Easy math puzzles

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October 18th, 2020 at 5:59:28 PM permalink

Quote: unJon
Quote: gordonm888
Quote: Gialmere

I see that this week's FiveThirtyEight column containing the election puzzle also has a similar Price is Right puzzle. Note the slight rule changes...

Officials answers to be released next week.

DUPLICATE

The item is "a randomly selected value between 0 and 100." Thus if we limit ourselves to two digits beyond the decimal point, there are a total of 99.99 distinct possible choices.

If the contestant A picks 33.33 and the contestant B picks 66.67 then the contestant C has these choices:

1. pick 0.00 or 0.01 and win 33.32 percent of the time; A wins 33.34% of time, B wins 33.33
2. pick 33.32 and win 33.32 percent of the time A wins 33.34, B wins 33.33
3. pick 33.34 and win 33.33 percent of the time: A wins 33.33, B wins 33.33 (optimal for C)
4. pick 66.68 and win 33.32 percent of the time A wins 66.66; B wins 0.001

If the 1st contestant picks 33.33 and the 2nd contestant picks 66.66 then the third contestant has these choices:

1. pick 0.00 or 0.01 and win 33.32 percent of the time (because the price cannot be zero); A wins 33.33% of time, B wins 33.33
2. pick 33.32 and win 33.32 percent of the time: A wins 33.33; B wins 33.34
3. pick 33.34 and win 33.32 percent of the time: A wins 33.33; B wins 33.34
4. pick 66.67 and win 33.33 percent of the time; A wins 66.5 and B wins 0.01 (optimal for C)

If A chooses 33.32 and B chooses 66.65 then C chooses 66.66 and A wins 66.64; B wins 0.01; C wins 33.34
If A chooses 33.32 and B chooses 66.66 then C chooses 33.33 and A wins 33.32; B wins 33.34; C wins 33.33
if A chooses 33.34 and B chooses 66.67 then C would choose either 0 or 33.33: A wins 33.33, B wins 33.33, C wins 33.33
if A chooses 33.34 and B chooses 33.33 then C would choose 33.35: A wins 0.01, B wins 33.33, C wins 66.65

IT is OPTIMAL for A to chooses 33.33; it is OPTIMAL for B to then choose either 66.67 or 66.66. In both these cases, A will win 33.33/99.99 or 1/3 of the time.

I don�t think that works:

If A picks 33.33 then B can guarantee 33.32 by picking 0 and C will pick 33.34. Leaving A with 0.00.

I think A wants to pick 66.68.

If B picks 0, then B will get only 33.32 after C picks 33.34.
On the other hand, if B picks 66.66, then B can get at least 33.33.
Remember that all three players are playing to maximize their own chance of winning.

unJon

unJon

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October 18th, 2020 at 6:09:10 PM permalink

Quote: ThatDonGuy
Quote: unJon
Quote: gordonm888
Quote: Gialmere

I see that this week's FiveThirtyEight column containing the election puzzle also has a similar Price is Right puzzle. Note the slight rule changes...

Officials answers to be released next week.

DUPLICATE

The item is "a randomly selected value between 0 and 100." Thus if we limit ourselves to two digits beyond the decimal point, there are a total of 99.99 distinct possible choices.

If the contestant A picks 33.33 and the contestant B picks 66.67 then the contestant C has these choices:

1. pick 0.00 or 0.01 and win 33.32 percent of the time; A wins 33.34% of time, B wins 33.33
2. pick 33.32 and win 33.32 percent of the time A wins 33.34, B wins 33.33
3. pick 33.34 and win 33.33 percent of the time: A wins 33.33, B wins 33.33 (optimal for C)
4. pick 66.68 and win 33.32 percent of the time A wins 66.66; B wins 0.001

If the 1st contestant picks 33.33 and the 2nd contestant picks 66.66 then the third contestant has these choices:

1. pick 0.00 or 0.01 and win 33.32 percent of the time (because the price cannot be zero); A wins 33.33% of time, B wins 33.33
2. pick 33.32 and win 33.32 percent of the time: A wins 33.33; B wins 33.34
3. pick 33.34 and win 33.32 percent of the time: A wins 33.33; B wins 33.34
4. pick 66.67 and win 33.33 percent of the time; A wins 66.5 and B wins 0.01 (optimal for C)

If A chooses 33.32 and B chooses 66.65 then C chooses 66.66 and A wins 66.64; B wins 0.01; C wins 33.34
If A chooses 33.32 and B chooses 66.66 then C chooses 33.33 and A wins 33.32; B wins 33.34; C wins 33.33
if A chooses 33.34 and B chooses 66.67 then C would choose either 0 or 33.33: A wins 33.33, B wins 33.33, C wins 33.33
if A chooses 33.34 and B chooses 33.33 then C would choose 33.35: A wins 0.01, B wins 33.33, C wins 66.65

IT is OPTIMAL for A to chooses 33.33; it is OPTIMAL for B to then choose either 66.67 or 66.66. In both these cases, A will win 33.33/99.99 or 1/3 of the time.

I don�t think that works:

If A picks 33.33 then B can guarantee 33.32 by picking 0 and C will pick 33.34. Leaving A with 0.00.

I think A wants to pick 66.68.

If B picks 0, then B will get only 33.32 after C picks 33.34.
On the other hand, if B picks 66.66, then B can get at least 33.33.
Remember that all three players are playing to maximize their own chance of winning.

But

If A picks 33.33 and B picks 66.66, then how is B guaranteed anything? C picks 66.7, locking in 33.33 and B is hosed.

Maybe I�m thinking about it wrong.

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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October 18th, 2020 at 6:28:06 PM permalink

Quote: unJon
Quote: ThatDonGuy
Quote: unJon
Quote: gordonm888
Quote: Gialmere

I see that this week's FiveThirtyEight column containing the election puzzle also has a similar Price is Right puzzle. Note the slight rule changes...

Officials answers to be released next week.

DUPLICATE

The item is "a randomly selected value between 0 and 100." Thus if we limit ourselves to two digits beyond the decimal point, there are a total of 99.99 distinct possible choices.

If the contestant A picks 33.33 and the contestant B picks 66.67 then the contestant C has these choices:

1. pick 0.00 or 0.01 and win 33.32 percent of the time; A wins 33.34% of time, B wins 33.33
2. pick 33.32 and win 33.32 percent of the time A wins 33.34, B wins 33.33
3. pick 33.34 and win 33.33 percent of the time: A wins 33.33, B wins 33.33 (optimal for C)
4. pick 66.68 and win 33.32 percent of the time A wins 66.66; B wins 0.001

If the 1st contestant picks 33.33 and the 2nd contestant picks 66.66 then the third contestant has these choices:

1. pick 0.00 or 0.01 and win 33.32 percent of the time (because the price cannot be zero); A wins 33.33% of time, B wins 33.33
2. pick 33.32 and win 33.32 percent of the time: A wins 33.33; B wins 33.34
3. pick 33.34 and win 33.32 percent of the time: A wins 33.33; B wins 33.34
4. pick 66.67 and win 33.33 percent of the time; A wins 66.5 and B wins 0.01 (optimal for C)

If A chooses 33.32 and B chooses 66.65 then C chooses 66.66 and A wins 66.64; B wins 0.01; C wins 33.34
If A chooses 33.32 and B chooses 66.66 then C chooses 33.33 and A wins 33.32; B wins 33.34; C wins 33.33
if A chooses 33.34 and B chooses 66.67 then C would choose either 0 or 33.33: A wins 33.33, B wins 33.33, C wins 33.33
if A chooses 33.34 and B chooses 33.33 then C would choose 33.35: A wins 0.01, B wins 33.33, C wins 66.65

IT is OPTIMAL for A to chooses 33.33; it is OPTIMAL for B to then choose either 66.67 or 66.66. In both these cases, A will win 33.33/99.99 or 1/3 of the time.

I don�t think that works:

If A picks 33.33 then B can guarantee 33.32 by picking 0 and C will pick 33.34. Leaving A with 0.00.

I think A wants to pick 66.68.

If B picks 0, then B will get only 33.32 after C picks 33.34.
On the other hand, if B picks 66.66, then B can get at least 33.33.
Remember that all three players are playing to maximize their own chance of winning.

But
If A picks 33.33 and B picks 66.66, then how is B guaranteed anything? C picks 66.7, locking in 33.33 and B is hosed.

Maybe I�m thinking about it wrong.

UnJon, I agree with your objection. That branch of my solution should be eliminated.

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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October 19th, 2020 at 8:13:32 AM permalink

A, B, C, and D are all different positive integers.

A < B < C < D

Find the largest D so that

1/A + 1/B + 1/C + 1/D = 1

Have you tried 22 tonight? I said 22.

Wizard
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October 19th, 2020 at 9:31:40 AM permalink

I get 42.

1/2 + 1/3 + 1/7 + 1/42 = 1

I won't give a solution unless the answer is confirmed, but it was pretty much trial and error.

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

charliepatrick

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October 19th, 2020 at 11:26:47 AM permalink

Quote: Wizard

I get 42.

1/2 + 1/3 + 1/7 + 1/42 = 1

I won't give a solution unless the answer is confirmed, but it was pretty much trial and error.

I think you can prove it.

(i) A=2.
If A wasn't 2 then the first largest value of the first three would be 1/3+1/4+1/5 = 47/60 leaving 13/60 for 1/D. This is greater than 1/C (1/5) so this cannot be a solution.

(ii) B=3 or 4.
Similar logic that 1/2+1/5 = 3/10. As 1/C+1/D=3/10, 1/C>3/20. The only value left is 1/6 (1/7<3/20) and this doesn't create a solution.
Also 1/2+1/6 leaves 1/3, so 1/C >1/6 which contadicts 1/6 being 1/B.

(iii) 1/2 + 1/4 (do this one first for interest).
Now find 1/C + 1/D which add up to 1/4. The values for 1/C can be between 1/5 and 1/7 (as 1/8 leads to 1/D=1/8).
1/5 gives : 1/2 1/4 1/5 1/20 (10 5 4 1 / 20)
1/6 gives : 1/2 1/4 1/6 1/12 (6 3 2 1 / 12)
1/7 doesn't work

(iv) 1/2 + 1/3 .
Now find 1/C + 1/D which add up to 1/6. The values for 1/C can be between 1/7 and 1/11 (as 1/12 leads to 1/D=1/12).
1/7 gives : 1/2 1/3 1/7 1/42 (21 14 6 1 / 42)
1/8 gives : 1/2 1/3 1/8 1/24 (12 8 3 1 / 24)
1/9 gives : 1/2 1/3 1/9 1/18 (9 6 2 1 / 18)
1/10 and 1/11 don't work

Also note that 1/N - 1/(N+1) = 1/(N (N+1)) so 1/N = 1/(N+1) + 1/(N (N+1)). In this case N=6 gives 1/6 = 1/7+1/42.

Gialmere

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October 19th, 2020 at 4:27:15 PM permalink

Quote: Wizard

I get 42.

1/2 + 1/3 + 1/7 + 1/42 = 1

I won't give a solution unless the answer is confirmed, but it was pretty much trial and error.

Quote: charliepatrick
I think you can prove it.

(i) A=2.
If A wasn't 2 then the first largest value of the first three would be 1/3+1/4+1/5 = 47/60 leaving 13/60 for 1/D. This is greater than 1/C (1/5) so this cannot be a solution.

(ii) B=3 or 4.
Similar logic that 1/2+1/5 = 3/10. As 1/C+1/D=3/10, 1/C>3/20. The only value left is 1/6 (1/7<3/20) and this doesn't create a solution.
Also 1/2+1/6 leaves 1/3, so 1/C >1/6 which contadicts 1/6 being 1/B.

(iii) 1/2 + 1/4 (do this one first for interest).
Now find 1/C + 1/D which add up to 1/4. The values for 1/C can be between 1/5 and 1/7 (as 1/8 leads to 1/D=1/8).
1/5 gives : 1/2 1/4 1/5 1/20 (10 5 4 1 / 20)
1/6 gives : 1/2 1/4 1/6 1/12 (6 3 2 1 / 12)
1/7 doesn't work

(iv) 1/2 + 1/3 .
Now find 1/C + 1/D which add up to 1/6. The values for 1/C can be between 1/7 and 1/11 (as 1/12 leads to 1/D=1/12).
1/7 gives : 1/2 1/3 1/7 1/42 (21 14 6 1 / 42)
1/8 gives : 1/2 1/3 1/8 1/24 (12 8 3 1 / 24)
1/9 gives : 1/2 1/3 1/9 1/18 (9 6 2 1 / 18)
1/10 and 1/11 don't work

Also note that 1/N - 1/(N+1) = 1/(N (N+1)) so 1/N = 1/(N+1) + 1/(N (N+1)). In this case N=6 gives 1/6 = 1/7+1/42.

Correct!

The fun little twist in this one is that it first asks for the largest number and then quickly turns it on its head by shoving it into a fraction and asking for the smallest value. Clever tweak.

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

Have you tried 22 tonight? I said 22.

Wizard
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October 19th, 2020 at 4:36:17 PM permalink

If you ask this again, I think you should specify D can't be infinity. Because 1/2 + 1/3 + 1/6 + 1/infinity = 1.

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

Gialmere

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October 19th, 2020 at 4:52:11 PM permalink

Is infinity an integer?

Have you tried 22 tonight? I said 22.

Wizard
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October 19th, 2020 at 5:05:12 PM permalink

Quote: Gialmere
Is infinity an integer?

Good question. If the answer is no, it would imply there is a largest integer, which is a contradiction in terms.

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

ThatDonGuy

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October 19th, 2020 at 6:32:35 PM permalink

Quote: Wizard
If you ask this again, I think you should specify D can't be infinity. Because 1/2 + 1/3 + 1/6 + 1/infinity = 1.

Uhhhhh...I don't think you can define 1/infinity = 0, as that would mean infinity times 0 = 1.

Gialmere

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October 20th, 2020 at 8:12:57 AM permalink

It's toughie Tuesday...

A young boy visiting his grandparent's farm sits in a square field and plays with a laser measuring tool.

He shoots the laser at various targets and finds that he is 13 meters from one of the corner posts of the field, 17 meters from the corner post diagonally opposite that one, and 20 meters from a third corner post.

What is the area of the field?

Assume the land is flat.

Have you tried 22 tonight? I said 22.

charliepatrick

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October 20th, 2020 at 9:37:03 AM permalink

Quote: Gialmere
...What is the area of the field?...

I haven't checked my sums but using the cosine forumla and that the other angle is 90-a then it leads to
Cos(a)=(400+x²-289)/40x
Cos(90-a)=Sin(a)=(400+x²-169)/40x
As Cos²+sin²=1 ...
[ 12321+222x²+x⁴ + 53361+462x²+x⁴ ] / 1600x² = 1
leads to 2x⁴-916x²+65682=0 or 1 458 32841

x²=[ 458 +- SQRT(209764-131364) ] / 2 = 229 +- SQRT(19600) = 229 +- 140 = 369 or 89.
89 is too small (as you couldn't fit in the long lines), hence the answer is 369.

Wizard
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October 20th, 2020 at 10:21:46 AM permalink

Answer is 369.

However, I had to use goal seek to get it.

It involved finding the solution to: sqrt(289-b^2) + sqrt(400-b^2) = b + sqrt(b^2-231)

I can't think of any pure way to solve that.

The goal seek answer is b = 15.61737607.

It is easy to get from there the side of the square is sqrt(369).

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

ThatDonGuy

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October 20th, 2020 at 12:25:34 PM permalink

Let ABCD be the square, and X the location of the boy; AX = 17, BX = 20, and CX = 13.
(Exercise for the reader: show that DX = sqrt(58), and this would be true if ABCD was any rectangle and not necessarily a square.)

Let E, F, and G be points on sides AB, BC, and CD respectively, such that XE is perpendicular to AB, XF is perpendicular to BC, and XG is perpendicular to CD.
Let s be the length of AB, and a and b be the lengths of AE and BF, respectively; BE = CG = s - a, and CF = s - b
Apply the Pythagorean Theorem three times:
(s - a)^2 + b^2 = 289
a^2 + b^2 = 400
a^2 + (s - b)^2 = 169

289 = b^2 + (s - a)^2 = b^2 + a^2 - 2as + s^2 = 400 - 2as + s^2, so a = (111 + s^2) / 2s
169 = a^2 + (s - b)^2 = a^2 + b^2 - 2bs + s^2 = 400 - 2bs + s^2, so b = (231 + s^2) / 2s
400 = a^2 + b^2 = ((111 + s^2) / 2s)^2 + ((231 + s^2) / 2s)^2
Let x = s^2 (and note that x is the area of the field, so it is the solution):
400 = ((111 + x) / 2 sqrt(x)))^2 + ((231 + x) / 2 sqrt(x))^2 = ( (111 + x)^2 + (231 + x)^2) / 4x
(111 + x)^2 + (231 + x)^2 = 1600 x
x^2 + 222 x + 111^2 + x^2 + 462 x + 231^2 = 1600 x
2 x^2 - (1600 - 222 - 462) x + (111^2 + 231^2) = 0
x^2 - 458 x + (111^2 + 231^2) / 2 = 0
Quadratic Equation: x = 229 +/- sqrt(458^2 - 2 * (111^2 + 231^2)) / 2 = 229 +/- 140 = 369 or 159
However, if x = 159, then a = 270 / 2 sqrt(159) and b = 390 / 2 sqrt(159), but a^2 + b^2 is not 400
(Also note that if x = 159, s < 10 sqrt(2), which means both a and b < 10 sqrt(2), but then BX could not be 20)
Therefore, the area is 369.

Gialmere

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October 20th, 2020 at 5:30:14 PM permalink

Quote: charliepatrick
I haven't checked my sums but using the cosine forumla and that the other angle is 90-a then it leads to
Cos(a)=(400+x²-289)/40x
Cos(90-a)=Sin(a)=(400+x²-169)/40x
As Cos²+sin²=1 ...
[ 12321+222x²+x⁴ + 53361+462x²+x⁴ ] / 1600x² = 1
leads to 2x⁴-916x²+65682=0 or 1 458 32841

x²=[ 458 +- SQRT(209764-131364) ] / 2 = 229 +- SQRT(19600) = 229 +- 140 = 369 or 89.
89 is too small (as you couldn't fit in the long lines), hence the answer is 369.

Quote: Wizard

Answer is 369.

However, I had to use goal seek to get it.

It involved finding the solution to: sqrt(289-b^2) + sqrt(400-b^2) = b + sqrt(b^2-231)

I can't think of any pure way to solve that.

The goal seek answer is b = 15.61737607.

It is easy to get from there the side of the square is sqrt(369).

Quote: ThatDonGuy

Let ABCD be the square, and X the location of the boy; AX = 17, BX = 20, and CX = 13.
(Exercise for the reader: show that DX = sqrt(58), and this would be true if ABCD was any rectangle and not necessarily a square.)

Let E, F, and G be points on sides AB, BC, and CD respectively, such that XE is perpendicular to AB, XF is perpendicular to BC, and XG is perpendicular to CD.
Let s be the length of AB, and a and b be the lengths of AE and BF, respectively; BE = CG = s - a, and CF = s - b
Apply the Pythagorean Theorem three times:
(s - a)^2 + b^2 = 289
a^2 + b^2 = 400
a^2 + (s - b)^2 = 169

289 = b^2 + (s - a)^2 = b^2 + a^2 - 2as + s^2 = 400 - 2as + s^2, so a = (111 + s^2) / 2s
169 = a^2 + (s - b)^2 = a^2 + b^2 - 2bs + s^2 = 400 - 2bs + s^2, so b = (231 + s^2) / 2s
400 = a^2 + b^2 = ((111 + s^2) / 2s)^2 + ((231 + s^2) / 2s)^2
Let x = s^2 (and note that x is the area of the field, so it is the solution):
400 = ((111 + x) / 2 sqrt(x)))^2 + ((231 + x) / 2 sqrt(x))^2 = ( (111 + x)^2 + (231 + x)^2) / 4x
(111 + x)^2 + (231 + x)^2 = 1600 x
x^2 + 222 x + 111^2 + x^2 + 462 x + 231^2 = 1600 x
2 x^2 - (1600 - 222 - 462) x + (111^2 + 231^2) = 0
x^2 - 458 x + (111^2 + 231^2) / 2 = 0
Quadratic Equation: x = 229 +/- sqrt(458^2 - 2 * (111^2 + 231^2)) / 2 = 229 +/- 140 = 369 or 159
However, if x = 159, then a = 270 / 2 sqrt(159) and b = 390 / 2 sqrt(159), but a^2 + b^2 is not 400
(Also note that if x = 159, s < 10 sqrt(2), which means both a and b < 10 sqrt(2), but then BX could not be 20)
Therefore, the area is 369.

Correct!

Very good.

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

Have you tried 22 tonight? I said 22.

Wizard
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October 20th, 2020 at 7:12:43 PM permalink

G,

Can you expand on how that image helps?

W

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

Gialmere

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October 20th, 2020 at 7:59:49 PM permalink

Quote: Wizard
G,

Can you expand on how that image helps?

W

Sure. Here is the official solve (although there are special cases and alternate proofs).

[Note they used an ant instead of a boy.]

Have you tried 22 tonight? I said 22.

Wizard
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October 21st, 2020 at 2:10:21 AM permalink

Quote: Gialmere
Sure. Here is the official solve (although there are special cases and alternate proofs).

I would have never have thought of that.

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

Wizard
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October 21st, 2020 at 9:25:49 AM permalink

Can anyone think of a direct way to solve for a, b, c, and d, given:

a^2 + b^2 = 289
a^2 + c^2 = 400
c^2 + d^2 = 169
b+c = a+d

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

ThatDonGuy

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October 21st, 2020 at 12:30:46 PM permalink

Somebody check my numbers here...

231 = a^2 - d^2 = (a + d)(a - d) = (c + b)(a - d) = (c + b)(a - (b + c - a)) = (c + b)(2a - b - c)
111 = c^2 - b^2 = (c + b)(c - b)

111 * 231 = 111 (c + b)(2a - b - c)
231 * 111 = 231 (c + b)(c - b)
111 (c + b)(2a - b - c) = 231 (c + b)(c - b)
Assuming c + b <> 0, 111 (2a - b - c) = 231 (c - b)
222 a - 111 b - 111 c = 231 c - 231 b
222 a + 120 b = 342 c
c = 111/171 a + 60/171 b
d = b - a + c = 231/171 b - 60/171 a

171 c = 111 a + 60 b
171 d = -60 a + 231 b

171^2 c^2 = 111^2 a^2 + 2 * 111 * 60 ab + 60^2 b^2
171^2 d^2 = 60^2 a^2 - 2 * 60 * 231 ab + 231^2 b^2

171^2 * 169 = (111^2 + 60^2) a^2 - 2 * 60 * (231-111) ab + (60^2 + 231^2) b^2

(60^2 + 231^2) b^2 - ((120 * 120) a) b + ((111^2 + 60^2) a^2 - 171^2 * 169) = 0

b = (120^2 a) / (2 (60^2 + 231^2)) +/- sqrt(120^4 a^2 - 4 (60^2 + 231^2)((111^2 + 60^2) a^2 - 171^2 * 169)) / (2 (60^2 + 231^2))
Substitute for b in 171 c = 111 a + 60 b and 171 d = -60 a + 231 b
You now have b, c, and d in terms of a

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October 21st, 2020 at 2:03:05 PM permalink

Quote: ThatDonGuy

Somebody check my numbers here...

231 = a^2 - d^2 = (a + d)(a - d) = (c + b)(a - d) = (c + b)(a - (b + c - a)) = (c + b)(2a - b - c)
111 = c^2 - b^2 = (c + b)(c - b)

111 * 231 = 111 (c + b)(2a - b - c)
231 * 111 = 231 (c + b)(c - b)
111 (c + b)(2a - b - c) = 231 (c + b)(c - b)
Assuming c + b <> 0, 111 (2a - b - c) = 231 (c - b)
222 a - 111 b - 111 c = 231 c - 231 b
222 a + 120 b = 342 c
c = 111/171 a + 60/171 b
d = b - a + c = 231/171 b - 60/171 a

171 c = 111 a + 60 b
171 d = -60 a + 231 b

171^2 c^2 = 111^2 a^2 + 2 * 111 * 60 ab + 60^2 b^2
171^2 d^2 = 60^2 a^2 - 2 * 60 * 231 ab + 231^2 b^2

171^2 * 169 = (111^2 + 60^2) a^2 - 2 * 60 * (231-111) ab + (60^2 + 231^2) b^2

(60^2 + 231^2) b^2 - ((120 * 120) a) b + ((111^2 + 60^2) a^2 - 171^2 * 169) = 0

b = (120^2 a) / (2 (60^2 + 231^2)) +/- sqrt(120^4 a^2 - 4 (60^2 + 231^2)((111^2 + 60^2) a^2 - 171^2 * 169)) / (2 (60^2 + 231^2))
Substitute for b in 171 c = 111 a + 60 b and 171 d = -60 a + 231 b
You now have b, c, and d in terms of a

Thank you. I hate to push my luck, but with four equations and four unknowns, can you get to exact values?

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

Ace2

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October 21st, 2020 at 3:35:27 PM permalink

A couple of degenerate gamblers decide to make a bet:

They will roll a standard single die until all six numbers have appeared at least once

Gambler A wins if it takes an even number of rolls and gambler B wins if it takes an odd number of rolls. It doesn�t matter who rolls.

What is the probability that gambler A wins?

Looking for an exact (fractional) answer

Last edited by: Ace2 on Oct 21, 2020

It�s all about making that GTA

gordonm888
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October 21st, 2020 at 4:39:00 PM permalink

I think I can do this without integrals.

The first number arrives on the first roll, roll #1 which is odd.

O

The second number will now arrive on either an even or odd roll:

OE
OO

The probability of OE is 6/7 and the probability of OO is 1/7.

The third number will be either: OEE, OEO, OOE or OOO. The probability of the third number having a different oddness/eveness than the 2 number is 3:1 or a probability of 3/4 of flipping the oddness and 1/4 of preserving it.

so: OEE = 6/7*1/4; OEO= 6/7 *3/4; OOE = 1/7 *3/4 and OOO = 1/7*1/4.

The fourth number will be rolled and the chance of it flipping "oddness" of the third number is 2:1 or 2/3; the chance of preserving the oddness of the third number is 1/3.

OEEE= 6/7*1/4*1/3
OEEO= 6/7*1/4*2/3
OEOE= 6/7 *3/4 *2/3
OEOO= 6/7 *3/4*1/3
OOEE = 1/7 *3/4*1/3
OOEO= 1/7 *3/4*2/3
OOOE = 1/7*1/4*2/3
OOOO = 1/7*1/4*1/3

The fifth number will be rolled and the chance of it flipping "oddness" of the fourth number is 3:2 or 3/5; the chance of preserving the oddness of the third number is 2/5. Therefore:

OEEEE= 6/7*1/4*1/3 *2/5
OEEEO= 6/7*1/4*1/3*3/5
OEEOE= 6/7*1/4*2/3*3/5
OEEOO= 6/7*1/4*2/3*2/5
OEOEE= 6/7 *3/4 *2/3 *2/5
OEOEO= 6/7 *3/4 *2/3 *3/5
OEOOE= 6/7 *3/4 *1/3 *3/5
OEOOO= 6/7 *3/4 *1/3 * 2/5
OOEEE = 1/7 *3/4*1/3 * 2/5
OOEEO = 1/7 *3/4*1/3 * 3/5
OOEOE= 1/7 *3/4*2/3 * 3/5
OOEOO= 1/7 *3/4*2/3 * 2/5
OOOEE = 1/7*1/4*2/3 *2/5
OOOEO = 1/7*1/4*2/3 *3/5
OOOOE = 1/7*1/4*1/3 * 3/5
OOOOO = 1/7*1/4*1/3 * 2/5

OKAY, for the roll that produces the last of the 6 numbers, it will have a 6:5 chance of flipping the oddness/evenness of the fifth role. So, that's 6/11 probability of flipping the oddness and 5/11 probability of preserving it.

So each of the 16 sequences above has a 6/11 probability of resulting in a flipped "parity" and a 7/11 probability of preserving the parity or oddness of the previous

I have an on-line poker game scheduled in 2 minutes. So I cnnot complete this. But if you understood what I have been babbling about, you should understand how yo calculate the requested number

Last edited by: gordonm888 on Oct 21, 2020

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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October 21st, 2020 at 6:47:58 PM permalink

Quote: gordonm888
I think I can do this without integrals.

The first number arrives on the first roll, roll #1 which is odd.

O

The second number will now arrive on either an even or odd roll:

OE
OO

The probability of OE is 6/7 and the probability of OO is 1/7.

The third number will be either: OEE, OEO, OOE or OOO. The probability of the third number having a different oddness/eveness than the 2 number is 3:1 or a probability of 3/4 of flipping the oddness and 1/4 of preserving it.

so: OEE = 6/7*1/4; OEO= 6/7 *3/4; OOE = 1/7 *3/4 and OOO = 1/7*1/4.

The fourth number will be rolled and the chance of it flipping "oddness" of the third number is 2:1 or 2/3; the chance of preserving the oddness of the third number is 1/3.

OEEE= 6/7*1/4*1/3
OEEO= 6/7*1/4*2/3
OEOE= 6/7 *3/4 *2/3
OEOO= 6/7 *3/4*1/3
OOEE = 1/7 *3/4*1/3
OOEO= 1/7 *3/4*2/3
OOOE = 1/7*1/4*2/3
OOOO = 1/7*1/4*1/3

The fifth number will be rolled and the chance of it flipping "oddness" of the fourth number is 3:2 or 3/5; the chance of preserving the oddness of the third number is 2/5. Therefore:

OEEEE= 6/7*1/4*1/3 *2/5
OEEEO= 6/7*1/4*1/3*3/5
OEEOE= 6/7*1/4*2/3*3/5
OEEOO= 6/7*1/4*2/3*2/5
OEOEE= 6/7 *3/4 *2/3 *2/5
OEOEO= 6/7 *3/4 *2/3 *3/5
OEOOE= 6/7 *3/4 *1/3 *3/5
OEOOO= 6/7 *3/4 *1/3 * 2/5
OOEEE = 1/7 *3/4*1/3 * 2/5
OOEEO = 1/7 *3/4*1/3 * 3/5
OOEOE= 1/7 *3/4*2/3 * 3/5
OOEOO= 1/7 *3/4*2/3 * 2/5
OOOEE = 1/7*1/4*2/3 *2/5
OOOEO = 1/7*1/4*2/3 *3/5
OOOOE = 1/7*1/4*1/3 * 3/5
OOOOO = 1/7*1/4*1/3 * 2/5

OKAY, for the roll that produces the last of the 6 numbers, it will have a 6:5 chance of flipping the oddness/evenness of the fifth role. So, that's 6/11 probability of flipping the oddness and 5/11 probability of preserving it.

So each of the 16 sequences above has a 6/11 probability of resulting in a flipped "parity" and a 7/11 probability of preserving the parity or oddness of the previous

I have an on-line poker game scheduled in 2 minutes. So I cnnot complete this. But if you understood what I have been babbling about, you should understand how yo calculate the requested number

You don�t need calculus for this

It�s all about making that GTA

ThatDonGuy

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October 21st, 2020 at 8:23:41 PM permalink

Quote: Wizard
Thank you. I hate to push my luck, but with four equations and four unknowns, can you get to exact values?

After some serious number crunching, and assuming all four variables are positive, I get:

b = 171/60 sqrt(400 - a^2) - 111/60 a

However, something tells me that a is the root of a quartic equation.
Substituting for b in a^2 + b^2 = 289 gets:
289 = a^2 + (171/60 sqrt{400 - a^2} - 111/60 a)^2

Wizard
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October 21st, 2020 at 8:39:23 PM permalink

Quote: ThatDonGuy
After some serious number crunching, ...

Thank you, again! I don't want to push you too hard.

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

charliepatrick

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October 21st, 2020 at 9:53:23 PM permalink

Quote: Ace2
A couple of degenerate gamblers decide to make a bet:

They will roll a standard single die until all six numbers have appeared at least once

Gambler A wins if it takes an even number of rolls and gambler B wins if it takes an odd number of rolls. It doesn�t matter who rolls.

What is the probability that gambler A wins?

Looking for an exact (fractional) answer

Look at number of misses before you move from one state to the other. An even number is 6/(n+6), and odd n/(n+6). Now look at all 32 ways of getting Even or Ofd number of misses starting with EEEEE EEEEO ... OOOOO.
The denominator will be 7*8*9*10*11 = 55440.
The numerators are the sums of all the permutations.
Even=27780 Odd=27660.

ThatDonGuy

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Posts: 6658

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October 22nd, 2020 at 10:22:40 AM permalink

Quote: ThatDonGuy
After some serious number crunching

After some more serious number crunching, it turns out that each value is the square root of a rational number

a^2 + b^2 = 289
a^2 + c^2 = 400
c^2 + d^2 = 169
a + d = b + c

Assume all of the numbers > 0

a^2 - d^2 = 231 = (a + d)(a - d) = (c + b)(a - d)
c^2 - b^2 = 111 = (c + b)(c - b)

111 * 231 = 111 (c + b) (a - d)
231 * 111 = 231 (c + b) (c - b)
If c + b = 0, then b = -c -> 289 = a^2 + b^2 = a^2 + (-c)^2 = a^2 + c^2, but this contradicts a^2 + c^2 = 400
Therefore, c <> b
111 (a - d) = 231 (c - b)
111 a - 111 d = 231 c - 231 b
111 a - 111 (b + c - a) = 231 c - 231 b
222 a + (231 - 111) b = (231 + 111) c
c = 111/171 a + 60/171 b

a^2 + (111/171 a + 60/171 b)^2 = 400

171^2 a^2 + (111 a + 60 b)^2 = 171^2 * 400
171^2 a^2 + 111^2 a^2 + 120 * 111 ab + 3600 b^2 = 171^2 * 400
171^2 a^2 + 111^2 a^2 + 120 * 111 ab + 3600 (289 - a^2) = 171^2 * 400
(171^2 + 111^2 - 3600) a^2 + 3600 * (289 - 57^2) = -120 * 111 ab
b = (3600 * (57^2 - 289) - (171^2 + 111^2 - 3600) a^2) / (120 * 111 a)

289 - a^2 = (3600 * (57^2 - 289) - (171^2 + 111^2 - 3600) a^2)^2 / (120^2 * 111^2 a^2)
Let x = a^2:
289 - x = (3600 * (57^2 - 289) - (171^2 + 111^2 - 3600) x)^2 / (120^2 * 111^2 x)
(120^2 * 111^2 x) (289 - x) = (3600 * (57^2 - 289) - (171^2 + 111^2 - 3600) x)^2

120^2 * 111^2 * 289 x - 120^2 * 111^2 * x^2 = (171^2 + 111^2 - 3600)^2 x^2 - 7200 (57^2 - 289) (171^2 + 111^2 - 3600) x + 3600^2 (57^2 - 289)^2

((171^2 + 111^2 - 3600)^2 + 120^2 * 111^2) x^2 - (7200 (57^2 - 289) (171^2 + 111^2 - 3600) + 120^2 * 111^2 * 289) x + 3600^2 (57^2 - 289)^2 = 0

This has two solutions: 243.90243902439 and 287.640449438202
However, if x = 287.640449438202, then
b^2 = 289 - x = 1.3595505617979 -> b < 2
c^2 = 400 - x = 112.359550561798 -> c < 11
b + c < 13, but a > 13, which means d < 0

a^2 =
[
(7200 (57^2 - 289) (171^2 + 111^2 - 3600) + 120^2 * 111^2 * 289)
- sqrt{((7200 (57^2 - 289) (171^2 + 111^2 - 3600) + 120^2 * 111^2 * 289))^2 - 4 ((171^2 + 111^2 - 3600)^2 + 120^2 * 111^2) 3600^2 (57^2 - 289)^2}
]
/ (2 ((171^2 + 111^2 - 3600)^2 + 120^2 * 111^2))
= 10,000 / 41

a = 100 / sqrt(41)
b = 43 / sqrt(41)
c = 80 / sqrt(41)
d = 23 / sqrt(41)

Last edited by: ThatDonGuy on Oct 22, 2020

Ace2

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October 22nd, 2020 at 11:31:48 AM permalink

Quote: charliepatrick
Look at number of misses before you move from one state to the other. An even number is 6/(n+6), and odd n/(n+6). Now look at all 32 ways of getting Even or Ofd number of misses starting with EEEEE EEEEO ... OOOOO.
The denominator will be 7*8*9*10*11 = 55440.
The numerators are the sums of all the permutations.
Even=27780 Odd=27660.

That is the correct answer.

I hadn�t thought of that method. Here�s mine

As we know, the expected rolls of a single die to get all six numbers is 6/6 + 6/5 + 6/4 + 6/3 + 6/2 + 6/1 = 14.7.

If you roll a pair of dice each time, the calculation (expected rolls to get all six numbers) is slightly more difficult. There are three possibilities after the first roll: you can get zero, one or two new numbers. For instance, if you have already hit three numbers, then you have a 9/36 chance of getting no new numbers on the next roll, a 21/36 chance of getting one and a 6/36 chance of getting two. Set up a set of linear equations for states 1 - 5, and the expected number of rolls is 70219 / 9240 =~ 7.5995 rolls.

Double that number is 15.199 rolls. It�s not 14.7 because we are rolling two dice per trial and only looking at the result after an even number of dice are rolled. So if, for instance, the sixth number was rolled on the eleventh die, we won�t see it until the twelfth is rolled. Therefore, every time the sixth number is hit on an odd dice count, one roll is added to get to even. That means 15.199 - 14.7 = 0.4989 of the wins happen on an odd roll, and 1 - 0.4989 =~ 0.5101 = 463 of 924 wins happen on an even roll

It�s all about making that GTA

Ace2

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October 22nd, 2020 at 11:31:57 AM permalink

Duplicate

This posted five times

It�s all about making that GTA

Ace2

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October 22nd, 2020 at 11:32:01 AM permalink

Duplicate

It�s all about making that GTA

Ace2

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Posts: 2706

Joined: Oct 2, 2017

October 22nd, 2020 at 11:32:18 AM permalink

Duplicate

It�s all about making that GTA

ThatDonGuy

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October 22nd, 2020 at 11:40:56 AM permalink

Quote: Ace2
Duplicate

It looks like the "recent threads" list isn't updating properly - also, when a reply is posted, the screen refreshes to a blank screen

gordonm888
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October 22nd, 2020 at 1:41:06 PM permalink

Quote: charliepatrick
Look at number of misses before you move from one state to the other. An even number is 6/(n+6), and odd n/(n+6). Now look at all 32 ways of getting Even or Ofd number of misses starting with EEEEE EEEEO ... OOOOO.
The denominator will be 7*8*9*10*11 = 55440.
The numerators are the sums of all the permutations.
Even=27780 Odd=27660.

This is exactly what i was doing in my earlier post. I just ran out of time and didn't calculate the final probabilities.

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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October 22nd, 2020 at 1:42:25 PM permalink

Quote: charliepatrick
Look at number of misses before you move from one state to the other. An even number is 6/(n+6), and odd n/(n+6). Now look at all 32 ways of getting Even or Ofd number of misses starting with EEEEE EEEEO ... OOOOO.
The denominator will be 7*8*9*10*11 = 55440.
The numerators are the sums of all the permutations.
Even=27780 Odd=27660.

This is exactly what i was doing in my earlier post. I just ran out of time and didn't calculate the final probabilities.

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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October 22nd, 2020 at 3:15:07 PM permalink

Getting back to the square plot problem...I just noticed something: the three sums of squares are not only squares themselves, but hypotenuses in Pythagorean triples:

289 = 17^2 = 15^2 + 8^2
400 = 20^2 = 16^2 + 12^2
169 = 13^2 = 12^2 + 5^2

gordonm888
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October 23rd, 2020 at 4:14:22 PM permalink

The game is Pick a Pair (played with a 52 card deck, no jokers.)

Rules: To start a hand, you are dealt two cards to the left and two cards on the right.

You�re must choose one of the two cards on the right. The card that you pick is added to your 5-card hand, while the non-selected card is discarded.

The two cards on the left are automatically part of your hand and can�t be changed.

Once you've selected a card to join the two on the left, two more cards are automatically dealt. You make the best poker hand from your 5 cards and are paid off as below:

Hand	Payout
Royal Flush	1000
Str. Flush	200
4 of a kind	100
Full House	18
Flush	15
Straight	11
3 of a kind	5
2 pair	3
1 Pair: 9's or better	2

Question: You are dealt:

Left: 9c-8d; Right: Ts-8h

Since you'll be unable to have 3 starting cards of the same suit, we can drop the suit designators.

To play optimally, which card on the right do you keep and which card do you discard? Do you draw two cards to 9-8-8? or two cards to T-9-8?

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

charliepatrick

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October 23rd, 2020 at 4:41:14 PM permalink

		T98		988
Cards	Perms	Odds	Pays	Odds	Pays
Pair(A-J)	24	2	48	2	48
Pair(T)	3	2	6	2	6
Pair(9)	3	2	6	3	9
Pair(8)	1	5	5	100	100
Pair(7-2)	36	0	0	3	108
QJ	16	11	176	0	0
J7	16	11	176	0	0
76	16	11	176	0	0
T9	9	3	27	3	27
T8	6	3	18	5	30
98	6	3	18	18	108
Tx	120	2	240	0	0
9x	120	2	240	3	360
8x	80	0	0	5	400
xy	672	0	0	0	0
TOTALS	1128		1136		1196
			1.007		1.060

Last edited by: OnceDear on Oct 24, 2020

charliepatrick

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October 23rd, 2020 at 4:46:32 PM permalink

Ignore my previous post (as the formatting goes wrong with the table)

		T98		988
Cards	Perms	Odds	Pays	Odds	Pays
Pair(A-J)	24	2	48	2	48
Pair(T)	3	2	6	2	6
Pair(9)	3	2	6	3	9
Pair(8)	1	5	5	100	100
Pair(7-2)	36	0	0	3	108
QJ	16	11	176	0	0
J7	16	11	176	0	0
76	16	11	176	0	0
T9	9	3	27	3	27
T8	6	3	18	5	30
98	6	3	18	18	108
Tx	120	2	240	0	0
9x	120	2	240	3	360
8x	80	0	0	5	400
xy	672	0	0	0	0
TOTALS	1128		1136		1196
			1.007		1.060

So the best play is to keep the pair.

ThatDonGuy

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Thanked by

October 23rd, 2020 at 5:57:48 PM permalink

Quote: charliepatrick
Ignore my previous post (as the formatting goes wrong with the table)
T98 988
Cards Perms Odds Pays Odds Pays

Pair(A-J)
24
2
48
2
48

Pair(T)
3
2
6
2
6

Pair(9)
3
2
6
3
9

Pair(8)
1
5
5
100
100

Pair(7-2)
36
0
0
3
108

QJ
16
11
176
0
0

J7
16
11
176
0
0

76
16
11
176
0
0

T9
9
3
27
3
27

T8
6
3
18
5
30

98
6
3
18
18
108

Tx
120
2
240
0
0

9x
120
2
240
3
360

8x
80
0
0
5
400

xy
672
0
0
0
0

TOTALS
1128
1136
1196
1.007 1.060

So the best play is to keep the pair.

I get slightly different numbers. For example, with the pair of 9s, if you hold 8/9/10, that is three of a kind, and if you hold 8/8/9, that is a full house, but you are counting the first one as a pair and the second one as either a pair or three of a kind depending on which table you are looking at.

gordonm888
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October 23rd, 2020 at 6:15:55 PM permalink

I get different math as well. (ex: Pairs TT-AA with 988 make 2 pair and pay more)

Last edited by: gordonm888 on Oct 23, 2020

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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October 23rd, 2020 at 7:09:45 PM permalink

For those interested, the Riddler at 538 has posted the answers to last week's Election and Price is Right puzzles. They're too lengthy to post here so...

Election & Price is Right puzzle answers.

Just scroll down to the solutions to last weeks section. [This week features basketball puzzles.]

Have you tried 22 tonight? I said 22.

charliepatrick

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October 23rd, 2020 at 7:21:15 PM permalink

Thanks - I'm being distracted by a baseball match!
I'm still assuming it's like a fruit machine or poker machine that, as an example, a good pair returns 2 units, but a loser returns 0 (rather than at a table game where they would take your wager for -1).

		T98		988
Cards	Perms	Odds	Pays	Odds	Pays
Pair(A-J)	24	2	48	3	72
Pair(T)	3	5	15	3	9
Pair(9)	3	5	15	18	54
Pair(8)	1	5	5	100	100
Pair(7-2)	36	0	0	3	108
QJ	16	11	176	0	0
J7	16	11	176	0	0
76	16	11	176	0	0
T9	9	3	27	3	27
T8	6	3	18	5	30
98	6	3	18	18	108
Tx	120	2	240	0	0
9x	120	2	240	3	360
8x	80	0	0	5	400
xy	672	0	0	0	0
TOTALS	1128		1154		1268
			1.023		1.124

gordonm888
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October 24th, 2020 at 9:28:08 AM permalink

Quote: charliepatrick
Thanks - I'm being distracted by a baseball match!
I'm still assuming it's like a fruit machine or poker machine that, as an example, a good pair returns 2 units, but a loser returns 0 (rather than at a table game where they would take your wager for -1).
T98 988
Cards Perms Odds Pays Odds Pays

Pair(A-J)
24
2
48
3
72

Pair(T)
3
5
15
3
9

Pair(9)
3
5
15
18
54

Pair(8)
1
5
5
100
100

Pair(7-2)
36
0
0
3
108

QJ
16
11
176
0
0

J7
16
11
176
0
0

76
16
11
176
0
0

T9
9
3
27
3
27

T8
6
3
18
5
30

98
6
3
18
18
108

Tx
120
2
240
0
0

9x
120
2
240
3
360

8x
80
0
0
5
400

xy
672
0
0
0
0

TOTALS
1128
1154
1268
1.023 1.124

CORRECT

Nice presentation format, calculation is transparent!

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

Gialmere

Threads: 45
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October 25th, 2020 at 1:54:16 PM permalink

Bunco is a game played with three dice. Points are scored based on the number of the round you're in. The game is 100% luck and is typically seen at ladies' social gatherings and charity fundraisers, often with prizes given for various achievements.

Your spouse has dragged you to a neighborhood Bunco night since there are some cash prizes being given our for the game. To your surprise, you have the most points at the end of the first round and therefore get a special bonus roll of the three dice in an attempt to win cash.

Since it's the first round, if one of the dice is a 1, you win $1. Two 1s wins you $2. Three 1s (a first round Bunco) wins you $21. Also, if you roll trips in 2-6 (a first round Bonco), you win $5.

On average, how much would you expect to win in this type of bonus roll and what would be the standard deviation?

Have you tried 22 tonight? I said 22.

Ace2

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October 25th, 2020 at 2:47:45 PM permalink

$0.699 +/- 1.66

It�s all about making that GTA

Gialmere