Thread Rating:

Poll

16 votes (50%)
12 votes (37.5%)
5 votes (15.62%)
2 votes (6.25%)
10 votes (31.25%)
3 votes (9.37%)
6 votes (18.75%)
5 votes (15.62%)
10 votes (31.25%)
7 votes (21.87%)

32 members have voted

Wizard
Administrator
Wizard
Joined: Oct 14, 2009
  • Threads: 1387
  • Posts: 23294
December 30th, 2020 at 9:40:01 PM permalink
Proof that 1=2:

Let a=b
a^2 = ab (multiplying both sides by a)
a^2 - b^2 = ab - b^2 (subtracting b^2 from each side)
(a+b)*(a-b) = b*(a-b)
a+b = b (divide both side by a-b)
2b = b (substitute b for a)
2 = 1 (divide both sides by b)

Where is the error in my logic?
It's not whether you win or lose; it's whether or not you had a good bet.
ssho88
ssho88
Joined: Oct 16, 2011
  • Threads: 47
  • Posts: 552
December 31st, 2020 at 12:38:41 AM permalink
Quote: Wizard

Proof that 1=2:

Let a=b
a^2 = ab (multiplying both sides by a)
a^2 - b^2 = ab - b^2 (subtracting b^2 from each side)
(a+b)*(a-b) = b*(a-b)
a+b = b (divide both side by a-b)
2b = b (substitute b for a)
2 = 1 (divide both sides by b)

Where is the error in my logic?





Divide by ZERO !

charliepatrick
charliepatrick
Joined: Jun 17, 2011
  • Threads: 34
  • Posts: 2424
December 31st, 2020 at 12:43:47 AM permalink
Quote: ssho88

Divide by ZERO !

He even did it twice!
2b=b implies b=0 not 2=1!
Wizard
Administrator
Wizard
Joined: Oct 14, 2009
  • Threads: 1387
  • Posts: 23294
December 31st, 2020 at 7:54:38 AM permalink
Quote: ssho88



Divide by ZERO !



Correct!
It's not whether you win or lose; it's whether or not you had a good bet.
Wizard
Administrator
Wizard
Joined: Oct 14, 2009
  • Threads: 1387
  • Posts: 23294
December 31st, 2020 at 8:37:35 AM permalink
You have a six-sided red die with 3 sides labeled red, 1 labeled blue and 2 labeled green.
You have a six-sided blue die with 2 sides labeled red, 2 labeled blue and 2 labeled green.
You have a six-sided green die with 1 side labeled red, 4 labeled blue and 1 labeled green.

Whatever you roll on any die shall determine which die you roll next.

You start out with the red die.

You keep rolling forever.

What ratio of rolls are done with each die?
It's not whether you win or lose; it's whether or not you had a good bet.
ssho88
ssho88
Joined: Oct 16, 2011
  • Threads: 47
  • Posts: 552
December 31st, 2020 at 9:50:40 AM permalink
Quote: Wizard

You have a six-sided red die with 3 sides labeled red, 1 labeled blue and 2 labeled green.
You have a six-sided blue die with 2 sides labeled red, 2 labeled blue and 2 labeled green.
You have a six-sided green die with 1 side labeled red, 4 labeled blue and 1 labeled green.

Whatever you roll on any die shall determine which die you roll next.

You start out with the red die.

You keep rolling forever.

What ratio of rolls are done with each die?





R : B : G = 6 : 7 : 5

Just my wild guess.

Wizard
Administrator
Wizard
Joined: Oct 14, 2009
  • Threads: 1387
  • Posts: 23294
December 31st, 2020 at 9:55:46 AM permalink
Quote: ssho88



R : B : G = 6 : 7 : 5

Just my wild guess.



I disagree.
It's not whether you win or lose; it's whether or not you had a good bet.
ThatDonGuy
ThatDonGuy
Joined: Jun 22, 2011
  • Threads: 105
  • Posts: 4925
December 31st, 2020 at 4:33:36 PM permalink

Define a "run" as a sequence of consecutive die rolls that start with a red die and end with a red side being rolled
(so the next die to be rolled would be red).

Since the number of rolls is infinite, the ratio of rolls of each die is the ratio of the expected number
of rolls each die within a run.

Let R, B, and G be the expected number of rolls remaining in the run if the current die being rolled is red, blue, or green, respectively.
Let (r,b,g) be the number of red, blue, and green rolls in the current run.

R = (1, 0, 0) + 1/6 B + 1/3 G

If the current die is blue, the expected number of blue-side rolls before a non-blue side is rolled is
2/3 x (0 + 1/3 x 1 + (1/3)^2 x 2 + ...)
= 2/3 x 1/3 x (1 + 1/3 x 2 + (1/3)^2 x 3 + ...)
= 2/9 x (1 + 1/3 + (1/3)^2 + ...)^2
= 2/9 x (3/2)^2
= 1/2
1/2 of the time, the first non-blue side is red, and 1/2 of the time, it is green
B = (0, 1, 0) + (0, 1/2, 0) + 1/2 G = (0, 3/2, 0) + 1/2 G

If the current die is green, the expected number of green-side rolls before a non-green side is rolled is
5/6 x (0 + 1/6 x 1 + (1/6)^2 x 2 + ...)
= 5/6 x 1/6 x (1 + 1/6 x 2 + (1/6)^2 x 3 + ...)
= 5/36 x (1 + 1/6 + (1/6)^2 + ...)^2
= 5/36 x (6/5)^2
= 1/5
1/5 of the time, the first non-green side is red, and 4/5 of the time, it is blue
G = (0, 0, 1) + (0, 0, 1/5) + 4/5 B = (0, 0, 6/5) + 4/5 B
= (0, 0, 6/5) + 4/5 (0, 3/2, 0) + 2/5 G
3/5 G = (0, 6/5, 6/5)
G = (0, 2, 2)

B = (0, 3/2, 0) + 1/2 G = (0, 3/2, 0) + 1/2 (0, 2, 2) = (0, 5/2, 1)

R = (1, 0, 0) + 1/6 B + 1/3 G = (1, 0, 0) + 1/6 (0, 5/2, 3) + 1/3 (0, 2, 1)
= (1, 13/12, 5/6)
The ratio of red : blue : green = 12 : 13 : 10

charliepatrick
charliepatrick
Joined: Jun 17, 2011
  • Threads: 34
  • Posts: 2424
December 31st, 2020 at 6:19:46 PM permalink
Let R(x)=probability of being Red on turn x; similarly B(x) G(x).
Then the next roll, it could be red having come from Red, Blue or Green.
R(x+1)=R(x)*3/6 + B(x)*2/6 + G(x)*1/6.
The chances of being Red (or other colours) on one turn should be the same as on the next turn, i.e. R(x+1) should be the same as R(x).
For here on using R G and B.

Consequently multiplying both sides by 6.
(1) 6R = 3R+2B+G; 3R=2B+G
(2) 6B = R+2B+4G;
(3) 6G = 2R+2B+1G

Using (1) 3R=2B+G
(2)x3 18B=3R+6B+12G = 8B+13G.
10B=13G.

(1)x5: 15R=10B+5G=18G
10R=12G.

So the ratio is Green 10: Red 12: Blue 13.
Wizard
Administrator
Wizard
Joined: Oct 14, 2009
  • Threads: 1387
  • Posts: 23294
Thanks for this post from:
charliepatrick
December 31st, 2020 at 7:17:03 PM permalink
Charlie & Don -- I agree!

Hope you enjoyed the problem. I found it pretty difficult too and subject to error.

This one is definitely beer worthy, to both of you.

Happy new year!
It's not whether you win or lose; it's whether or not you had a good bet.

  • Jump to: