## Poll

 I love math! 21 votes (46.66%) Math is great. 14 votes (31.11%) My religion is mathology. 6 votes (13.33%) Women didn't speak to me until I was 30. 2 votes (4.44%) Total eclipse reminder -- 04/08/2024 12 votes (26.66%) I steal cutlery from restaurants. 3 votes (6.66%) I should just say what's on my mind. 6 votes (13.33%) Who makes up these awful names for pandas? 5 votes (11.11%) I like to touch my face. 12 votes (26.66%) Pork chops and apple sauce. 10 votes (22.22%)

45 members have voted

MichaelBluejay
Joined: Sep 17, 2010
• Posts: 1589
November 23rd, 2023 at 9:51:39 AM permalink
From the weekly Math column in my local paper:

"You have sealed the ninth of nine boxes that weigh about the same, only to discover that your diamond ring has accidentally fallen into one of the packages. You don't want to unwrap every box. Can you work out how to find the box containing the ring by using the balance scale just twice?"

I failed.

solution
Presidential Election polls and odds: https://2605.me/p
ThatDonGuy
Joined: Jun 22, 2011
• Posts: 5947
Thanks for this post from:
November 23rd, 2023 at 10:05:48 AM permalink
Quote: MichaelBluejay

From the weekly Math column in my local paper:

"You have sealed the ninth of nine boxes that weigh about the same, only to discover that your diamond ring has accidentally fallen into one of the packages. You don't want to unwrap every box. Can you work out how to find the box containing the ring by using the balance scale just twice?"

I failed.

solution

Number the boxes 1-9
Weigh 1, 2, 3 on one side and 4, 5, 6 on the other
If the sides don't balance, the heavier side has the ring; if they do, then one of 7, 8, 9 does
For the second weighing, put one of the three boxes in the "heavy three" on one side, and another on the other
If they don't balance, the heavier side has the ring; if they do, the box you didn't weigh does

DogHand
Joined: Sep 24, 2011
• Posts: 1314
Thanks for this post from:
November 23rd, 2023 at 10:11:23 AM permalink
Quote: MichaelBluejay

From the weekly Math column in my local paper:

"You have sealed the ninth of nine boxes that weigh about the same, only to discover that your diamond ring has accidentally fallen into one of the packages. You don't want to unwrap every box. Can you work out how to find the box containing the ring by using the balance scale just twice?"

I failed.

solution

MichaelBluejay,

"... nine boxes that weigh about the same... "

If the eight non-ring boxes do not have EXACTLY the same weight, the solution fails.

In real life, variations in the amount of tape used to seal each box will invalidate the given solution anyway. After all, how much does a diamond ring actually weigh?

Dog Hand

P.S. I enjoyed the puzzle anyway and had no trouble finding the solution, despite my qualms ;-)
TinMan
Joined: Nov 17, 2009
• Posts: 441
November 23rd, 2023 at 10:33:00 AM permalink
(1) Balance cans 123 vs 456

(2A) If they balance out, balance 7 vs 8. If they balance out, 9 has the ring. If they don’t balance, the heavier one has the ring.

(2B) If they don’t balance, take the 3 cans on the heavy side. Let’s say it’s 123. Balance 1 vs 2. If they balance 3 has the ring. If one is heavier that has the ring.
Last edited by: unnamed administrator on Nov 23, 2023
If anyone gives you 10,000 to 1 on anything, you take it. If John Mellencamp ever wins an Oscar, I am going to be a very rich dude.
ThatDonGuy
Joined: Jun 22, 2011
• Posts: 5947
Thanks for this post from:
November 23rd, 2023 at 10:50:12 AM permalink

Use this:

[spoiler=Spoiler Button Text]
Spoiler Goes here
[/spoiler]

to create this:

Spoiler goes here
Dieter
Joined: Jul 23, 2014
• Posts: 5016
November 23rd, 2023 at 12:03:28 PM permalink
Quote: DogHand

After all, how much does a diamond ring actually weigh?

(snip!)
The diamonds don't weigh much (often less than half a gram, unless... but if you have a diamond boulder on your finger, are you really sealing packages?)
The diamonds usually come attached to several grams of gold.
May the cards fall in your favor.
Wizard
Joined: Oct 14, 2009
• Posts: 25961
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November 23rd, 2023 at 12:14:07 PM permalink
That is one of my favorites. I tell it with nine pearls and one of them is fake, weighing a little less.
“Extraordinary claims require extraordinary evidence.” -- Carl Sagan
Wizard
Joined: Oct 14, 2009
• Posts: 25961
November 23rd, 2023 at 1:46:25 PM permalink
I would like to wish all my math puzzle friends a happy Thanksgiving! May we all be thankful for the good things and people we have.

Here is my next puzzle.

sqrt(1-x) = 1-x2

Find x.
“Extraordinary claims require extraordinary evidence.” -- Carl Sagan
charliepatrick
Joined: Jun 17, 2011
• Posts: 2933
November 23rd, 2023 at 3:57:40 PM permalink
Quote: Wizard

I would like to wish all my math puzzle friends a happy Thanksgiving! May we all be thankful for the good things and people we have.

Here is my next puzzle.

sqrt(1-x) = 1-x2

Find x.

Squaring gives (1-x)=(1-x2)2=1-2x2+x4
Subtracting (1-x) from both sides and simplifying gives x(x3-2x+1)=0=x(x-1)(x2+ax-1) and it transpires a=1.
So the solutions are 0, 1, (SQRT(5)-1)/2, (-SQRT(5)-1)/2.
Wizard
Joined: Oct 14, 2009
• Posts: 25961
Thanks for this post from:
November 23rd, 2023 at 4:20:58 PM permalink
Quote: charliepatrick

Squaring gives (1-x)=(1-x2)2=1-2x2+x4
Subtracting (1-x) from both sides and simplifying gives x(x3-2x+1)=0=x(x-1)(x2+ax-1) and it transpires a=1.
So the solutions are 0, 1, (SQRT(5)-1)/2, (-SQRT(5)-1)/2.

Correct!
“Extraordinary claims require extraordinary evidence.” -- Carl Sagan
Gialmere
Joined: Nov 26, 2018
• Posts: 2804
November 24th, 2023 at 4:28:39 PM permalink
Here's a belated Thanksgiving puzzle from the Stanford math department in 1947. Assume no tax.

My grandfather’s papers included an old invoice:

72 turkeys \$-67.9-

The first and last digits are illegible.

What are the missing digits, and what was the price of one turkey?
Have you tried 22 tonight? I said 22.
charliepatrick
Joined: Jun 17, 2011
• Posts: 2933
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November 24th, 2023 at 5:18:52 PM permalink
A turkey is \$5.11 costing \$367.92. Personally I just divided n6790 by 72 until it was just under a whole number.
ThatDonGuy
Joined: Jun 22, 2011
• Posts: 5947
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November 24th, 2023 at 6:54:17 PM permalink
Quote: Gialmere

My grandfather’s papers included an old invoice:

72 turkeys \$-67.9-

The first and last digits are illegible.

What are the missing digits, and what was the price of one turkey?

A multiple of 72 is also a multiple of 8 and 9.
Multiples of 8 have their last three digits as multiples of 8; the only multiple of 8 between 790 and 799 inclusive is 792, so the last digit is 2.
Multiples of 9 have the sum of their digits be a multiple of 9; 6 + 7 + 9 + 2 = 24, so the first digit has to be 3.
The missing digits are 3 (first digit) and 2 (last digit), and the price of one turkey is \$367.92 / 72 = \$5.11.

Wizard
Joined: Oct 14, 2009
• Posts: 25961
November 25th, 2023 at 2:57:11 AM permalink
A craps electronic table game awards points upon bets resolved, including the odds. The game has double odds. As usual, when laying odds, the player may lay up to the amount that would result in a win 2x the don't pass/come bet (in other words 4x on a 4 or 10, 3x on a 5 or 9, and 2.4x on a 6 or 8).

There are four players. All make \$1 pass, don't pass, come and don't come bets and back them up the full way on the odds.

• Alice makes one pass bet only
• Bob makes one don't pass bet only
• Cathy makes a pass or don't pass bet every roll
• David makes a don't pass or don't come bet every

What is the average wager resolved per throw of each player?
“Extraordinary claims require extraordinary evidence.” -- Carl Sagan
aceside
Joined: May 14, 2021
• Posts: 373
November 25th, 2023 at 8:08:46 AM permalink
Hi Wizard, I believe your online Blackjack Hand Calculator is not working as it used to be. It doesn’t allow me to type in the shoe composition numbers because these input boxes are greyed out now. Could you please check into this? Thank you in advance!
ThatDonGuy
Joined: Jun 22, 2011
• Posts: 5947
November 25th, 2023 at 8:22:10 AM permalink
Quote: Wizard

• Cathy makes a pass or don't pass bet every roll

Should that be, "Cathy makes a pass or come bet every roll"? How can Cathy make a pass or don't pass bet on "every roll"?

Also, for Cathy and David, should I only be counting the rolls until their original comeout is resolved, or keep going until they seven out and have to pass the dice?
Gialmere
Joined: Nov 26, 2018
• Posts: 2804
November 25th, 2023 at 9:24:44 AM permalink
Quote: charliepatrick

A turkey is \$5.11 costing \$367.92. Personally I just divided n6790 by 72 until it was just under a whole number.

Quote: ThatDonGuy

Quote: Gialmere

My grandfather’s papers included an old invoice:

72 turkeys \$-67.9-

The first and last digits are illegible.

What are the missing digits, and what was the price of one turkey?

A multiple of 72 is also a multiple of 8 and 9.
Multiples of 8 have their last three digits as multiples of 8; the only multiple of 8 between 790 and 799 inclusive is 792, so the last digit is 2.
Multiples of 9 have the sum of their digits be a multiple of 9; 6 + 7 + 9 + 2 = 24, so the first digit has to be 3.
The missing digits are 3 (first digit) and 2 (last digit), and the price of one turkey is \$367.92 / 72 = \$5.11.

Correct!!

Good show.
---------------------------------------

I need enough for 12 people and maybe 2 police officers.

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

Still in play...

Quote: Wizard

A craps electronic table game awards points upon bets resolved, including the odds. The game has double odds. As usual, when laying odds, the player may lay up to the amount that would result in a win 2x the don't pass/come bet (in other words 4x on a 4 or 10, 3x on a 5 or 9, and 2.4x on a 6 or 8).

There are four players. All make \$1 pass, don't pass, come and don't come bets and back them up the full way on the odds.

• Alice makes one pass bet only
• Bob makes one don't pass bet only
• Cathy makes a pass or don't pass bet every roll
• David makes a don't pass or don't come bet every

What is the average wager resolved per throw of each player?

Have you tried 22 tonight? I said 22.
Wizard
Joined: Oct 14, 2009
• Posts: 25961
November 25th, 2023 at 10:05:19 AM permalink
I'd like to add to my problem that a 12 after a don't pass or don't come bet should count as a bet resolved.
“Extraordinary claims require extraordinary evidence.” -- Carl Sagan
Ace2
Joined: Oct 2, 2017
• Posts: 2664
November 25th, 2023 at 11:55:16 AM permalink

Should be

A) (1/3 + 2/3 * 3) * 165/557 =~ 0.691
B) (1/3 + 1/6 * 5 + 2/9 * 4 + 5/18 * 3.4) * 165/557 =~ 0.889
C) (1/3 + 2/3 * 3) = 7/3
D) 1/3 + 1/6 * 5 + 2/9 * 4 + 5/18 * 3.4) = 3
It’s all about making that GTA
Wizard
Joined: Oct 14, 2009
• Posts: 25961
November 25th, 2023 at 12:31:33 PM permalink
Quote: Ace2

Should be

A) (1/3 + 2/3 * 3) * 165/557 =~ 0.691
B) (1/3 + 1/6 * 5 + 2/9 * 4 + 5/18 * 3.4) * 165/557 =~ 0.889
C) (1/3 + 2/3 * 3) = 7/3
D) 1/3 + 1/6 * 5 + 2/9 * 4 + 5/18 * 3.4) = 3

Thank for working these out. We agree on C and D. I'm not saying you're wrong on the other two. I think I'll run a simulation to get a third opinion. I'm not sure where you get the fraction in your answer for A and B.
“Extraordinary claims require extraordinary evidence.” -- Carl Sagan
Ace2
Joined: Oct 2, 2017
• Posts: 2664
November 25th, 2023 at 1:01:21 PM permalink
Quote: Wizard

Quote: Ace2

Should be

A) (1/3 + 2/3 * 3) * 165/557 =~ 0.691
B) (1/3 + 1/6 * 5 + 2/9 * 4 + 5/18 * 3.4) * 165/557 =~ 0.889
C) (1/3 + 2/3 * 3) = 7/3
D) 1/3 + 1/6 * 5 + 2/9 * 4 + 5/18 * 3.4) = 3

Thank for working these out. We agree on C and D. I'm not saying you're wrong on the other two. I think I'll run a simulation to get a third opinion. I'm not sure where you get the fraction in your answer for A and B.

1 + 1/6 * 4 + 2/9 * 3.6 + 5/18 * 36/11 = 557/165 which is the average number of rolls to resolve a Pass/DP bet

165/557 is a fraction used so often in craps problems that I have it memorized. If, for instance, you bet DP only and assume 100 rolls per hour then you can expect 100 * 165/557 =~ 30 bets resolved per hour. I just use 0.3 for calculations in my head
It’s all about making that GTA
Wizard
Joined: Oct 14, 2009
• Posts: 25961
November 25th, 2023 at 2:29:17 PM permalink
Quote: Ace2

165/557 is a fraction used so often in craps problems that I have it memorized. If, for instance, you bet DP only and assume 100 rolls per hour then you can expect 100 * 165/557 =~ 30 bets resolved per hour. I just use 0.3 for calculations in my head

Ah, thank you. That helped me find a mistake in my work. We now agree.
“Extraordinary claims require extraordinary evidence.” -- Carl Sagan
Ace2
Joined: Oct 2, 2017
• Posts: 2664
November 25th, 2023 at 3:06:56 PM permalink
Quote: Wizard

A craps electronic table game awards points upon bets resolved, including the odds. The game has double odds. As usual, when laying odds, the player may lay up to the amount that would result in a win 2x the don't pass/come bet (in other words 4x on a 4 or 10, 3x on a 5 or 9, and 2.4x on a 6 or 8).

There are four players. All make \$1 pass, don't pass, come and don't come bets and back them up the full way on the odds.

• Alice makes one pass bet only
• Bob makes one don't pass bet only
• Cathy makes a pass or don't pass bet every roll
• David makes a don't pass or don't come bet every

What is the average wager resolved per throw of each player?

May I suggest the following extra credit problems.

Cathy plays 3-point Molly instead of always coming. What is her average wager resolved per roll?

What is her per-roll expectation and standard deviation?

*3-point Molly means you play the passline and make come bets until you have wagers on three numbers total, including the passline point
It’s all about making that GTA
Wizard
Joined: Oct 14, 2009
• Posts: 25961
Thanks for this post from:
November 26th, 2023 at 5:32:28 AM permalink
This was a puzzle on the last Survivor episode. Players had 3 minutes to do it. I didn't time it, but I think it took me about double that. On the show one of three players solved it in time.

Add three + symbols and one - symbol for this equation to be true:

9 8 7 6 5 4 3 2 1 = 100
“Extraordinary claims require extraordinary evidence.” -- Carl Sagan
aceside
Joined: May 14, 2021
• Posts: 373
November 26th, 2023 at 5:59:34 AM permalink

9 8 -7 6 +5 4 +3 +2 1 = 100

6 min
charliepatrick
Joined: Jun 17, 2011
• Posts: 2933
November 26th, 2023 at 6:10:57 AM permalink
I got it within a minute by initially seeing that 65-43-21 = 1 and the trick was to use similar logic. For instance this combined with something like 98 and 7 could get close. Playing around I saw 76 was close to 54+21 and could also create 1, leaving 98+3-1 which would be 100. This gives 98 + 3 - (76 - 54 - 21) or 98-76+52+3+21.
Wizard
Joined: Oct 14, 2009
• Posts: 25961
November 26th, 2023 at 6:22:37 AM permalink
Quote: Ace2

Cathy plays 3-point Molly instead of always coming. What is her average wager resolved per roll?

What is her per-roll expectation and standard deviation?

*3-point Molly means you play the passline and make come bets until you have wagers on three numbers total, including the passline point

I get 1.668579 unit wagers resolved per throw. I admit this was done by simulation. I'd be very interested in your method of an exact calculation.
“Extraordinary claims require extraordinary evidence.” -- Carl Sagan
ThatDonGuy
Joined: Jun 22, 2011
• Posts: 5947
November 26th, 2023 at 8:08:23 AM permalink
Quote: Wizard

This was a puzzle on the last Survivor episode. Players had 3 minutes to do it. I didn't time it, but I think it took me about double that. On the show one of three players solved it in time.

Add three + symbols and one - symbol for this equation to be true:

9 8 7 6 5 4 3 2 1 = 100

Took me about 90 seconds, but then again, I have seen my share of these type of problems:

98 - 76 + 54 + 3 + 21 = 100

Ace2
Joined: Oct 2, 2017
• Posts: 2664
November 26th, 2023 at 9:29:22 AM permalink
Quote: Wizard

Quote: Ace2

Cathy plays 3-point Molly instead of always coming. What is her average wager resolved per roll?

What is her per-roll expectation and standard deviation?

*3-point Molly means you play the passline and make come bets until you have wagers on three numbers total, including the passline point

I get 1.668579 unit wagers resolved per throw. I admit this was done by simulation. I'd be very interested in your method of an exact calculation.

On average, a shooter takes 1.5 rolls to establish a point and then 6 rolls to roll a seven. If playing a 2-point molly, he’d always have 2 bets on the table (pass + come or point) except for come out rolls. So wagers increase by 6 / (1.5 + 6) = 80% relative to betting PL only.

For a 3-point molly, we need to calculate how often the shooter, after establishing a point, will roll (not necessarily make) at least one additional point before sevening out. If, for example, he establishes a point of 5 or 9, then there is a (24-4)/(30-4) = 10/13 chance he will roll at least one additional point during his turn. The probabilities for points 4&10 and 6&8 are 7/9 and 19/25 respectively. Therefore, that weighed probability is 1/4 * 7/9 + 1/3 * 10/13 + 5/12 * 19/25 = 449/585.

As previously shown, 6 of 7.5 bets will make it to a 2-point molly. Of those 6, 449/585 will make it to a 3-point molly. Therefore, ((449/585 + 1) * 6) / 7.5 = 4,136 / 2,925 =~ 141% increase in wagers relative to betting PL only

So playing double odds and 3-point Molly, Cathy’s average wager per roll will be 7/3 * 165/557 * (4136/2925 + 1) = 543697 / 325845 =~ 1.6686 units

Note, I previously posted the 3-point Molly increase calculation here: https://wizardofvegas.com/forum/off-topic/general/34651-craps-3-point-molly/#post768558
It’s all about making that GTA
Wizard
Joined: Oct 14, 2009
• Posts: 25961
November 26th, 2023 at 4:06:27 PM permalink
Quote: Ace2

On average, a shooter takes 1.5 rolls to establish a point and then 6 rolls to roll a seven. If playing a 2-point molly, he’d always have 2 bets on the table (pass + come or point) except for come out rolls. So wagers increase by 6 / (1.5 + 6) = 80% relative to betting PL only.

For a 3-point molly, we need to calculate how often the shooter, after establishing a point, will roll (not necessarily make) at least one additional point before sevening out. If, for example, he establishes a point of 5 or 9, then there is a (24-4)/(30-4) = 10/13 chance he will roll at least one additional point during his turn. The probabilities for points 4&10 and 6&8 are 7/9 and 19/25 respectively. Therefore, that weighed probability is 1/4 * 7/9 + 1/3 * 10/13 + 5/12 * 19/25 = 449/585.

As previously shown, 6 of 7.5 bets will make it to a 2-point molly. Of those 6, 449/585 will make it to a 3-point molly. Therefore, ((449/585 + 1) * 6) / 7.5 = 4,136 / 2,925 =~ 141% increase in wagers relative to betting PL only

So playing double odds and 3-point Molly, Cathy’s average wager per roll will be 7/3 * 165/557 * (4136/2925 + 1) = 543697 / 325845 =~ 1.6686 units

Note, I previously posted the 3-point Molly increase calculation here: https://wizardofvegas.com/forum/off-topic/general/34651-craps-3-point-molly/#post768558

That is very good stuff, thank you! You have humbled me once again.

p.s. At least my simulator worked properly.
Last edited by: Wizard on Nov 26, 2023
“Extraordinary claims require extraordinary evidence.” -- Carl Sagan
Wizard
Joined: Oct 14, 2009
• Posts: 25961
November 27th, 2023 at 4:17:13 AM permalink
Here are my average bets per roll for various odds and max points.

0X Odds

Points Do Don’t
1 0.296230 0.296230
2 0.533214 0.533214
3 0.715104 0.715104
Max 1.000000 1.000000

1X Odds

Points Do Don’t
1 0.493716 0.592460
2 0.888689 1.066427
3 1.191840 1.430208
Max 1.666667 2.000000

2X Odds

Points Do Don’t
1 0.691203 0.888689
2 1.244165 1.599641
3 1.668576 2.145311
Max 2.333333 3.000000

3-4-5X Odds

Points Do Don’t
1 1.036804 1.481149
2 1.866248 2.666068
3 2.502863 3.575519
Max 3.500000 5.000000
“Extraordinary claims require extraordinary evidence.” -- Carl Sagan
Gialmere
Joined: Nov 26, 2018
• Posts: 2804
November 27th, 2023 at 7:02:17 AM permalink

It's Easy Monday. Here's another trilogy of tiny teasers...

What is the best full house? Suppose you are playing poker and a genie offers to arrange the deal so that you receive the full house of your choice.

What hand should you specify?

You’re about to roll five regular dice.

Which is more likely, rolling no sixes or rolling exactly one six?

Who mates in 1?
Have you tried 22 tonight? I said 22.
chevy
Joined: Apr 15, 2011
• Posts: 138
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November 27th, 2023 at 7:29:09 AM permalink
Quote: Gialmere

It's Easy Monday. Here's another trilogy of tiny teasers...

What is the best full house? Suppose you are playing poker and a genie offers to arrange the deal so that you receive the full house of your choice.

What hand should you specify?

You’re about to roll five regular dice.

Which is more likely, rolling no sixes or rolling exactly one six?

Who mates in 1?

A, A, A, 10, 10. where one of the 10's is the suit of the missing Ace? to minimize on hands that will beat this full house?

The same?
no 6.....(5/6)^5
exactly one 6.......5 * (1/6) * (5/6)^4

White?

charliepatrick
Joined: Jun 17, 2011
• Posts: 2933
November 27th, 2023 at 7:30:33 AM permalink
Interesting as I get it as whoever moves next. White can N-g6; while Black can BxP (discovered check)
charliepatrick
Joined: Jun 17, 2011
• Posts: 2933
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November 27th, 2023 at 7:35:04 AM permalink
This one relies on the fact that no sixes gives each die five options, so the answer = 55=3125.
One die being six works out as four die (non-sixes) which each have the five options, but there's the possibility of five die that could have been the six. So it's also 3125.
Five sixes = 1
Four sixes = 25
Three sixes = 250
Two sixes = 1250
One six = 3125
No sixes = 3125
Total = 7776.
Joeman

Joined: Feb 21, 2014
• Posts: 2375
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November 27th, 2023 at 7:36:09 AM permalink
I would want A's full of 9's. All A's full hands can only be beaten by 4oak or higher. Having the 9's blocks more straight flush possibilities than the highest full house (AAAKK). Actually, any pair of 6's thru 9's would eliminate the same number of SF possibilities, but blocking the higher ones is preferable since opponents are more likely to fold lower cards.

Both Black & White can mate in 1, depending on whose turn it is.

White can win by either moving his knight to G6, directly attacking Black's king, or by moving his bishop either way for a discovered checkmate.

Black can win by taking White's pawn with his bishop for a discovered checkmate.
"Dealer has 'rock'... Pay 'paper!'"
charliepatrick
Joined: Jun 17, 2011
• Posts: 2933
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November 27th, 2023 at 7:51:24 AM permalink
Obviously (see below) you want the best full house, so will want AAAxx. For winning purposes (against any given hand) it doesn't matter which pair you have. The objective is therefore to reduce the chances of your opponents getting a winning hand.

You can't do anything about quads, since any rank of quad will beat you and you have to leave eleven, but you can reduce the number of possible straight flushes.

The Aces have removed AKQJT and A2345 in three of the suits. Therefore your best chances are now to remove as many others as possible. I get it that any pair, 9 thru 6, would get rid of five, in their suit, so are most useful. (e.g. 9 KQJT9 QJT98 JT987 T9876 98765.) (A ten or five of the same suit as an Ace would only manage four, since one has already been eliminated from the Ace.)

What is interesting in whether something like TTT55, which gets rid of more SFs, would let in too many Full Houses. The answer is yes because there are, for instance, 4*6 ways to get AAAKK. Thus the primary aim is to have AAAxx.

So the answer is AAAxx where x is 9 8 7 or 6, the actual suits used being irrelevant.
gordonm888
Joined: Feb 18, 2015
• Posts: 4809
November 27th, 2023 at 8:14:09 AM permalink

The key is to understand that:

1. with 9 numbers and 4 digits you are mostly adding or subtracting two-digit numbers
2. The high digits at the beginning of the sequence cannot be added, i.e., 98+77 yields a sum too high to be reduced to a number that can reach 100. Thus, one should explore sequences in which the minus sign is used first to reduce 4 high digits to a relatively small sum.

Therefore:

98 - 76 + 54 + 3 + 21 = 100

Both can mate in one move.
White: N to G6 mate
Black: BxF6 mate
Last edited by: gordonm888 on Nov 27, 2023
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
aceside
Joined: May 14, 2021
• Posts: 373
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November 27th, 2023 at 8:29:55 AM permalink
I haven’t fully understood the best full house question. Do you mean Texas Hold’Em poker or the Mississippi Stud type of poker? The former has five community cards for every player to share, but the latter has only three community cards. It looks like it is for Texas Hold’Em.
Gialmere
Joined: Nov 26, 2018
• Posts: 2804
November 27th, 2023 at 8:41:50 AM permalink
Quote: aceside

I haven’t fully understood the best full house question. Do you mean Texas Hold’Em poker or the Mississippi Stud type of poker. The former has five community cards for every player to share.

I don't think it matters, but assume you're playing simple 5-card draw.
-------------------------
In other news...

The Chess puzzle remains unsolved.
Have you tried 22 tonight? I said 22.
unJon
Joined: Jul 1, 2018
• Posts: 4344
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November 27th, 2023 at 8:47:56 AM permalink
Quote: Gialmere

Quote: aceside

I haven’t fully understood the best full house question. Do you mean Texas Hold’Em poker or the Mississippi Stud type of poker. The former has five community cards for every player to share.

I don't think it matters, but assume you're playing simple 5-card draw.
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In other news...

The Chess puzzle remains unsolved.

Clever re chess:

Black mates in 1. It must be blacks turn because white must have just moved as there is no last legal move black could have made given the board setup.
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gordonm888
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November 27th, 2023 at 9:10:57 AM permalink

I don't see any possible way that this chess configuration could exist. Specifically, it is almost impossible for the White King to have moved into its position on E6 -where it is now incapable of moving.

Also, why hasn't White checkmated Black on its previous move by either N-F6 or B|G8-H7? One of those Mate possibilities must have existed on the prior move, right?

Also, Black must have been able to CheckMate White on its previous move with BxP.

So, the situation is impossible/

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Wizard
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November 27th, 2023 at 9:25:56 AM permalink

No sixes = (5/6)^5 = 40.1878%
One six = 5*5^4/6^5 = 40.1878%

Interesting. I would have expected that with six dice, but not five.
“Extraordinary claims require extraordinary evidence.” -- Carl Sagan
Wizard
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November 27th, 2023 at 9:28:37 AM permalink

Both.
White plays Ng6.
Black plays g6.
“Extraordinary claims require extraordinary evidence.” -- Carl Sagan
Wizard
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November 27th, 2023 at 9:33:34 AM permalink

This question was poorly phrased, if I may say so. It depends on what form of poker.

My interpretation is that it's simple five-card stud. Your opponents gets five random cards and that's it.

That said, my answer is aces high full house, including a pair of 6's to 9's. For example AAA77.

The only hands that will beat you are a four of a kind and straight flush. Nothing you can do about the four of a kinds. However, the middle pairs serve as a blocker to straight flushes.
“Extraordinary claims require extraordinary evidence.” -- Carl Sagan
Mental
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November 27th, 2023 at 10:44:16 AM permalink
Quote: Gialmere

It's Easy Monday. Here's another trilogy of tiny teasers...

What is the best full house? Suppose you are playing poker and a genie offers to arrange the deal so that you receive the full house of your choice.

What hand should you specify?

You’re about to roll five regular dice.

Which is more likely, rolling no sixes or rolling exactly one six?

Who mates in 1?

You need to specify what kind of poker. Assuming it is 5-card stud, then there are multiple answers. 6h 6s As Ad Ac or As 9d Ad 9c Ac are equally good, right? Both are beaten by 1 RF, 23 SF, and 473 Quads.

The same.

Both white and black.
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Ace2
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November 27th, 2023 at 11:06:31 AM permalink
Quote: Wizard

No sixes = (5/6)^5 = 40.1878%
One six = 5*5^4/6^5 = 40.1878%

Interesting. I would have expected that with six dice, but not five.

You might be thinking in terms of a Poisson distribution.

If, for instance, you receive an average of four spams per day (one every six hours) then the probability of receiving both zero spams and one spam in any six-hour period is 1/e

In a binomial distribution, such as rolling these dice, that equilibrium is attained at 1/p - 1 rolls, not 1/p rolls
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Wizard
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November 27th, 2023 at 11:27:15 AM permalink
Quote: Ace2

You might be thinking in terms of a Poisson distribution.

If, for instance, you receive an average of four spams per day (one every six hours) then the probability of receiving both zero spams and one spam in any six-hour period is 1/e

In a binomial distribution, such as rolling these dice, that equilibrium is attained at 1/p - 1 rolls, not 1/p rolls

Indeed, I was thinking of Poisson. I didn't know that about discrete events.
“Extraordinary claims require extraordinary evidence.” -- Carl Sagan
rawtuff
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November 27th, 2023 at 3:39:08 PM permalink
A drunken person arrives at home and wants to get in.
There are 10 keys on his keyring and only 1 fits. He randomly pulls a key and tries to open the door.
If it doesn't fit he returns the key to the ring and randomly pulls another key again and again until he finds the one that fits.
Under these circumstances, on which try is he most likely to open the door?
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Mental
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November 27th, 2023 at 3:43:22 PM permalink
Quote: rawtuff

A drunken person arrives at home and wants to get in.
There are 10 keys on his keyring and only 1 fits. He randomly pulls a key and tries to open the door.
If it doesn't fit he returns the key to the ring and randomly pulls another key again and again until he finds the one that fits.
Under these circumstances, on which try is he most likely to open the door?

How drunk?
He is most likely to succeed on the first try. He always makes the first try. He almost never makes the 100th try. He either gets in or passes out before the later tries.
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