability at any time t that any single card has not been drawn. The complement of that to the 13th power (1-1/e^(t/52))^13 is the probability that all 13 cards of any one suit have been drawn at least once. The complement of that to the 4th power (1-(1-1/e^(t/52))^13)^4 is the probability that none of the 4 suits meet that condition (of all 13 cards being drawn at least once).

The formula works due to one of the most useful properties in probability: the expected time for an event to happen is equal to the sum of the probabilities over all time that it has not happened yet. This is basically the geometric series 1 + r + r^2 + r^3...= 1 / (1-r) carrying over from the binomial distribution to the exponential one.
It’s all about making that GTA
Wizard
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June 12th, 2021 at 9:42:46 PM permalink
A circle of radius 1 is inscribed in an equilateral triangle. What is the area of the triangle?
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
Ace2
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June 12th, 2021 at 10:21:15 PM permalink
Quote: Wizard

A circle of radius 1 is inscribed in an equilateral triangle. What is the area of the triangle?

3 * 3^.5 or about 5.2
It’s all about making that GTA
gordonm888
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June 13th, 2021 at 12:02:48 AM permalink
What is wrong with this picture below?

So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
aceside
• Posts: 544
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June 13th, 2021 at 3:52:18 AM permalink
Quote: gordonm888

What is wrong with this picture below?

I can solve this one. It’s 32,00 beans.
aceside
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Joined: May 14, 2021
June 13th, 2021 at 3:56:05 AM permalink
Quote: Ace2

Quote: ThatDonGuy

47,070,163,786,203,595,393,879,102,723 / 1,237,940,039,285,380,274,899,124,224, which is about 38.

I disagree. My answer is significantly higher than yours.

I did a basic simulation in excel and there is only about a 1 in 3,000 chance of success with that many draws

How do you do simulations in excel? I’d like to learn this skill for my simulating some blackjack side bets.
ThatDonGuy
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June 13th, 2021 at 6:46:54 AM permalink
Quote: Wizard

A circle of radius 1 is inscribed in an equilateral triangle. What is the area of the triangle?

Let A, B, C be the vertices of the triangle, O the center of the circle,
and D, E, F be the points where the circle is tangent to sides AB, AC, and BC, respectively.
OD = OE = OF = the radius of the circle, which is 1.
AEO and AFO are both right angles.
EO = FO = 1 and AO = AO, so right triangles AEO and AFO are congruent (Hypotenuse-Leg), which means angle OAE = angle OAF = 1/2 of 60 degrees = 30 degrees, and AOE is a 30-60-90 triangle. with the side opposite the 30-degree angle having length 1, so AE has length sqrt(3).
Using the same methods on COD and COE, CE also has length sqrt(3), so the length of AC = 2 sqrt(3).
The altitude of the triangle from B to AC divides it into two 30-60-90 triangles with the side opposite the 30-degree angle = sqrt(3), so the altitude = 3, and the area = 1/2 x 2 sqrt(3) x 3 = 3 sqrt(3).

charliepatrick
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June 13th, 2021 at 7:30:18 AM permalink
Quote: ThatDonGuy

"Not a Wild Guess"...

I agree but here's a slightly different method of proof.
Assuming AB is the lower side of the triangle and D is the tangent, then consider triangle OAD. The height is 1, as it's the radius. One angle (BAO) is half BAC, so is 30 degrees. Thus the hypotenuse is 2, and the base is SQRT(4-1)=SQRT(3). So the area of that triangle (1/2 b h) is SQRT(3)/2. There are six of these triangles (which make up the ABC triangle) so the total area is 3 SQRT(3).
Ace2
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June 13th, 2021 at 10:19:17 AM permalink
I looked up a equilateral triangle on wikipedia where it says:

radius of an inscribed circle = 3^.5 * (side) / 6 = 1
area of triangle = 3^.5 * (side)^2 / 4

Very easy to solve with that
It’s all about making that GTA
Wizard
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June 13th, 2021 at 4:29:42 PM permalink
Quote: Ace2

3 * 3^.5 or about 5.2

I agree!
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
ThatDonGuy
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June 14th, 2021 at 8:09:32 AM permalink
Quote: Ace2

You can also get the answer by integrating the following formula over all t:

(1-(1-1/e^(t/52))^13)^4 dt

Okay, I'll ask: how do you simplify this?
Gialmere
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June 14th, 2021 at 8:10:43 AM permalink
It's easy Monday...

What is the missing number in this sequence?

20, 22, 24, 26, 30, 33, 40, 44, 120, ___, 11000
Have you tried 22 tonight? I said 22.
charliepatrick
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Thanked by
June 14th, 2021 at 8:23:02 AM permalink
Ace2
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June 14th, 2021 at 8:48:10 AM permalink
Quote: ThatDonGuy

Okay, I'll ask: how do you simplify this?

What do you mean by simplify?
It’s all about making that GTA
Ace2
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Joined: Oct 2, 2017
June 14th, 2021 at 8:48:13 AM permalink
Quote: ThatDonGuy

Okay, I'll ask: how do you simplify this?

What do you mean by simplify?
It’s all about making that GTA
Ace2
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Joined: Oct 2, 2017
June 14th, 2021 at 8:48:15 AM permalink
Quote: ThatDonGuy

Okay, I'll ask: how do you simplify this?

What do you mean by simplify?
It’s all about making that GTA
gordonm888
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Joined: Feb 18, 2015
June 14th, 2021 at 9:40:55 AM permalink
Quote: Ace2

You can also get the answer by integrating the following formula over all t:

(1-(1-1/e^(t/52))^13)^4

Quote: ThatDonGuy

Okay, I'll ask: how do you simplify this?

Quote: Ace2

What do you mean by simplify?

I think he means that the analytical expression to be integrated has 2^13^4 = 5E15 terms if you straightforwardly expand it. So, can you analytically simplify the integration or must you do it numerically?
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
ThatDonGuy
• Posts: 6471
Joined: Jun 22, 2011
June 14th, 2021 at 9:56:20 AM permalink
Quote: Ace2

What do you mean by simplify?

I mean, is there a way to get from the integral to the rational number result without using approximation?

I assumed that, since you expressed the answer as a rational number, that there was a way to do it.
Ace2
• Posts: 2706
Joined: Oct 2, 2017
June 14th, 2021 at 12:10:09 PM permalink
Quote: ThatDonGuy

I mean, is there a way to get from the integral to the rational number result without using approximation?

I assumed that, since you expressed the answer as a rational number, that there was a way to do it.

I use https://www.integral-calculator.com/ for the integration of something like this. If you put that formula in, it shows all the steps (several pages) and it does come out to the rational answer. To check the answer for reasonableness, I either do a simulation or a numerical integration in excel. In this case, if you do a numerical integration with spacing of 0.1 the first four digits of the answer have already converged.
It’s all about making that GTA
Wizard
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Joined: Oct 14, 2009
June 14th, 2021 at 12:43:07 PM permalink
Quote: Ace2

You can also get the answer by integrating the following formula over all t:

(1-(1-1/e^(t/52))^13)^4 dt

Which gives us the answer of 712830140335392780521 / 6621889966337599800 =~ 108 draws

Starting from the middle, 1/e^(t/52) is the probability at any time t that any single card has not been drawn. The complement of that to the 13th power (1-1/e^(t/52))^13 is the probability that all 13 cards of any one suit have been drawn at least once. The complement of that to the 4th power (1-(1-1/e^(t/52))^13)^4 is the probability that none of the 4 suits meet that condition (of all 13 cards being drawn at least once).

The formula works due to one of the most useful properties in probability: the expected time for an event to happen is equal to the sum of the probabilities over all time that it has not happened yet. This is basically the geometric series 1 + r + r^2 + r^3...= 1 / (1-r) carrying over from the binomial distribution to the exponential one.

I'm finally catching up with another classic Ace2 calculus problem. I plan to make this a future Ask the Wizard question. Here are some notes for now on my solution. I welcome all comments and corrections.

Integral calculator: https://www.integral-calculator.com/

Q: A card is draw with replacement from a deck of 52 cards. What is the expected number of draws needed until every card of any one rank has been drawn? Please use calculus for your solution.

Hint: The expected time for an event to happen equals the sum over all time that it hasn't happned yet. This is true for both discreet and continuous variables.

Solution: Instead of a card being drawn exactly once per unit of time, the answer will be the same if a card is drawn with a random period of time between draws if that average time follows an exponential distribution with mean of 1.

The time between any given card being drawn will have a mean of 52. Given the properties of the exponential distribution, the probability the card will not have been drawn after t units of time is exp(-t/52).

After t units of time, the probability any specific card will have been drawn at least once is 1-exp(-t/52).

After t units of time, the probability 13 specific cards will have been drawn at least once is (1-exp(-t/52))^13.

After t units of time, at least one of 13 specific cards will cards will NOT have been drawn is 1-(1-exp(-t/52))^13.

After t units of time, four groups of 13 cards each, having no overlapping cards, all will have at least one missing card is (1-(1-exp(-t/52))^13)^4.

Putting this equation into an integral calculator, being careful to set to set the bounds of integration from 0 to infinity, yields 712830140335392780521 / 6621889966337599800 =~ 107.6475362712258
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
DogHand

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Thanked by
June 14th, 2021 at 12:52:44 PM permalink
Quote: Gialmere

It's easy Monday...

What is the missing number in this sequence?

20, 22, 24, 26, 30, 33, 40, 44, 120, ___, 11000

With thanks to Charlie for the hint, I get...

220 = 24 in base 3. Each number is 24 in based decreasing from 12 to 2.

Dog Hand