## Poll

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1 vote (25%) | |||

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**4 members have voted**

March 8th, 2020 at 10:39:46 AM
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There is a square dartboard of dimensions 1 by 1. A dart is thrown at it such that it can land anywhere with equal probability. Let the coordinates of where is lands be (x,y), where both x and y are uniformly and independently distributed from 0 to 1.

Let z = round(x/y)*. In other words, z = x/y, rounded to the nearest integer.

What is the probability that z is even?

Usual rules:

Notes:

* corrected. Previous formula incorrect.

Let z = round(x/y)*. In other words, z = x/y, rounded to the nearest integer.

What is the probability that z is even?

Usual rules:

- Please don't just plop a URL to a solution elsewhere until a winner here has been declared.
- All those who have won a beer previously are asked to not post answers or solutions for 24 after this posting. Past winners who must chime in early, may PM me.
- Beer to the first satisfactory answer and solution, subject to rule 2.
- Please put answers and solutions in spoiler tags.

Notes:

* corrected. Previous formula incorrect.

Last edited by: Wizard on Mar 8, 2020

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

March 8th, 2020 at 3:07:42 PM
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Looks trivial with a computer, and difficult but possible by hand. I'll shut up until tomorrow because I already won.

March 8th, 2020 at 3:11:00 PM
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I'll have a go, but I suspect it involves calculus (integration) .I'm rubbish at that, though.Quote:Wizard

What is the probability that z is even?

Take care out there.
Spare a thought for the newly poor who were happy in their world just a few days ago, but whose whole way of life just collapsed..

March 8th, 2020 at 3:45:37 PM
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Quote:OnceDearI'll have a go, but I suspect it involves calculus (integration) .I'm rubbish at that, though.

I solved it without calculus.

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

March 8th, 2020 at 4:49:04 PM
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Quote:WizardThere is a square dartboard of dimensions 1 by 1. A dart is thrown at it such that it can land anywhere with equal probability. Let the coordinates of where is lands be (x,y), where both x and y are uniformly and independently distributed from 0 to 1.

Let z = (int)(x/y). In other words, z = x/y, rounded to the nearest integer.

What is the probability that z is even?

I object to misdirection tactic: The Int() function does not round to the nearest integer: It always rounds down!

Function Round(,0) rounds nearest!

I'm going to progressively break down the target into squares and hit the centre of each square with a dart.

1 x 1 array of squares hit in the middle.

x=y therefore z=1 = odd with 0% probability of even.

2 x 2 array each hit in the middle

x1=0.25, y1=0.25, z=1 is odd

x2=0.25, y2=0.75, z=0 is even

x3=0.75, y3=0.25, z=3 is odd

x4=0.75, y4=0.75, z=1 is odd

z is even 1/4 of the time therefore 25% probability

3x3 array

x1=1/6, y1=1/6, z=1 is odd

x2=1/6, y2=3/6, z=0 is even

x3=1/6, y3=5/6, z=0 is even

x4=3/6, y1=1/6, z=3 is odd

x5=3/6, y2=3/6, z=1 is odd

x6=3/6, y3=5/6, z=0 is even

x7=5/6, y1=1/6, z=5 is odd

x8=5/6, y2=3/6, z=1 is odd

x9=5/6, y3=5/6, z=1 is odd

z is even 3/9 of the time therefore 33.3% probability

4 x 4 array

Trust me and Excel

z is even 6/16 of the time therefore 37.5% probability

HMMM Are we converging on 50%?

10 x 10 array

Trust me and Excel

z is even 43/100 of the time therefore 43% probability

HMMM Are we converging on 50%?

100 x 100 array

Trust me and Excel

z is even 4606/10000 of the time therefore 46.06% probability

HMMM Are we converging on 50%?

1000 x 1000 array

Trust me and Excel

z is even 464220/1000000 of the time therefore 46.42% probability

HMMM Are we converging SLOWLY on 50%

I'm going to guess 50% will be the eventual answer.

Damned if I can do integration though

1 x 1 array of squares hit in the middle.

x=y therefore z=1 = odd with 0% probability of even.

2 x 2 array each hit in the middle

x1=0.25, y1=0.25, z=1 is odd

x2=0.25, y2=0.75, z=0 is even

x3=0.75, y3=0.25, z=3 is odd

x4=0.75, y4=0.75, z=1 is odd

z is even 1/4 of the time therefore 25% probability

3x3 array

x1=1/6, y1=1/6, z=1 is odd

x2=1/6, y2=3/6, z=0 is even

x3=1/6, y3=5/6, z=0 is even

x4=3/6, y1=1/6, z=3 is odd

x5=3/6, y2=3/6, z=1 is odd

x6=3/6, y3=5/6, z=0 is even

x7=5/6, y1=1/6, z=5 is odd

x8=5/6, y2=3/6, z=1 is odd

x9=5/6, y3=5/6, z=1 is odd

z is even 3/9 of the time therefore 33.3% probability

4 x 4 array

Trust me and Excel

z is even 6/16 of the time therefore 37.5% probability

HMMM Are we converging on 50%?

10 x 10 array

Trust me and Excel

z is even 43/100 of the time therefore 43% probability

HMMM Are we converging on 50%?

100 x 100 array

Trust me and Excel

z is even 4606/10000 of the time therefore 46.06% probability

HMMM Are we converging on 50%?

1000 x 1000 array

Trust me and Excel

z is even 464220/1000000 of the time therefore 46.42% probability

HMMM Are we converging SLOWLY on 50%

I'm going to guess 50% will be the eventual answer.

Damned if I can do integration though

Last edited by: OnceDear on Mar 8, 2020

Take care out there.
Spare a thought for the newly poor who were happy in their world just a few days ago, but whose whole way of life just collapsed..

March 8th, 2020 at 6:30:56 PM
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P(z=even) = 0.4646 ?

March 8th, 2020 at 6:37:36 PM
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Quote:OnceDearI object to misdirection tactic: The Int() function does not round to the nearest integer: It always rounds down!

You're absolutely right. I meant z=round(x/y).

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

March 8th, 2020 at 6:38:42 PM
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Quote:ssho88P(z=even) = 0.4646 ?

I agree with this answer to four digits. However, to get full credit, and the beer, I need to see an expression of the answer as well as a solution. In other words, show your work.

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

March 8th, 2020 at 9:02:53 PM
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Can't show the step by step solution, just pure simulation results. LOL

March 8th, 2020 at 9:04:08 PM
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Fun problem.

I get (5-pi)/4

Not trying to win a beer, just having fun.

I heart Crystal Math.