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Two points are dropped at random onto the unit interval, creating three sub intervals this way.

Find the probability that the central sub interval will be two times shorter than the right one.

I tried to solve it in the following way:

the length of the central interval $= y - x$

the length of the right interval $= 1 - y$

so, the favorable outcome would satisfy the following inequality,

$y-x \le \frac{1}{2} \cdot (1-y)$

$\Rightarrow 3y - 2x \le 1$

enter image description here

Now, how can I advance further to find the probability?

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    $\begingroup$ It's like in your previous question - the area of interest divided by the area of possibility. $\endgroup$ – Ami Tavory Oct 16 '16 at 6:23
  • $\begingroup$ I assume "two times shorter" means "half as long." Do you mean "half as long" or "less than or equal to half as long"? $\endgroup$ – jbowman Nov 30 '16 at 20:34
  • $\begingroup$ By expressing the event in terms of the interval lengths, you made the problem harder, because now you have to figure out their joint distribution. Since you already know the joint distribution of the points--and it's a very simple one--try expressing the event in terms of their coordinates. In the diagram they will comprise two triangles, each of base $1/3$ and height $1$. $\endgroup$ – whuber Dec 2 '16 at 21:49
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You look at the random variable

$$Z = 3Y - 2X \tag{1}$$

and you want to determine the probability

$$P(Z\leq 1 \mid Y > X) \tag{2}$$

Using Bayes' formula we have

$$P(Z\leq 1 \mid Y > X)= \frac {P(Z\leq 1 , Y > X)}{P(Y > X)} \tag{3}$$

$Y$ and $X$ are independent continuous uniform $U(0,1)$ random variables, so

$$P(Y > X) = \int_0^1\int_x^1dydx = \int_0^1(1-x)dx = 1-\frac 12 x^2\Big |^1_0 = 1/2 \tag{4}$$

For the numerator, we have

$$P(Z\leq 1 , Y > X) = P(3Y - 2X\leq 1 , Y > X) = P\left(Y \leq \frac {1+2X}{3} , Y > X\right)$$

$$= P\left (X \leq Y \leq \frac {1+2X}{3}\right)= \int_0^1\int_x^{\frac {1+2x}{3}}dydx = \int_0^1\left (\frac {1+2x}{3}-x\right)dx$$

$$=\frac 13 \int_0^1dx - \frac 13\int_0^1xdx = \frac 13 - \frac 16 = 1/6 \tag{5}$$

Inserting $(4)$ and $(5)$ into $(3)$ we get

$$P(Z\leq 1 \mid Y > X) = \frac {1/6}{1/2} = 1/3 \tag{6}$$

This is the mathematical approach that is equivalent to the geometric argument provided in @MikeP answer.

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I think the critical piece is to determine the universe of potential events, which isn't just [0,1] [0,1]. Since you've set the problem up with y being greater than x, the universe of possibilities is just the area above the y=x line (i.e. 1/2 unit area). Of that, the area below the line you plotted is what satisfies the even of interest. Since it starts at y=1/3, the area of the triangle from y=x to that is 1/2*1/3*1 = 1/6. Dividing 1/6 by 1/2 gives you the answer of 1/3.

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Another (!) way of solving it is as follows. Note that the size of the leftmost interval is irrelevant; all you care about is the combined rightmost and central intervals. This combined interval has some length $L$, and, by construction, is divided into two sub-intervals by having a uniform($0,L$) variable, label it $x$, "dropped" into it. The leftmost of those two sub-intervals (which corresponds to the original middle interval) is $\leq$ half as long as the right interval (which corresponds to the original right interval) if $x \leq L/3$, which will obviously happen $1/3$ of the time.

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    $\begingroup$ This description is hard to follow. After referring to three intervals, you write "this interval has some length $L$..." Which interval would "this" refer to? None of the three you just mentioned is subdivided. $\endgroup$ – whuber Dec 2 '16 at 21:35
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    $\begingroup$ Thank you for the clarification. There seems to be a gap in the argument: how do you know that the the larger of two independent uniform variables is uniformly distributed between the smaller and $1$? $\endgroup$ – whuber Dec 2 '16 at 22:03
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Somewhat hesitantly I suggest zero.

If we think of throwing the two markers (points) one after the other (a perfectly reasonable procedure), then once the first one has landed the second one must land in one out of two positions out of infinite possibilities (presuming we're dealing in fractional lengths) to make the ratio exactly 1 : 2


On the other hand, if we interpret the question to mean half or less than half, I, also hesitantly, suggest 5/12.

Thinking of the markers (points) landing one after the other again, let the first one land a distance X from the right hand end of the interval (i.e. from 1). The second one can land either side of the first one.

If it lands on the right hand side of the first one, the chance that the middle interval is a half or less than a half of the right hand one is 1/3. All in all the chance of this happening is therefore X/3.

If it lands on the left hand side of the first one, the chance that the middle interval is half or less than half the right hand one is X/2 divided by 1-X. But then the chance of the second one landing to the left of the first one is 1-X. Therefore the chance of landing to the left and leading to the desired outcome is X/2.

Adding together we get X/2 plus X/3, i.e. 5X/6. So, we need the expected value of this as X varies between 0 and 1. if we presume all values of X are equally likely we end up with 5/12

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  • $\begingroup$ This erroneous argument highlights the value of using rigorous analysis when dealing with probabilities. (The diagram in the question clearly shows the chance is $1/3$, not $5/12$, and all other answers provide convincing and varied arguments on behalf of $1/3$.) $\endgroup$ – whuber Dec 2 '16 at 21:40
  • $\begingroup$ Er, thanks whuber. I am familiar with the double integral method, so can see where 1/3 comes from. However, aside from saying that that is the correct method, can you point me at what is wrong with the method I used? I'm struggling to see why it doesn't come to the right answer. $\endgroup$ – John Dec 3 '16 at 9:11
  • $\begingroup$ "If it lands on the left hand side of the first one": to see the error in that calculation, draw a picture of that situation when $X \gt 2/3$ (that is, the first point is near the left hand side). You're basically trying to do the integral informally--which is fine--but it helps immensely to let a diagram of the event guide you. $\endgroup$ – whuber Dec 3 '16 at 17:04
  • $\begingroup$ Ah yes, thanks. However, this will only increase my final answer, taking it further away from 1/3, since landing on the left hand side for X > 2/3 makes it certain that the middle interval will be less than half of the right hand one. There must be more than one error in my informal method. Unless 1/3 isn't right? I don't doubt that the calculation of the double integral is correct, but is the setup right? 1/3 is what we get when we imagine the second one landing to the right of the first one. Landing to the left changes things. Just a thought. I'll ponder more. Thanks again. $\endgroup$ – John Dec 3 '16 at 19:32
  • $\begingroup$ Once you are convinced the correct answer is $1/3$, I believe you'll have no trouble identifying the flaw in your current argument. $\endgroup$ – whuber Dec 3 '16 at 22:59

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