## 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**

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?

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.

December 31st, 2020 at 12:38:41 AM
permalink

Quote:WizardProof 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 !

December 31st, 2020 at 12:43:47 AM
permalink

He even did it twice!Quote:ssho88Divide by ZERO !

2b=b implies b=0 not 2=1!

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.

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?

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.

December 31st, 2020 at 9:50:40 AM
permalink

Quote:WizardYou 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.

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.

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

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.

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.

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!

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.