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46 members have voted

MichaelBluejay
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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
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ThatDonGuy
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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
link to original post



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
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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
link to original post


MichaelBluejay,

The adverb that gave me pause is "about":

"... 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
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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
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November 23rd, 2023 at 10:50:12 AM permalink
TinMan, please use spoilers when posting your answers here

Use this:

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

to create this:

Spoiler goes here
Dieter
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November 23rd, 2023 at 12:03:28 PM permalink
Quote: DogHand


After all, how much does a diamond ring actually weigh?
link to original post



(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
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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.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
Wizard
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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.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
charliepatrick
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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.
link to original post

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
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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.

link to original post



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

link to original post



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
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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?
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
aceside
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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
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November 25th, 2023 at 8:22:10 AM permalink
Quote: Wizard

  • Cathy makes a pass or don't pass bet every roll
  • link to original post


    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
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    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.

    link to original post


    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?

    link to original post



    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.


    link to original post


    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?
    link to original post

    Have you tried 22 tonight? I said 22.
    Wizard
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    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.
    "For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
    Ace2
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    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
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    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

    link to original post



    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.
    "For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
    Ace2
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    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

    link to original post



    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.
    link to original post

    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
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    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
    link to original post



    Ah, thank you. That helped me find a mistake in my work. We now agree.
    "For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
    Ace2
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    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?
    link to original post

    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
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    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
    "For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
    aceside
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    November 26th, 2023 at 5:59:34 AM permalink

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

    6 min
    charliepatrick
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    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
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    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
    link to original post



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

    link to original post



    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
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    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
    link to original post



    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.
    link to original post

    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
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    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
    link to original post



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

    link to original post





    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
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    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
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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
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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.
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    charliepatrick
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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
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    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
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    aceside
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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
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    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.
    link to original post


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

    The Chess puzzle remains unsolved.
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    unJon
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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.
    link to original post


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

    The Chess puzzle remains unsolved.
    link to original post



    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.
    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 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/

    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 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.
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    November 27th, 2023 at 9:28:37 AM permalink

    Both.
    White plays Ng6.
    Black plays g6.
    "For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
    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.
    "For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
    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?

    link to original post


    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.

    link to original post

    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
    It’s all about making that GTA
    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
    link to original post



    Indeed, I was thinking of Poisson. I didn't know that about discrete events.
    "For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
    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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    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?
    link to original post

    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.
    This forum is more enjoyable after I learned how to use the 'Block this user' button.
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