[Guest]

1) You have a stick of length 1 unit. You cut it into three pieces randomly. What is the probability that the three pieces form a triangle?

2) A tosses coins n+1 times, and B tosses coin n times. What is the probability that A has more head than B?

[Guest]

The only way to not be able to form a triangle is to have two pieces combine to be shorter (or exactly equal depending on how you want to think about it) than the third one. Considering that these are all coming from a single stick the way to get this occurrence is to have one of our cuts be more than half of the whole stick. First we’ll mark one cut. Here we can see the stick and where we are going to put our first cut.

___|____________

Now if we are truly cutting randomly we can see that we will have a 50% chance of cutting to the left of the center for our second cut. This will give us a 50% chance of no triangle. However we also have chance of going so far to the right that we get a bad triangle and this is a little more difficult to figure. Imaging 100 cutting places if we cut on spot 1 this leaves 49 bad spots. Cut spot 2 leaves 48 bad and so on until spot 49 leaves 1 bad spot. I hope that we can see that if we sum all of these bad spots and average them we will end up with an average of 25% bad spots too far too the right. So this leaves a total of 75% bad spots. (If the first cut is too the right of center just flip the stick.)

75% bad means 25% good. There is a 25% chance that you will be able to form a triangle. This is of course assuming two completely random cuts, i.e. you have no info about the first cut when marking the second cut.

[Yoruichi-san]

With help from my friend Leroy...

1. 1/4

No piece can be larger than 1/2, and no piece can be smaller than the difference between the other two. So, calling the location of the two cuts x and y, which are both randomly distributed over the interval [0.1], we can draw a square which represents the possible x and y. Then divide this square into fourths with lines at x=1/2 and y=1/2. The condition that no piece can be greater than 1/2 rules out the square x>1/2,y>1/2, and the square x<1/2,y<1/2. Now we are left with two fourths.

The last condition becomes |x-y|<1/2, i.e., the maximum difference is when one piece is almost 1/2 and the other piece is infinitesimally small. Rearranging, this condition is mathematically equivalent to y<x+1/2 and y>x-1/2 which cuts away half of the two remaining squares, leaving two triangles of area 1/8 each. So total probability is 2*1/8=1/4.

2. 1/2

For a coin toss, there are 2 possible outcomes (heads or tails). For n coin tosses, each toss can be heads or tails, so the number of different possible outcomes is 2^n. For two people (A and B) each tossing n coins, there are (2^n)*(2^n) = 2^(2n) possible outcomes.

There are three possible cases considering the number of heads of A vs. B. A>B, A=B, A<B. Since the probability of heads is constant for both, the number of cases of A>B equals the number of cases of A<B, lets call this x. Call the number of cases of A=B y. So there are x+y+x=2^(2n) total possible outcomes.

Now we add the n+1th coin toss.Now we have a total of (2^(n+1)*2^n)=2^(2n+1) possible outcomes. For A to have more heads, then if we get heads for the n+1th toss, we need originally A>B or A=B, so x+y cases. If it is tails, we need the original count to be A>B, so x cases. So A gets more heads in a total of (x+y)+x=2^(2n) cases.

So the probability of A getting more heads is 2^(2n)/2^(2n+1)=1/2.

[Guest]

A good way to think of this is imagine the coin is tossed a million times for B and a million and one for A. We know that as numbers get large averages become more and more true to form. This means that the first million tosses for each will average out to both have the exact same amount of heads. And the one extra toss for A will be heads %50 percent of the time.

[Guest]

1) You have a stick of length 1 unit. You cut it into three pieces randomly. What is the probability that the three pieces form a triangle?

100% poss to form a triangle, just not end, if overlap or passing of ends are allowed

[Prime]

I do not see any way to solve the stick problem without differential equations. So here it goes:

The only way not to be able to construct a triangle is to have one of the cuts greater than 1/2 of the total stick. There are no other conditions.

Order the cuts as following: first cut cuts off the first triangle side from the right edge, and the second cut divides the remainder on the left into two.

The probability for the first cut to land in the right half of the stick is P1 = 1/2. (The halves are of equal length.)

The second cut then must land somewhere in the right half, then neither segment exceeds 1/2 of the total length. The probability of that depends on where the first cut was made.

Suppose, the first cut was made at point k, where k is a fraction of the total length. The probability of a "good" second cut is the ratio of the length of the segment where we want that cut (1/2 of the total length – the left half) to the length of the remaining segment (1 – k). So for each individual k, P_k = (1/2)/(1 – k). Where 0 <= k <= 1/2.

The average of that function is the probability we seek. The average of a function in a given interval is the integral of that function divided by the length of the interval. The interval for k is from 0 to 1/2, as already stated. Thus the probability for the second "good" cut is:

1/(1/2) * Integral ((1/2)/(1-k))dk = 2*(1/2)*Integral(1/(1-k))dk = -ln(1-k).

Now plugging in the function boundaries for the k, we get:

P₂ = -ln(1 – 1/2) – (-ln(1 - 0)) = -ln(1/2) = ln(2) ~= 0.6931.

Multiplying the two probabilities for the "good" first and second cuts we get the answer: (1/2)*ln(2) ~= 0.3466, or just under 35%.

So the probability to cut the stick is about 35%.

A very economical coin problem solution has been already provided by Yoruichi-san. I would like to stress one point implicit therein. Some may be under impression that with equal number of coin tosses, probability for A to have more heads than B is 1/2 and vice versa. It is not, as evident from Y’s solution.

[Guest]

There are two different ways to interpret the stick problem:

A. One cut is made whose position is a random variable uniformly distributed on (0,1). A second cut is made whose position is also uniformly distributed on (0,1).

B. One cut is made whose position is a random variable, call it K, uniformly distributed on (0,1). A second cut is made whose position is uniformly distributed on (k,1).

These two interpretations give different solutions.

This is the way YS did it, thinking of the problem as a square divided into four quadrants.

The answer is 1/4.

This is the way Prime looked at it, although I think he made a mistake. After the first cut at k (which must be <1/2), you have to make the next cut (call it j) so that each segment is also <1/2. In particular, this means that the distance from k must be less than 1/2, so j <k+1/2. It also must be within 1/2 of 1, so j>1/2.

This leaves j the interval (1/2,k+1/2), which has a length of k. The length of the interval from which j is chosen is 1-k. This means that, for any k, the probability of j being "good" is k/(1-k).

To solve this, we integrate that probability over 0,1/2 (because this is where k is "good"). Int(0,1/2,k/(1-k)dk) comes out to -1/2-ln1/2, which is about .19.

This is lower than in interpretation A. One way to think about it is that in A, the second cut is more likely to fall in the section where it forms a triangle since it can go in either. In option B, though, if k>1/2, there is no chance j can fall in that bigger window to make a good triangle.

I don't think there is any reason to say one interpretation is more correct than the other. My first instinct was A, but I think B is reasonable as well.

[bonanova]

The first question was posed on April 6 2008 and answered [p=1/4] here.

The interpretation was that two points were chosen at random on the line segment and the cuts were made at the points.

The coin toss problem sounds familiar but I didn't find it in search.

[Prime]

There are two different ways to interpret the stick problem:

A. One cut is made whose position is a random variable uniformly distributed on (0,1). A second cut is made whose position is also uniformly distributed on (0,1).

B. One cut is made whose position is a random variable, call it K, uniformly distributed on (0,1). A second cut is made whose position is uniformly distributed on (k,1).

These two interpretations give different solutions.

This is the way YS did it, thinking of the problem as a square divided into four quadrants.

The answer is 1/4.

This is the way Prime looked at it, although I think he made a mistake. After the first cut at k (which must be <1/2), you have to make the next cut (call it j) so that each segment is also <1/2. In particular, this means that the distance from k must be less than 1/2, so j <k+1/2. It also must be within 1/2 of 1, so j>1/2.

This leaves j the interval (1/2,k+1/2), which has a length of k. The length of the interval from which j is chosen is 1-k. This means that, for any k, the probability of j being "good" is k/(1-k).

To solve this, we integrate that probability over 0,1/2 (because this is where k is "good"). Int(0,1/2,k/(1-k)dk) comes out to -1/2-ln1/2, which is about .19.

This is lower than in interpretation A. One way to think about it is that in A, the second cut is more likely to fall in the section where it forms a triangle since it can go in either. In option B, though, if k>1/2, there is no chance j can fall in that bigger window to make a good triangle.

I don't think there is any reason to say one interpretation is more correct than the other. My first instinct was A, but I think B is reasonable as well.

Yes, I did mess up the equation for the interpretation B (where you make first random cut on the whole stick, and the second on the remainder). It should be, like you show Int(0,1/2,k/(1-k)dk).

Interestingly, for the interpretation A an integral works as well and it evaluates 1/4:

If the distance between two cuts is x, then the probability for each individual distance is 2(1-x). Whereas probability for each individual x to built triangle is again x/(1-x). Muliplying the two, we get instanteneous probability 2x. The integral of that with x in the interval from 0 to 1/2 evaluates to 1/4.

[Yoruichi-san]

The first question was posed on April 6 2008 and answered [p=1/4] here.

The interpretation was that two points were chosen at random on the line segment and the cuts were made at the points.

The coin toss problem sounds familiar but I didn't find it in search.

Lol...well, thanks for not pointing this out three months ago and ruining my fun in solving it for myself!