Easy math puzzles

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November 3rd, 2020 at 5:26:05 PM permalink

Quote: ThatDonGuy

Instead of adding 20 balls of the new color, add N balls, where N - 1 is prime

Each color is equally likely to be the first ball drawn
The probability that the second ball is the same as the first is (B - 1) / (BC - 1)

Instead of adding 20 balls of the new color, add N balls; there are now BC + N balls
The probability that the first ball is one of the originals is BC / (BC + N);
the probability that the second ball is the same as the first is (B - 1) / (BC + N - 1)
The probability that the first ball is one of the new ones is N / (BC + N);
the probability that the second ball is the same as the first is (N - 1) / (BC + N - 1)
The overall probability that the first two balls are the same color is
(BC (B - 1) + N (N - 1)) / ((BC + N)(BC + N - 1))

These are equal when
(B - 1) / (BC - 1) = (BC (B - 1) + N (N-1)) / ((BC + N)(BC + N - 1))
(B - 1)(BC + N)(BC + N - 1) = (BC (B - 1) + N (N-1)) (BC - 1)
(B - 1) ((BC)^2 + 2 BCN + N^2 - BC - N) = (B^2 C - BC + N^2 - N) (BC - 1)
B^3 C^2 + 2 B^2CN + BN^2 - B^2C - BN - (BC)^2 - 2 BCN - N^2 + BC + N = B^3 C^2 - (BC)^2 + BCN^2 - BCN - B^2 C + BC - N^2 + N
2 B^2CN + BN^2 - BN - BCN = BCN^2
2 BCN + N^2 - N - CN = CN^2
2 BC + N - 1 - C = CN
C (N + 1 - 2B) = N - 1

N - 1 is prime, so either C = 1 or N + 1 - 2B = 1
If C = 1, then N + 1 - 2B = N - 1, so B = 1, but that means there is only one ball at the start
Therefore, N + 1 - 2B = 1 and C = N - 1
BC = (N / 2) (N - 1)

For N = 20, there are 10 x 19 = 190 balls at the start

Correct!

Very good.
--------------------------------

They hurt.

Have you tried 22 tonight? I said 22.

ThatDonGuy

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November 4th, 2020 at 9:10:29 AM permalink

Here are a couple from me:

On the English TV series Countdown (also known as Letters and Numbers in Australia), some of the rounds involve two contestants being shown six numbers and having 30 seconds to work out a way to get some or all of the numbers to get to a given three-digit number, using only the four basic operations and parentheses. Note that not all of the numbers have to be used, but no number can be used more than once, and numbers cannot be concatenated (i.e. 2 and 7 cannot be combined into 27 or 72).
For example, if the six numbers are 1, 3, 4, 6, 8, and 25, and the "target" is 243, this is 25 x (8 + 1) + (6 x 3); note the 4 is not used.
Note that, on the show, you only get 30 seconds.

1. Make 952 from 3, 6, 25, 50, 75, 100

2. Make 813 from 1, 10, 25, 50, 75, 100

For the first, I got 953: (3 + 6) x 100 + 50 + 75 / 25

For the second, I got 823: (10 - 1) x 100 - 75 - 50 / 25

However, there is a way to get 952 in the first one and 813 in the second one.

charliepatrick

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November 4th, 2020 at 11:38:55 AM permalink

Quote: ThatDonGuy
...Countdown...1. Make 952 from 3, 6, 25, 50, 75, 100...

I thought I recognised it but didn't look this up before working it out but did remembered the end division. (( 3 * (100+6) * 75 )-50 ) /25 ( 3*106*75 = 23850. 23800/25 = 952 ).

The video is here > https://www.youtube.com/watch?v=6mCgiaAFCu8

You may not know this but originally Countdown was made in Leeds, Yorkshire and my sister won the first series. So many years later, when Richard and Carol were still presenting it, I took her to a recording and we sat in the audience. During the recording Carol spotted Joyce, said hello and had the camera pan in.

Thanks for asking the question as I have now googled and found this early video with Joyce where Richard (the other girl presenter isn't Carol) explains how the numbers bit works. In the real show they used an eletronic random number generator, so it wasn't always possible to solve - you get 5/7/10 points for being close, closer, correct. In recent series I think the "computer" choses random numbers that have a solution.
https://www.youtube.com/watch?v=5OwSHIwZ4ak

charliepatrick

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November 4th, 2020 at 3:23:30 PM permalink

(75+1)*10 + 50 + (100/25) = 760+50+4

Wizard
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November 4th, 2020 at 3:57:00 PM permalink

Quote: Gialmere
The ball bin at the front counter of a miniature golf course contains a number of colored golf balls, with equal numbers of each color. Adding 20 balls of a new color to the bin would not change the probability of the counterperson blindly drawing (without replacement) two balls of the same color.

190.

19 colors and 10 balls each. Probability before and after is 0.047619.

My method was pretty much trial and error. Interested to know if there is a more satisfying solution.

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

charliepatrick

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November 4th, 2020 at 4:25:26 PM permalink

Just happened to multiply 65*25 and saw the result.
(75-10)*25 = 1625
1625+1 = 1626
100/50 = 2
1626/2 = 813

With these "four from the top" there's nearly always a division somewhere!

charliepatrick

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November 4th, 2020 at 4:43:02 PM permalink

Quote: Gialmere
...The ball bin at the front counter of a miniature golf course contains a number of colored golf balls, with equal numbers of each color. Adding 20 balls of a new color to the bin would not change the probability of the counterperson blindly drawing (without replacement) two balls of the same color. Before the extra balls are added, how many golf balls are in the bin?...

I wish I'd read equal numbers of each colour. Looking at three colours I found that 4, 6 and 9 work, and suspect there are other solutions for different numbers of colours.

ThatDonGuy

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November 5th, 2020 at 7:21:23 AM permalink

Quote: charliepatrick
I wish I'd read equal numbers of each colour. Looking at three colours I found that 4, 6 and 9 work, and suspect there are other solutions for different numbers of colours.

So far, I have checked all combinations where there are 20 or fewer balls of each color, up to 17 different colors

Solutions where all of the colors have a different number of balls:
4, 6, 9
2, 3, 4, 7, 8, 9, 11, 13
2, 3, 5, 6, 7, 10, 11, 13

Total solutions:
0 have 2 colors
1 has 3 colors
1 has 4 colors
0 have 5 colors
0 have 6 colors
25 have 7 colors
75 have 8 colors
86 have 9 colors
71 have 10 colors
41 have 11 colors
26 have 12 colors
10 have 13 colors
7 have 14 colors
3 have 15 colors
1 has 16 colors: 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 5, 6, 20
0 have 17 colors

Last edited by: ThatDonGuy on Nov 5, 2020

Gialmere

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November 5th, 2020 at 10:01:08 AM permalink

You are on The Price is Right and playing a new version of the clock game.

In this version, you have 30 seconds to guess how many dollar bills are inside a box: a randomly selected amount from $1 to $1000. If you correctly guess the amount, you win the money.

Like the original game, if you guess incorrectly the host will tell you if your next guess should be higher or lower. Unfortunately, since the new version has just the one prize, you are only allowed to make 9 guesses.

If you want to get the maximum value out of this game, what should your first guess be?

tick-tock tick-tock

Have you tried 22 tonight? I said 22.

charliepatrick

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November 5th, 2020 at 10:50:40 AM permalink

Quote: Gialmere
...If you want to get the maximum value out of this game, what should your first guess be?...

745.

At each round you can guess a number. If it happens to be correct you win. It not you are told whether the number is higher or lower. You start with a large range and graduyally narrow it down, until (worst case scenario) you find the correct answer.

However this puzzle only allows you nine rounds. Therefore you can only cover a range of the numbers. It seems logical that you cover ther highest numbers. So the range where you can win will be 1000 down to some number.

It's easiest to work backwards. Thus on the penultimate round (i.e. 2 more guesses) you need to have narrowed down to range of three. So if you choose correct (the middle number of the three) you win, otherwise you know from the reply what the correct answer is. (e.g. knowing it's 1000-998 you guess 999.

On the round before that (3 more guesses) you need to have a guess where if you're wrong it narrows it down to three consecutive numbers. (They don't have to be consecutive, but your best strategy is to go for the highest numbers). Thus your range is seven numbers and you guess the middle one.

Counting backwards one of the worst case scenarios is 1000 (until the last round you'd always be low).
1 - Last round - correct guess (say) 1000. (1000-2⁰+1 )
2 - Penultimate - range=3 (1000-998)- guess = 999 (1000-2¹+1 )
3 - range=7 (1000-994) - guess 997 (1000-2²+1 ).
4 - range=15 (1000-986) - guess 993 (1000-2³+1 )
5 - range=31 (1000-970) - guess 985 (1000-2⁴+1 )
::
9 - range=511 (1000-490) - guess 745 (1000-2⁸+1 )

Gialmere

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November 5th, 2020 at 5:42:29 PM permalink

Quote: charliepatrick
745.

At each round you can guess a number. If it happens to be correct you win. It not you are told whether the number is higher or lower. You start with a large range and graduyally narrow it down, until (worst case scenario) you find the correct answer.

However this puzzle only allows you nine rounds. Therefore you can only cover a range of the numbers. It seems logical that you cover ther highest numbers. So the range where you can win will be 1000 down to some number.

It's easiest to work backwards. Thus on the penultimate round (i.e. 2 more guesses) you need to have narrowed down to range of three. So if you choose correct (the middle number of the three) you win, otherwise you know from the reply what the correct answer is. (e.g. knowing it's 1000-998 you guess 999.

On the round before that (3 more guesses) you need to have a guess where if you're wrong it narrows it down to three consecutive numbers. (They don't have to be consecutive, but your best strategy is to go for the highest numbers). Thus your range is seven numbers and you guess the middle one.

Counting backwards one of the worst case scenarios is 1000 (until the last round you'd always be low).
1 - Last round - correct guess (say) 1000. (1000-2⁰+1 )
2 - Penultimate - range=3 (1000-998)- guess = 999 (1000-2¹+1 )
3 - range=7 (1000-994) - guess 997 (1000-2²+1 ).
4 - range=15 (1000-986) - guess 993 (1000-2³+1 )
5 - range=31 (1000-970) - guess 985 (1000-2⁴+1 )
::
9 - range=511 (1000-490) - guess 745 (1000-2⁸+1 )

Correct!

Yes. You have to forget about the chump change end of the spread.

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

Have you tried 22 tonight? I said 22.

ThatDonGuy

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November 5th, 2020 at 8:58:21 PM permalink

Quote: ThatDonGuy

So far, I have checked all combinations where there are 20 or fewer balls of each color, up to 17 different colors

Solutions where all of the colors have a different number of balls:
4, 6, 9
2, 3, 4, 7, 8, 9, 11, 13
2, 3, 5, 6, 7, 10, 11, 13

Total solutions:
0 have 2 colors
1 has 3 colors
1 has 4 colors
0 have 5 colors
0 have 6 colors
25 have 7 colors
75 have 8 colors
86 have 9 colors
71 have 10 colors
41 have 11 colors
26 have 12 colors
10 have 13 colors
7 have 14 colors
3 have 15 colors
1 has 16 colors: 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 5, 6, 20
0 have 17 colors

Update:
1,482 have 18 colors
30,045 have 19 colors
153,398 have 20 colors

Last edited by: ThatDonGuy on Nov 6, 2020

Wizard
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November 6th, 2020 at 5:53:52 AM permalink

Quote: Gialmere
In this version, you have 30 seconds to guess how many dollar bills are inside a box: a randomly selected amount from $1 to $1000. If you correctly guess the amount, you win the money.

Like the original game, if you guess incorrectly the host will tell you if your next guess should be higher or lower. Unfortunately, since the new version has just the one prize, you are only allowed to make 9 guesses.

If you want to get the maximum value out of this game, what should your first guess be?

I think we can quit putting this one in spoiler tags.

Aren't my chances the same 511/1000, with a strategy of starting at 500 and cutting the possible range in half every time? Sometimes you will have an even number to cut, so will have to pick arbitrarily.

For example, let's say the number is 523.

1. 500 - higher
2. 750 - lower
3. 625 - lower
4. 562 - lower
5. 531 - lower
6. 515 - higher
7. 522 - higher
8. 526 - lower
9. 524 - wrong

However, I might have hit it, depending on which way I went in situations where the range was even.

I think I would need to do a simulation to verify/disprove my theory.

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

charliepatrick

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November 6th, 2020 at 6:30:30 AM permalink

Quote: Wizard
...Aren't my chances the same 511/1000, with a strategy of starting at 500 and cutting the possible range in half every time? Sometimes you will have an even number to cut, so will have to pick arbitrarily.

Yes your chances of being correct are the same, however the question suggested that you also wanted to maximise your expected winnings. This leads to an optimal starting position.

With your various picks, depending on High/Low etc, you (sort of) get a choice of which possibilities you can correctly find (so in your example you cannot find 523). Thus one of your 511 values (you can find) must be a lower value instead. Therefore this is not an optimum strategy.

The optimum strategy accepts you will not find the correct answer if it is less than 490, but all values 490 and over (there are 511 of them!) can be found.

unJon

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November 6th, 2020 at 9:32:28 AM permalink

It�s been a long time since I looked at a higher / lower problem, but I thought chopping things in half was not optimal. My recollection (likely wrong) was that dividing remaining unknowns into something like 1/3 and 2/3 worked better.

The race is not always to the swift, nor the battle to the strong; but that is the way to bet.

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November 6th, 2020 at 10:21:19 AM permalink

Quote: charliepatrick
Yes your chances of being correct are the same, however the question suggested that you also wanted to maximise your expected winnings. This leads to an optimal starting position.

Thank you. I overlooked the part about winning whatever the amount of money is. I assumed it was a flat win, although I clearly was lazy and didn't read the wording carefully.

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

charliepatrick

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November 6th, 2020 at 12:11:27 PM permalink

Quote: unJon
It�s been a long time since I looked at a higher / lower problem, but I thought chopping things in half was not optimal. My recollection (likely wrong) was that dividing remaining unknowns into something like 1/3 and 2/3 worked better.

I also think that's where you can put weights on either side of the scales - 1 3 9 ... work well.

Gialmere

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November 9th, 2020 at 8:03:18 AM permalink

It's easy Monday...

Place 10 bishops on a standard chessboard such that they attack all squares on the board including occupied ones.

Have you tried 22 tonight? I said 22.

ThatDonGuy

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November 10th, 2020 at 10:06:30 AM permalink

Quote: Gialmere

Place 10 bishops on a standard chessboard such that they attack all squares on the board including occupied ones.

Sorry if it's a little hard to read. Capital letters are the locations of the bishops (I skipped letter I as it was hard to read in the table); small letters are bishops that attack that square. Each bishop's square is attacked by another bishop.

g	a	h	a	f	b	f	c
b	g	A	h	b	F	c	h
j	a	g	a	g	c	h	d
a	j	B	G	C	H	d	f
k	b	j	b	g	c	h	e
b	k	c	J	D	g	c	h
g	c	K	d	j	E	g	c
c	k	d	k	e	g	e	J

charliepatrick

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November 10th, 2020 at 11:52:18 AM permalink

If you look at just the Black squares then they form seven SW-NE diagonals, see diagram. If there can be five bishops, one on each of the diagonals 2 thru 6 then that covers those squares. All these bishops also have to be on the 1 to 7 diagonals (the long SE-NW diagonals), two on one and three on the other. This is needed so another bishop attacks every square each bishop is on and, by being on the 1 to 7 diagonals, these will cover the all the 1 and 7 squares.

By symmetry the White squares can be covered with five "White" bishops.

. 1 . 2 . 3 . 4
1 . 2 . 3 . 4 .
. 2 . 3 . 4 . 5
2 . 3 . 4 . 5 .
. 3 . 4 . 5 . 6
3 . 4 . 5 . 6 .
. 4 . 5 . 6 . 7
4 . 5 . 6 . 7 .

This is one possible solution for the Black bishops.

. 1 . 2 . 3 . 4
1 . 2 . 3 . 4 .
. B . 3 . 4 . 5
2 . B . 4 . 5 .
. 3 . B . B . 6
3 . 4 . 5 . B .
. 4 . 5 . 6 . 7
4 . 5 . 6 . 7 .

Wizard
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November 10th, 2020 at 1:41:40 PM permalink

When are we going to get the Toughy Tuesday question?

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

Wizard
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November 10th, 2020 at 3:14:41 PM permalink

I think I'm going to have to pinch hit for G and post the Toughy Tuesday puzzle myself.

You call in the beer the Wizard owes you. However, the Wizard has beer at room temperature only, which we will say is 25 degrees (Celsius*,**). He puts a warm beer in the freezer to cool it down. The temperature of the freezer is always -5 degrees. Upon entry, the beer cools at a rate of 1 degree in 5 minutes. However, the rate the beer cools is proportional to the difference in temperature between the beer and freezer.

How long will it take to cool the beer to 4 degrees?

* All degrees in Celsius.
** Why do we capitalize Celsius?

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

ThatDonGuy

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November 10th, 2020 at 4:44:47 PM permalink

Quote: Wizard
** Why do we capitalize Celsius?

Presumably for the same reason we capitalize Fahrenheit and Kelvin - it's named after a person. Specifically, Anders Celsius.

Let T(t) be the difference between the beer temperature and the refrigerator temperature at time t (in minutes)
The rate of change of temperature when the difference is T is -T/150 degrees/minute
dT(t)/dt = -T(t) / 150, with the condition T(0) = 30

Let T(t) = 30 e^(-1/150 t)
T'(t) = 30 (-1/150 e^(-1/150 t)) = -T(t) / 150

The solution is the value of t for which T(t) = 9
30 e^(-1/150 t) = 9
e^(-1/150 t) = 0.3
-1/150 t = ln 0.3
t = -150 ln 0.3 = 180.5959 minutes

Last edited by: ThatDonGuy on Nov 10, 2020

gordonm888
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November 10th, 2020 at 5:22:24 PM permalink

Here's a problem for Tuesday:

In Texas Hold'em, what is the probability that the 5 common cards in the middle (the "board") will be the Nuts? The 5 common cards are the Nuts when there are no two hole cards that can improve the board so as to make a higher ranking poker hand.

So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.

ThatDonGuy

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November 10th, 2020 at 6:01:14 PM permalink

Quote: gordonm888
In Texas Hold'em, what is the probability that the 5 common cards in the middle (the "board") will be the Nuts? The 5 common cards are the Nuts when there are no two hole cards that can improve the board so as to make a higher ranking poker hand.

There are C(52,5) = 2,598,960 possible boards

Of these:
A Royal Flush is the Nuts (4)
A Straight Flush besides a Royal can be improved by having the next-higher card in your hand (e.g. if the board is 5-6-7-8-9 of spades, you have a better hand with the 10 of spades as a hole card)
Four Aces and a King is the Nuts (4)
Four 2s-Kings and an Ace is the Nuts (12 possible suits x 4 suits for the Ace = 48)
A full house can be quaded in your hand (e.g. A-A-A-8-8 becomes four Aces with an Ace in your hand)
A flush can be improved by getting a suited card higher than the lowest card on the board; the only way that is not possible is with a Royal
An Ace-high straight is the Nuts if no more than two cards are suited, as three or four suited cards can become a flush; there are 1024 possible A-K-Q-J-10 boards, but four are Royals, 4 (suits) x 5 (unsuited card) x 3 (suits for the unsuited card) = 120 have four suited, and 4 (suits) x 10 (sets of 2 unsuited cards) x 3 (suits for the higher unsuited card) x 3 (suits for the lower unsuited card) = 360 have three suited, so there are 1024 - (4 + 120 + 360) = 540 that are the Nuts
Any other straight can be made a higher straight with the next-higher card, the same way a Straight Flush can
Three of a kind or lower can always be improved by having the matching card (e.g. three Aces can become four Aces; Kings over Queens can become a full house; a pair of Jacks can become three (or four) Jacks; garbage can become a pair (or better).
The total number of Nuts boards = 4 + 4 + 48 + 540 = 606
The probability = 606 / 2,598,960, or about 1 / 4288.713

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November 10th, 2020 at 6:05:00 PM permalink

Quote: ThatDonGuy
Presumably for the same reason we capitalize Fahrenheit and Kelvin - it's named after a person. Specifically, Anders Celsius.

I'm sure you're right, but in quoting a temperature, the reference is not to Anders Celsius. I also think it's narcissistic to name things after yourself. Speaking of which, the word narcissistic is a reference to Narcissus, a character in Greek mythology, so why don't we capitalize narcissistic and all variants?

Let T(t) be the difference between the beer temperature and the refrigerator temperature at time t (in minutes)
The rate of change of temperature when the difference is T is -T/150 degrees/minute
dT(t)/dt = -T(t) / 150, with the condition T(0) = 30

Let T(t) = 30 e^(-1/150 t)
T'(t) = 30 (-1/150 e^(-1/150 t)) = -T(t) / 150

The solution is the value of t for which T(t) = 9
30 e^(-1/150 t) = 9
e^(-1/150 t) = 0.3
-1/150 t = ln 0.3
t = -150 ln 0.3 = 180.5959 minutes

I agree! I think that one was beer worthy, so add a luke-warm beer to my tab of what I owe you.

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

ThatDonGuy

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November 10th, 2020 at 6:23:41 PM permalink

Quote: Wizard
I'm sure you're right, but in quoting a temperature, the reference is not to Anders Celsius. I also think it's narcissistic to name things after yourself. Speaking of which, the word narcissistic is a reference to Narcissus, a character in Greek mythology, so why don't we capitalize narcissistic and all variants?

A temperature is not, for example, "ten Celsius" or "ten Celsiuses," but "ten degrees Celsius."

I assume that "narcissistic" is not capitalized because it is a condition and not a person's name. Had it been called "Narcissus syndrome," it probably would be capitalized, the way, say, "Down syndrome" and "Oedipus complex" are.

I don't think Celsius named the scale - it was called "centigrade" for a long time, wasn't it?

unJon

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November 10th, 2020 at 7:43:19 PM permalink

Quote: ThatDonGuy
A temperature is not, for example, "ten Celsius" or "ten Celsiuses," but "ten degrees Celsius."

I assume that "narcissistic" is not capitalized because it is a condition and not a person's name. Had it been called "Narcissus syndrome," it probably would be capitalized, the way, say, "Down syndrome" and "Oedipus complex" are.

I don't think Celsius named the scale - it was called "centigrade" for a long time, wasn't it?

Centigrade makes intuitive sense.

The race is not always to the swift, nor the battle to the strong; but that is the way to bet.

gordonm888
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November 10th, 2020 at 9:12:10 PM permalink

Quote: ThatDonGuy

There are C(52,5) = 2,598,960 possible boards

Of these:
A Royal Flush is the Nuts (4)
A Straight Flush besides a Royal can be improved by having the next-higher card in your hand (e.g. if the board is 5-6-7-8-9 of spades, you have a better hand with the 10 of spades as a hole card)
Four Aces and a King is the Nuts (4)
Four 2s-Kings and an Ace is the Nuts (12 possible suits x 4 suits for the Ace = 48)
A full house can be quaded in your hand (e.g. A-A-A-8-8 becomes four Aces with an Ace in your hand)
A flush can be improved by getting a suited card higher than the lowest card on the board; the only way that is not possible is with a Royal
An Ace-high straight is the Nuts if no more than two cards are suited, as three or four suited cards can become a flush; there are 1024 possible A-K-Q-J-10 boards, but four are Royals, 4 (suits) x 5 (unsuited card) x 3 (suits for the unsuited card) = 120 have four suited, and 4 (suits) x 10 (sets of 2 unsuited cards) x 3 (suits for the higher unsuited card) x 3 (suits for the lower unsuited card) = 360 have three suited, so there are 1024 - (4 + 120 + 360) = 540 that are the Nuts
Any other straight can be made a higher straight with the next-higher card, the same way a Straight Flush can
Three of a kind or lower can always be improved by having the matching card (e.g. three Aces can become four Aces; Kings over Queens can become a full house; a pair of Jacks can become three (or four) Jacks; garbage can become a pair (or better).
The total number of Nuts boards = 4 + 4 + 48 + 540 = 606
The probability = 606 / 2,598,960, or about 1 / 4288.713

Sorry. You are very close, but I believe you have one arithmetic mistake.

look again at the number of combinations for the Ace-high straight (no suit with>2 cards)

So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.

gordonm888
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November 10th, 2020 at 9:12:16 PM permalink

Quote: ThatDonGuy

There are C(52,5) = 2,598,960 possible boards

Of these:
A Royal Flush is the Nuts (4)
A Straight Flush besides a Royal can be improved by having the next-higher card in your hand (e.g. if the board is 5-6-7-8-9 of spades, you have a better hand with the 10 of spades as a hole card)
Four Aces and a King is the Nuts (4)
Four 2s-Kings and an Ace is the Nuts (12 possible suits x 4 suits for the Ace = 48)
A full house can be quaded in your hand (e.g. A-A-A-8-8 becomes four Aces with an Ace in your hand)
A flush can be improved by getting a suited card higher than the lowest card on the board; the only way that is not possible is with a Royal
An Ace-high straight is the Nuts if no more than two cards are suited, as three or four suited cards can become a flush; there are 1024 possible A-K-Q-J-10 boards, but four are Royals, 4 (suits) x 5 (unsuited card) x 3 (suits for the unsuited card) = 120 have four suited, and 4 (suits) x 10 (sets of 2 unsuited cards) x 3 (suits for the higher unsuited card) x 3 (suits for the lower unsuited card) = 360 have three suited, so there are 1024 - (4 + 120 + 360) = 540 that are the Nuts
Any other straight can be made a higher straight with the next-higher card, the same way a Straight Flush can
Three of a kind or lower can always be improved by having the matching card (e.g. three Aces can become four Aces; Kings over Queens can become a full house; a pair of Jacks can become three (or four) Jacks; garbage can become a pair (or better).
The total number of Nuts boards = 4 + 4 + 48 + 540 = 606
The probability = 606 / 2,598,960, or about 1 / 4288.713

Sorry. You are very close, but I believe you have one arithmetic mistake.

look again at the number of combinations for the Ace-high straight (no suit with>2 cards)

So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.

Ace2

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November 11th, 2020 at 6:45:07 AM permalink

Quote: unJon
Centigrade makes intuitive sense.

I disagree. IMO Centigrade would be more intuitive for Fahrenheit, since air temperature in most of the inhabited earth is almost always between zero and a hundred. This is the temperature we use the most on a daily basis.

In Celsius, the 0 - 100 range is from freezing to boiling water. Like most units in the metric system, it�s not really applicable to everyday practical use. If you are in a northern city, your air temperature will be a negative number for the entire winter. Not a good scale, even Kelvin would be better

It�s all about making that GTA

unJon

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November 11th, 2020 at 7:12:50 AM permalink

Quote: Ace2
I disagree. IMO Centigrade would be more intuitive for Fahrenheit, since air temperature in most of the inhabited earth is almost always between zero and a hundred. This is the temperature we use the most on a daily basis.

In Celsius, the 0 - 100 range is from freezing to boiling water. Like most units in the metric system, it�s not really applicable to everyday practical use. If you are in a northern city, your air temperature will be a negative number for the entire winter. Not a good scale, even Kelvin would be better

I was saying something different. The name centigrade makes intuitive sense to me what it means: 100 grades.

The race is not always to the swift, nor the battle to the strong; but that is the way to bet.

ThatDonGuy

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November 11th, 2020 at 7:20:54 AM permalink

Quote: gordonm888
Quote: ThatDonGuy

There are C(52,5) = 2,598,960 possible boards

Of these:
A Royal Flush is the Nuts (4)
A Straight Flush besides a Royal can be improved by having the next-higher card in your hand (e.g. if the board is 5-6-7-8-9 of spades, you have a better hand with the 10 of spades as a hole card)
Four Aces and a King is the Nuts (4)
Four 2s-Kings and an Ace is the Nuts (12 possible suits x 4 suits for the Ace = 48)
A full house can be quaded in your hand (e.g. A-A-A-8-8 becomes four Aces with an Ace in your hand)
A flush can be improved by getting a suited card higher than the lowest card on the board; the only way that is not possible is with a Royal
An Ace-high straight is the Nuts if no more than two cards are suited, as three or four suited cards can become a flush; there are 1024 possible A-K-Q-J-10 boards, but four are Royals, 4 (suits) x 5 (unsuited card) x 3 (suits for the unsuited card) = 120 have four suited, and 4 (suits) x 10 (sets of 2 unsuited cards) x 3 (suits for the higher unsuited card) x 3 (suits for the lower unsuited card) = 360 have three suited, so there are 1024 - (4 + 120 + 360) = 540 that are the Nuts
Any other straight can be made a higher straight with the next-higher card, the same way a Straight Flush can
Three of a kind or lower can always be improved by having the matching card (e.g. three Aces can become four Aces; Kings over Queens can become a full house; a pair of Jacks can become three (or four) Jacks; garbage can become a pair (or better).
The total number of Nuts boards = 4 + 4 + 48 + 540 = 606
The probability = 606 / 2,598,960, or about 1 / 4288.713

Sorry. You are very close, but I believe you have one arithmetic mistake.

look again at the number of combinations for the Ace-high straight (no suit with>2 cards)

What, you mean 5 x 4 x 3 isn't 120?
Next you'll be telling me that 540 + 48 + 4 + 4 isn't 606.

There are 1024 possible A-K-Q-J-10 boards, but four are Royals, 4 (suits) x 5 (unsuited card) x 3 (suits for the unsuited card) = 60 have four suited, and 4 (suits) x 10 (sets of 2 unsuited cards) x 3 (suits for the higher unsuited card) x 3 (suits for the lower unsuited card) = 360 have three suited, so there are 1024 - (4 + 60 + 360) = 600 that are the Nuts
The total number of Nuts boards = 4 + 4 + 48 + 600 = 656
The probability = 656 / 2,598,960, or about 1 / 3961.83

Gialmere

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November 11th, 2020 at 7:54:03 AM permalink

Here, I'm assuming that the redundant capital J at h1 is a typo...

Quote: ThatDonGuy

Sorry if it's a little hard to read. Capital letters are the locations of the bishops (I skipped letter I as it was hard to read in the table); small letters are bishops that attack that square. Each bishop's square is attacked by another bishop.

g a h a f b f c

b g A h b F c h

j a g a g c h d

a j B G C H d f

k b j b g c h e

b k c J D g c h

g c K d j E g c

c k d k e g e J

Quote: charliepatrick
If you look at just the Black squares then they form seven SW-NE diagonals, see diagram. If there can be five bishops, one on each of the diagonals 2 thru 6 then that covers those squares. All these bishops also have to be on the 1 to 7 diagonals (the long SE-NW diagonals), two on one and three on the other. This is needed so another bishop attacks every square each bishop is on and, by being on the 1 to 7 diagonals, these will cover the all the 1 and 7 squares.

By symmetry the White squares can be covered with five "White" bishops.
. 1 . 2 . 3 . 4
1 . 2 . 3 . 4 .
. 2 . 3 . 4 . 5
2 . 3 . 4 . 5 .
. 3 . 4 . 5 . 6
3 . 4 . 5 . 6 .
. 4 . 5 . 6 . 7
4 . 5 . 6 . 7 .
This is one possible solution for the Black bishops.
. 1 . 2 . 3 . 4
1 . 2 . 3 . 4 .
. B . 3 . 4 . 5
2 . B . 4 . 5 .
. 3 . B . B . 6
3 . 4 . 5 . B .
. 4 . 5 . 6 . 7
4 . 5 . 6 . 7 .

Correct!
------------------------------

Have you tried 22 tonight? I said 22.

Gialmere

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November 12th, 2020 at 10:08:02 AM permalink

Use the anagrams and clues to identify ten famous mathematicians.

1) DIME SEARCH "If I had a fulcrum, I could move the world."

2) SCARED SET He did not belong to this set, for he was a soldier before he was a mathematician.

3) A DACRON He could not have "obtained" a dacron shirt, but he seems to have "acquired" a solution of a cubic equation.

4) PANIER He was not a Frenchman, and his "bones" were not kept in a basket.

5) LE RUE He crossed bridges instead of streets.

6) MENU SALE He was one of the few of the first century AD who did original mathematical work of any ability.

7) WAS ILL He had much to do with the founding of the Royal Society in England.

8) ALL PACE He discovered a famous differential equation.

9) RED TOUGH Yet he was delighted to hear of the king's return, and even invented a rule.

10) ROME DIVE He was chosen to decide the controversy over the discovery of the calculus.

Have you tried 22 tonight? I said 22.

Gialmere

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November 14th, 2020 at 1:58:11 PM permalink

Quote: Gialmere

Use the anagrams and clues to identify ten famous mathematicians.

1) DIME SEARCH "If I had a fulcrum, I could move the world."

2) SCARED SET He did not belong to this set, for he was a soldier before he was a mathematician.

3) A DACRON He could not have "obtained" a dacron shirt, but he seems to have "acquired" a solution of a cubic equation.

4) PANIER He was not a Frenchman, and his "bones" were not kept in a basket.

5) LE RUE He crossed bridges instead of streets.

6) MENU SALE He was one of the few of the first century AD who did original mathematical work of any ability.

7) WAS ILL He had much to do with the founding of the Royal Society in England.

8) ALL PACE He discovered a famous differential equation.

9) RED TOUGH Yet he was delighted to hear of the king's return, and even invented a rule.

10) ROME DIVE He was chosen to decide the controversy over the discovery of the calculus.

1) ARCHIMEDES

2) DESCARTES

3) CARDANO

4) NAPIER

5) EULER

6) MENELAUS

7) WALLIS

8) LAPLACE

9) OUGHTRED

10) DE MOIVRE

Have you tried 22 tonight? I said 22.

gordonm888
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November 15th, 2020 at 5:32:15 AM permalink

Quote: ThatDonGuy
Quote: gordonm888
Quote: ThatDonGuy

There are C(52,5) = 2,598,960 possible boards

Of these:
A Royal Flush is the Nuts (4)
A Straight Flush besides a Royal can be improved by having the next-higher card in your hand (e.g. if the board is 5-6-7-8-9 of spades, you have a better hand with the 10 of spades as a hole card)
Four Aces and a King is the Nuts (4)
Four 2s-Kings and an Ace is the Nuts (12 possible suits x 4 suits for the Ace = 48)
A full house can be quaded in your hand (e.g. A-A-A-8-8 becomes four Aces with an Ace in your hand)
A flush can be improved by getting a suited card higher than the lowest card on the board; the only way that is not possible is with a Royal
An Ace-high straight is the Nuts if no more than two cards are suited, as three or four suited cards can become a flush; there are 1024 possible A-K-Q-J-10 boards, but four are Royals, 4 (suits) x 5 (unsuited card) x 3 (suits for the unsuited card) = 120 have four suited, and 4 (suits) x 10 (sets of 2 unsuited cards) x 3 (suits for the higher unsuited card) x 3 (suits for the lower unsuited card) = 360 have three suited, so there are 1024 - (4 + 120 + 360) = 540 that are the Nuts
Any other straight can be made a higher straight with the next-higher card, the same way a Straight Flush can
Three of a kind or lower can always be improved by having the matching card (e.g. three Aces can become four Aces; Kings over Queens can become a full house; a pair of Jacks can become three (or four) Jacks; garbage can become a pair (or better).
The total number of Nuts boards = 4 + 4 + 48 + 540 = 606
The probability = 606 / 2,598,960, or about 1 / 4288.713

Sorry. You are very close, but I believe you have one arithmetic mistake.

look again at the number of combinations for the Ace-high straight (no suit with>2 cards)

What, you mean 5 x 4 x 3 isn't 120?
Next you'll be telling me that 540 + 48 + 4 + 4 isn't 606.

There are 1024 possible A-K-Q-J-10 boards, but four are Royals, 4 (suits) x 5 (unsuited card) x 3 (suits for the unsuited card) = 60 have four suited, and 4 (suits) x 10 (sets of 2 unsuited cards) x 3 (suits for the higher unsuited card) x 3 (suits for the lower unsuited card) = 360 have three suited, so there are 1024 - (4 + 60 + 360) = 600 that are the Nuts
The total number of Nuts boards = 4 + 4 + 48 + 600 = 656
The probability = 656 / 2,598,960, or about 1 / 3961.83

CORRECT

So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.

Gialmere

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November 16th, 2020 at 8:32:23 AM permalink

It's easy Monday...

A bum in downtown Las Vegas can make 1 cigarette out of 6 cigarette butts.

If he scrounges up 72 butts off the street and from public ashtrays, how many cigarette can he make and smoke?

Have you tried 22 tonight? I said 22.

rsactuary

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November 16th, 2020 at 9:02:04 AM permalink

72/6 = 12 + 2 from the new butts = 14

Last edited by: unnamed administrator on Nov 16, 2020

charliepatrick

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November 16th, 2020 at 9:10:07 AM permalink

Did you know a Baker's Dozen could be 13 or occasionally 14. It originates from the severe penalties for underselling by weight, so bakers added additional loaves ("inbread") to ensure the weight exceeded the requirements.

The answer here is 14 for a similar reason. The bum can make 12 cigarattes, but then use the butts of those 12 to make 2 more.

ThatDonGuy

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November 16th, 2020 at 9:11:34 AM permalink

Quote: Gialmere
It's easy Monday...

A bum in downtown Las Vegas can make 1 cigarette out of 6 cigarette butts.

If he scrounges up 72 butts off the street and from public ashtrays, how many cigarette can he make and smoke?

Provided he has one butt as a "base", every 5 butts will make a new cigarette (base plus 5 new ones becomes a cigarette, which returns to its base once smoked).

72 butts = 1 base + 71 added, which is 14 cigarettes plus 2 butts left over.

Gialmere

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November 16th, 2020 at 7:50:20 PM permalink

Quote: rsactuary

72/6 = 12 + 2 from the new butts = 14

Quote: charliepatrick
Did you know a Baker's Dozen could be 13 or occasionally 14. It originates from the severe penalties for underselling by weight, so bakers added additional loaves ("inbread") to ensure the weight exceeded the requirements.

The answer here is 14 for a similar reason. The bum can make 12 cigarattes, but then use the butts of those 12 to make 2 more.

Quote: ThatDonGuy

Provided he has one butt as a "base", every 5 butts will make a new cigarette (base plus 5 new ones becomes a cigarette, which returns to its base once smoked).

72 butts = 1 base + 71 added, which is 14 cigarettes plus 2 butts left over.

Correct!
---------------------------------

Have you tried 22 tonight? I said 22.

Gialmere

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November 17th, 2020 at 8:01:46 AM permalink

It's toughie Tuesday. Time to drop dead...

Drop Dead is a game played with five standard dice.

You begin your turn by rolling all five dice. If none of them are a 2 or 5, you total the dice, add the sum to your point score and roll again.

If you do roll any 2s or 5s, your score for the roll is zero. All the dice showing a 2 or 5 are declared dead and set aside. You then roll again with the remaining dice.

Play continues with you either scoring points or removing dice.

Your turn ends when all your dice are eliminated whereupon you are said to have dropped dead. All your scoring rolls are added together for your final score. High score wins.

What is your expected score for this game?
--------------------------------

If I'm not being clear with the rules, this two minute video explains them...

Have you tried 22 tonight? I said 22.

ThatDonGuy

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November 17th, 2020 at 12:01:23 PM permalink

Dice where 2s and 5s "don't count" remind me of Scarney Dice (created by John Scarne), where the 2s and 5s actually say "Dead".

Hopefully I avoided any idiot math errors this time; the number appears to be confirmed by simulation.

Let E(n) be the expected score with N dice remaining
E(0) = 0

E(1) = 2/3 (E(1) + 7/2)
= 7

E(2) = 4/9 (E(2) + 7) + 4/9 E(1)
= 4/9 E(2) + 56/9
= 56/5

E(3) = 8/27 (E(3) + 21/2) + 12/27 E(2) + 6/27 E(1)
= 8/27 E(3) + 84/27 + 12/27 x 56/5 + 6/27 x 7
19 E(3) = 84 + 12 x 56/5 + 42
E(3) = (126 + 12 x 56/5) / 19
= (126 x 5 + 12 x 56) / 95
= 1302 / 95

E(4) = 16/81 (E(4) + 14) + 32/81 E(3) + 24/81 E(2) + 8/81 E(1)
= 16/81 E(4) + 224/81 + 32/81 x 1302/95 + 24/81 x 56/5 + 8/81 x 7
65 E(4) = 224 + 32 x 1302 / 95 + 24 x 56 / 5 + 56
E(4) = (280 x 475 + 41,664 x 5 + 1344 x 95) / (95 x 5 x 65)
= (280 x 95 + 41,664 + 1344 x 19) / (95 x 65)
= 3752 / 247

E(5) = 32/243 (E(5) + 35/2) + 80/243 E(4) + 80/243 E(3) + 40/243 E(2) + 10/243 E(1)
243 E(5) = 32 E(5) + 560 + 80 x 3752 / (19 x 13) + 80 x 1302 / (19 x 5) + 448 + 70
211 E(5) = 1078 + 80 x 3752 / (19 x 13) + 80 x 1302 / (19 x 5)
(211 x 19 x 65) E(5) = 1078 x 19 x 65 + 80 x 3752 x 5 + 80 x 1302 x 13
E(5) = 837,242 / 52,117, or about 16.065

gordonm888
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November 17th, 2020 at 12:22:38 PM permalink

Quote: ThatDonGuy
Dice where 2s and 5s "don't count" remind me of Scarney Dice (created by John Scarne), where the 2s and 5s actually say "Dead".

Hopefully I avoided any idiot math errors this time; the number appears to be confirmed by simulation.

Let E(n) be the expected score with N dice remaining
E(0) = 0

E(1) = 2/3 (E(1) + 7/2)
= 7

E(2) = 4/9 (E(2) + 7) + 4/9 E(1)
= 4/9 E(2) + 56/9
= 56/5

E(3) = 8/27 (E(3) + 21/2) + 12/27 E(2) + 6/27 E(1)
= 8/27 E(3) + 84/27 + 12/27 x 56/5 + 6/27 x 7
19 E(3) = 84 + 12 x 56/5 + 42
E(3) = (126 + 12 x 56/5) / 19
= (126 x 5 + 12 x 56) / 95
= 1302 / 95

E(4) = 16/81 (E(4) + 14) + 32/81 E(3) + 24/81 E(2) + 8/81 E(1)
= 16/81 E(4) + 224/81 + 32/81 x 1302/95 + 24/81 x 56/5 + 8/81 x 7
65 E(4) = 224 + 32 x 1302 / 95 + 24 x 56 / 5 + 56
E(4) = (280 x 475 + 41,664 x 5 + 1344 x 95) / (95 x 5 x 65)
= (280 x 95 + 41,664 + 1344 x 19) / (95 x 65)
= 3752 / 247

E(5) = 32/243 (E(5) + 35/2) + 80/243 E(4) + 80/243 E(3) + 40/243 E(2) + 10/243 E(1)
243 E(5) = 32 E(5) + 560 + 80 x 3752 / (19 x 13) + 80 x 1302 / (19 x 5) + 448 + 70
211 E(5) = 1078 + 80 x 3752 / (19 x 13) + 80 x 1302 / (19 x 5)
(211 x 19 x 65) E(5) = 1078 x 19 x 65 + 80 x 3752 x 5 + 80 x 1302 x 13
E(5) = 837,242 / 52,117, or about 16.065

OMG!!! Both an analytic solution and a confirming simulation! You are soooo fast. Very nicely done.

I was part way through this analytic solution - with at least one "idiot math error" - when this post popped into the thread. I am very impressed!

So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.

gordonm888
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November 17th, 2020 at 12:22:45 PM permalink

Quote: ThatDonGuy
Dice where 2s and 5s "don't count" remind me of Scarney Dice (created by John Scarne), where the 2s and 5s actually say "Dead".

Hopefully I avoided any idiot math errors this time; the number appears to be confirmed by simulation.

Let E(n) be the expected score with N dice remaining
E(0) = 0

E(1) = 2/3 (E(1) + 7/2)
= 7

E(2) = 4/9 (E(2) + 7) + 4/9 E(1)
= 4/9 E(2) + 56/9
= 56/5

E(3) = 8/27 (E(3) + 21/2) + 12/27 E(2) + 6/27 E(1)
= 8/27 E(3) + 84/27 + 12/27 x 56/5 + 6/27 x 7
19 E(3) = 84 + 12 x 56/5 + 42
E(3) = (126 + 12 x 56/5) / 19
= (126 x 5 + 12 x 56) / 95
= 1302 / 95

E(4) = 16/81 (E(4) + 14) + 32/81 E(3) + 24/81 E(2) + 8/81 E(1)
= 16/81 E(4) + 224/81 + 32/81 x 1302/95 + 24/81 x 56/5 + 8/81 x 7
65 E(4) = 224 + 32 x 1302 / 95 + 24 x 56 / 5 + 56
E(4) = (280 x 475 + 41,664 x 5 + 1344 x 95) / (95 x 5 x 65)
= (280 x 95 + 41,664 + 1344 x 19) / (95 x 65)
= 3752 / 247

E(5) = 32/243 (E(5) + 35/2) + 80/243 E(4) + 80/243 E(3) + 40/243 E(2) + 10/243 E(1)
243 E(5) = 32 E(5) + 560 + 80 x 3752 / (19 x 13) + 80 x 1302 / (19 x 5) + 448 + 70
211 E(5) = 1078 + 80 x 3752 / (19 x 13) + 80 x 1302 / (19 x 5)
(211 x 19 x 65) E(5) = 1078 x 19 x 65 + 80 x 3752 x 5 + 80 x 1302 x 13
E(5) = 837,242 / 52,117, or about 16.065

OMG!!! Both an analytic solution and a confirming simulation! You are soooo fast. Very nicely done.

I was part way through this analytic solution - with at least one "idiot math error" - when this post popped into the thread. I am very impressed!

So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.

Wizard
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November 17th, 2020 at 2:03:33 PM permalink

I agree with Don's answer.

Version 2: What is the answer if the player gets to keep all rolls that are not a 2 or 5? For example, if the first roll is a 1-2-3-4-5, then the player gets 1+3+4=8 points and whatever points he gets from a position of 3 dice?

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

ThatDonGuy

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November 17th, 2020 at 3:55:50 PM permalink

Quote: Wizard
Version 2: What is the answer if the player gets to keep all rolls that are not a 2 or 5? For example, if the first roll is a 1-2-3-4-5, then the player gets 1+3+4=8 points and whatever points he gets from a position of 3 dice?

This is equivalent to:
Roll each die separately, adding up the numbers until you roll a 2 or a 5 (without counting the last roll).
For each die, there is:
probability 1/3 of a 2 or 5 on the first roll; this is 0
probability 2/3 x 1/3 of a 2 or 5 on the second roll; this is 2/9 x 7/2
probability (2/3)^2 x 1/3 of a 2 or 5 on the third roll; this is 2/9 x 2/3 x (7/2 x 2)
probability (2/3)^3 x 1/3 of a 2 or 5 on the fourth roll; this is 2/9 x (2/3)^2 x (7/2 x 3)
and so on

The sum is 2/9 x 7/2 x (1 + 2/3 x 2 + (2/3)^2 x 3 + (2/3)^3 x 4 + ...)
= 7/9 x (1 + 2/3 + (2/3)^2 + ...)^2
= 7/9 x (1 / (1 - 2/3))^2
= 7

The total for five dice is 5 x 7 = 35

charliepatrick