Wikipedia:Reference desk/Archives/Mathematics/2015 December 3

= December 3 =

Revolver probability
Problem: given a (special) 18-shot revolver with 9 hollow point, 6 open tip and 3 ballistic tip bullets and the cylinder is spun before each shot trigger, what is the probability that there will be (exactly) one of each type of bullet remaining after 15 shots triggers? Any ideas on how I could approach this? Thanks. Sigarent (talk) 03:09, 3 December 2015 (UTC)

Firing 15 live shots followed by firing 3 dud shots has the same probability as firing 3 dud shots followed by firing 15 live shots. So what is the probability of firing 3 dud shots where each of the dud shots are one of each type of bullet? 175.45.116.61 (talk) 03:31, 3 December 2015 (UTC)

Doesn't that neglect the chance of firing on empty chambers, though? Or does it somehow not matter for this type of question? Answer to your question: HOB -> (9/18)(6/17)(3/16) = 9/272 and multiply by six for all HOB possibilities -> 27/136. sigarent (talk) 03:39, 3 December 2015 (UTC)


 * Given your question, how do you interpret the word "shot" in the phase "after 15 shots"? Does pulling a trigger on an empty cylinder count?  If so, then you need to multiply your result by the very low  probability that there will be 15 successful shots without an unsuccessful one.  (You can calculate that, right?)  If you don't count unsuccessful shots but instead spin the cylinder and try again, then you are done.  If unsuccessful shots don't count, but you spin the cylinder only once and keep pulling the cylinder until you have a successful shot before spinning again, then the problem becomes more complicated and the answer is dependent on the pattern in which the bullets are loaded. -- ToE 04:42, 3 December 2015 (UTC)


 * OK. It would appear I misstated the problem which refers to "triggers" not "shots", so a trigger need not necessarily result in a bullet being fired. Following your restatement of the problem (counting empty cylinders), we have: (9/18)(8/18)(7/18)(6/18)(5/18)(4/18)(3/18)(2/18)(6/18)(5/18)(4/18)(3/18)(2/18)(3/18)(2/18) = 175/753145430616 multiplied by the number of permutations of successful shot sequences (?). I can't mentally verify that, but am I following you right? sigarent (talk) 05:05, 3 December 2015 (UTC)


 * Giving your question, how do you interpret the word "one" in the phrase "there will be one of each type of bullet remaining"? ;^)  I initially interpreted it to mean "exactly one", but you might mean "at least one". -- ToE 05:24, 3 December 2015 (UTC)
 * "One of each type" as in exactly one H, one O, and one B. sigarent (talk) 05:26, 3 December 2015 (UTC)
 * Good. Then your problem requires that exactly three bullets remain after 15 triggers.  What is the probability that the initial 15 triggers will result in 15 successful shots? -- ToE 05:29, 3 December 2015 (UTC)

That doesn't look like my problem, but I think the answer to that would be: (1)(17/18)(16/18)(15/18)(14/18)(13/18)(12/18)(11/18)(10/18)(9/18)(8/18)(7/18)(6/18)(5/18)(4/18) = 14889875/94143178827. No? sigarent (talk) 05:36, 3 December 2015 (UTC)
 * Right. And you do agree that 15 shots in only 15 triggers is necessary (though not sufficient) for your desired condition, right?  If not then I'm misunderstanding the statement of the problem, but if so then you already calculated the probability of the three remaining bullets (once this has already happened) being the desired variety as 9C1 ⋅ 6C1 ⋅ 3C1 / 18C3 = 27/136, so what do you do with those two probabilities?  (I assume that you are familiar with the concept and notation of Combination that I just used.  Also, using Permutation notation, you could write your immediately preceding probability as 18P15 / 18^15.) -- ToE 05:48, 3 December 2015 (UTC)


 * If I still follow you, then we have: (((9 choose 1)(6 choose 1)(3 choose 1))/(18 choose 3))*((18 permute 15)/18^15) = 27/136 * 14889875/94143178827 = 8785875/22161728569728.
 * So in the case of 3 H, 2 O, 1 B, the (simpler calculation to see if this makes sense) would be ((3 choose 1)(2 choose 1)(1 choose 1))/(6 choose 3))) * ((6 permute 3)/6^3) = 1/36 but that I think is wrong, so I guess I am not quite following you. sigarent (talk) 13:28, 3 December 2015 (UTC)
 * I see two arithmetic errors. First in "27/136 * 14889875/94143178827 = 8785875/22161728569728".  I know you are reducing to lowest terms, so perhaps something when wrong there.  Second, I agree with your (3C1⋅2C1⋅1C1/6C3)(6P3/6^3) for the six shooter case, but I think your 1/36 came from accidentally calculating 6C3 instead of 6P3. -- ToE 15:57, 3 December 2015 (UTC)
 * Right you are on both counts. Thank you for the guidance! sigarent (talk) 17:32, 3 December 2015 (UTC)
 * My pleasure! For the record, the final probabilities are (9⋅6⋅3/18C3)(18P15/18^15) = 875875/27894275208 ≈ 3.14x10-5 ≈ 1/31847 for the original problem and (3·2·1/6C3)(6P3/6^3) = 1/6 for the six shooter.
 * These sorts of problems usually have several ways of looking at them which yield different but equivalent formulae. If you still aren't quite happy with what we have so far, try this approach on for size.  Number your 18 bullets.  You know that there are 9·6·3 combinations of three bullets with one of each type.  If you were to choose one such combination, what is the probability that you would be left with exactly that very combination after 15 triggers?  (15/18)(14/18)(13/18)...(2/18)(1/18) = 15!/18^15, right?  (Your first trigger needs to hit any but the chosen three, the second trigger needs to also miss one already fired round, and so on.)  So the answer is  9·6·3·15!/18^15, which you can verify is the same value we arrived at earlier. -- ToE 18:19, 3 December 2015 (UTC)
 * Remarkable. With so many ways to arrive at the answer, I probably should have pushed harder independently. Still, thanks for the help. sigarent (talk) 02:07, 4 December 2015 (UTC)

My brain is so tired after work and drinking whiskey.

9C8 * 6C5 * 3C2 * 9! * 6! * 3!/(18^15) == 175/4649045868 == 3.76421*10^-8

110.22.20.252 (talk) 12:23, 3 December 2015 (UTC)


 * Perhaps I've not had enough whiskey, but I understand that you want to know the probability that, after 15 random triggerings, exactly three rounds remain unfired (you have already correctly calculated that probability as 18P15/18^15 = 14889875/94143178827), and that those three unfired rounds include one each of the three possible varieties (you have already correctly calculated that, having already fired 15 random bullets, the probability that the remaining three include one of each variety is 9C1⋅6C1⋅3C1/18C3 = 27/136). I assume that you understand Conditional probability: P(A∩B)=P(A|B)P(B), but you are refraining from multiplying the two probabilities because you don't yet see the problem in those terms.  (Which is good.  Don't just accept a numerical answer unless you understand and agree with the process that leads there.)  Note that the resulting probability is significantly smaller than your 3.76E-8, so that it is practical to write a simulation in your favorite programming language to check that the modeled results are consistent with your calculated results.  (Coding a simulation can also lead to better understanding the math behind a problem.) -- ToE 14:26, 3 December 2015 (UTC)
 * I am not the 110 ip account above. Please respond to my last message, if you feel so inclined. Thanks. sigarent (talk) 15:19, 3 December 2015 (UTC)
 * Thanks. I thought that the whiskey bottle logged you out. -- ToE 15:57, 3 December 2015 (UTC)

There are 18 chambers and there are 15 trigger events. Each trigger event must choose a chamber that has not been used before. Thus 15 unique chambers must be chosen. Each type of bullet must be chosen (n-1) times where n is the quantity of the type of bullet. Thus 15! * (1/18)^15 * 9C8 * 6C5 * 3C2 / 18C15 = 875875/22761728569728 == 3.84801*10^-8

I think I should explain my reasoning. Suppose you have chosen 15 unique chambers. The probability of trigger1(chamber1) followed by trigger2(chamber2) and so on for 15 chambers are: (1/18)^15 Next the number of ways those 15 chambers can be sequenced is equal to the number of ways 15 runners can finished a race. Which is 15! So we have

15! * (1/18)^15

Next question is how many ways we can choose 15 chambers out of 18 chambers? The answer is 18C15

But not all those choices meets our demand that "Each type of bullet must be chosen (n-1) times where n is the quantity of the type of bullet."

In fact, our demand is a subset of the total number of ways to choose 15 chambers out of 18 chambers. So the fraction of the total ways is

9C8 * 6C5 * 3C2 / 18C15

So putting it all together to get the result we have

15! * (1/18)^15 * 9C8 * 6C5 * 3C2 / 18C15 = 875875/22761728569728 == 3.84801*10^-8

175.45.116.61 (talk) 23:42, 3 December 2015 (UTC)


 * This is actually a very helpful explanation, thank you. sigarent (talk) 02:07, 4 December 2015 (UTC)


 * I agree with most of your logic, but your final result is off by a factor of 18C15. (That is, the correct probability is 816 times more likely.)
 * 1. Yes, the probability of 15 random triggers firing a specific sequence of chambers (in order) is (1/18)^15 = 1/18^15 ≈ 1.482E-19.
 * 2. Yes, there are 15! distinct sequences in which 15 distinct chambers can be ordered, so the probability of 15 random triggers firing 15 specific chambers (in any order) and leaving three specific chambers unfired is 15!/18^15 ≈ 1.938E-7.
 * 3. Yes, there are 18C15 different choices of 15 chambers out of 18 possible, so the probability of 15 random triggers leaving exactly three unfired bullets (any bullets, not prespecified) is 18C15*15!/18^15 ≈ 1.582E-4.
 * 4. And yes, only 9C8*6C5*3C2 = 9*6*3 of those 18C15 different choices yield one remaining bullet of each type, and the desired fraction is 9*6*3/18C15 = 27/136 ≈ 0.1985.
 * 5. So you can multiply that fraction from #4 by the probability from #3 and get (9*6*3/18C15)*(18C15*15!/18^15) = (9*6*3/ 18C15 )*( 18C15 *15!/18^15) = 9*6*3*15!/18^15 = 875875/27894275208 ≈ 3.140E-5. (In words, the probability of having exactly three bullets remaining after 15 random triggers multiplied by the fraction, of all three bullet combinations, which have one bullet of each type.)
 * 6. Or you can just skip step #3 all together, and multiply 9*6*3 from #4 with the probability from #2 and get the same value: 9*6*3*15!/18^15 = 875875/27894275208 ≈ 3.140E-5. (In words, the probability of a specific group of three bullets remaining after 15 random triggers multiplied by the number of three bullet groups which include one bullet of each type.)
 * 7. But it does not make sense to multiply the fraction from #4 by the probability from #2. (In words, the probability of a specific group of three bullets remaining after 15 random triggers multiplied by the fraction, of all three bullet combinations, which have one bullet of each type.  That doesn't make sense.)  Suggesting that doing so is the answer is equivalent to stating that the probability of ending up with any of the 9*6*3 = 162 desired end results is less likely than ending up with one prespecified result.
 * -- ToE 16:23, 6 December 2015 (UTC)

Cubic roots for complex numbers, sum of square roots
1. Does anyone know how to solve algebraically only $$\sqrt[3]{2 \pm 11i} = x+yi$$ with the expression $$x^3-3xy^2 + (3x^2y-y^3)i = 2 \pm 11i$$
 * without:
 * 1. Any Guesses of the results at all
 * 2. Any Use of De Moivre's formula
 * Or is it algebraically Impossible by not using those two?

2. Is there no algebraic formula for $$\sqrt1+\sqrt2+\sqrt3+ \cdots +\sqrt{n}$$? — Preceding unsigned comment added by יהודה שמחה ולדמן (talk • contribs) 12:30, 3 December 2015 (UTC)


 * For the first question with integers you can try all the integer values for x and y for which x^2+y^2 is the cube root of that for the result. If that doesn't give a solution there is none without using square or cube roots. In general for a cubic with real coefficients a real root may not be expressible algebraically without using imaginary numbers as well the normal arithmetic operations and square and cube roots. The article Nested radical may be of interest, i can be considered a radical.


 * For the second that value will be a root of some large polynomial. You can get various powers till you get all the various different square roots on the right hand side formed by multiplying some of those roots together and then just solve that system of linear equations to give a polynomial in the root. So for square roots of 2, 3 and 5 on the right powers will give equations with square roots of 6, 10, 15 and 30 so seven in all plus one for the non square roots so we're talking about a polynomial of degree 8. Dmcq (talk) 15:56, 3 December 2015 (UTC)
 * Wha...t(!?)
 * Sorry - I didn't get a word of yours, a word!
 * May you please add a math expression to this response? Necessary for reading your point.
 * And if you know any mathematicians, I'd like to chat with them. Thanks. יהודה שמחה ולדמן (talk) 18:05, 3 December 2015 (UTC)
 * I though I had expressed myself simply and clearly the ideas are pretty elementary. I was trying to describe the ideas behind what I was saying not do a calculation. Saying 'If you know any mathematicians' is implying that I do not know what I am talking about. Even if you think for some reason that you know better than someone answering you I would ask that you try and avoid trying to insult people on the reference desks please. Dmcq (talk) 01:16, 4 December 2015 (UTC)
 * I can see how you read the OP's response as an insult, but I feel this was not his intention. I'm not quite sure what was his intention, though. -- Meni Rosenfeld (talk) 21:13, 6 December 2015 (UTC)
 * As to question 2, consider:


 * $$(x - \sqrt{1} - \sqrt{2} - \sqrt{3}) (x + \sqrt{1} - \sqrt{2} - \sqrt{3}) (x - \sqrt{1} + \sqrt{2} - \sqrt{3}) (x + \sqrt{1} + \sqrt{2} - \sqrt{3}) (x - \sqrt{1} - \sqrt{2} + \sqrt{3}) (x + \sqrt{1} - \sqrt{2} + \sqrt{3}) (x - \sqrt{1} + \sqrt{2} + \sqrt{3}) (x + \sqrt{1} + \sqrt{2} + \sqrt{3}) = x^{8} - 24 x^{6} + 128 x^{4} - 192 x^{2} + 64$$


 * (so that's all $$\scriptstyle{2^{n}}$$ possible arrangements of plus and minus signs on the square roots). So you can specify the number $$\scriptstyle{\sqrt{1} + \sqrt{2} + \sqrt{3}}$$ by saying it's the largest number $$\scriptstyle{x}$$ (out of eight of them) that satisfies


 * $$x^{8} - 24 x^{6} + 128 x^{4} - 192 x^{2} + 64 = 0$$


 * This is similar to a Swinnerton-Dyer polynomial except that those are constructed from the square roots of just the prime numbers rather than the roots of all the natural numbers. For the Swinnerton-Dyer polynomials at least, none of the solutions can be integers or rational numbers. I've previously asked here why this is so, but not yet got a complete answer. catslash (talk) 23:00, 3 December 2015 (UTC)


 * My original question was in fact slightly harder: why can't the Swinnerton-Dyer polynomials be factored over the integers? - which I think comes to why can't the sums of the square roots of the first n primes satisfy any polynomial of degree less than $$\scriptstyle{2^{n}}$$? I'm still keen to know the answer. --catslash (talk) 23:32, 3 December 2015 (UTC)
 * 1. Starting from the system of polynomial equations $$\{x^3 - 3 x y^2 - 2 = 0, 3 x^2 y - y^3 - 11 = 0\}$$ apply a Gröbner basis algorithm to transform it to $$\{64 x^9 - 96 x^6 - 3327 x^3 - 8 = 0, 64 x^7 - 112 x^4 - 3299 x + 198 y = 0\}$$. The first equation is a cubic equation in $$x^3$$, so it can be solved algebraically. Substitute the value of $$x$$ into the second equation to find $$y$$.
 * 2. Are you looking for an algebraic expression in $$n$$ similar to $$1^2 + 2^2 + 3^2 + \cdots + n^2 = \frac{n(n+1)(2n+1)}{6}$$, but for exponent $$\frac{1}{2}$$? I doubt that one exists because it would probably have been found and listed in the Faulhaber's formula article. Egnau (talk) 00:47, 4 December 2015 (UTC)

Note that if x=√1+√2+...+√n, then x2 satisfies an equation of degree 2n-1. Bo Jacoby (talk) 23:17, 4 December 2015 (UTC).

For example the equation for $$x=\sum_{k=1}^3 \sqrt k$$ was
 * $$x^8-2^3 3^1 x^6+2^7 x^4-2^6 3^1 x^2+2^6=0$$

Divide by 64
 * $$2^{-6}x^8-2^{-3} 3^1 x^6+2^1 x^4-3^1 x^2+1=0$$

Let $$x^2=-2y$$ to get
 * $$2^{-2}y^4+ 3^1 y^3+2^3 y^2+2^1 3^1 y+1= 0$$

This equation is of degree 4. Bo Jacoby (talk) 00:19, 6 December 2015 (UTC).


 * For the first question, let $$\lambda=\sqrt[3]{2+11i}$$ be the unknown root. Then $$x=\lambda^3$$ is a solution of the quadratic equation $$x^2-4x+125=0$$.  That is, $$\lambda$$ is a solution of the degree six equation $$\lambda^6-4\lambda^3+125=0$$.  Polynomial factorization gives $$\lambda^6-4\lambda^3+125=(\lambda^2-4\lambda+5)(\lambda^4+4\lambda^3+11\lambda^2+20\lambda+25)$$.  Setting the quadratic factor to zero and applying the quadratic formula gives the two solutions $$\lambda=2\pm i$$.   S ławomir  Biały  15:03, 7 December 2015 (UTC)

The equation x3−(2+11i) = 0 has the solution 2+i, but 2−i is not a solution. The factorization is x3−(2+11i) = (x−(2+i))(x2+(2+i)x+(3+4i))

The solutions to the quadratic equation x2+(2+i)x+(3+4i) = 0 are approximately −1.86603+1.23205i and −0.133975−2.23205i. Bo Jacoby (talk) 19:40, 7 December 2015 (UTC).


 * Right. This gives both conjugates $$\sqrt[3]{2\pm 11i}$$, because all polynomials are real here.   S ławomir  Biały  19:57, 7 December 2015 (UTC)

The factorization is $$x^3-2-11i= (x-2-i) (x+1-\frac{\sqrt 3}2 -\frac i 2 -i\sqrt 3) (x+1+\frac{\sqrt 3}2 -\frac i 2 +i\sqrt 3)$$ Bo Jacoby (talk) 20:50, 7 December 2015 (UTC).


 * Factorization of what? Was your post in reply to mine?  It seemed to be.   S ławomir  Biały  22:11, 7 December 2015 (UTC)

Sorry, mr. Biały, for me being unclear. I was elaborating on the factorization of the polynomial $$x^3-(2+11i)=(x-r_1)(x-r_2)(x-r_3)$$ showing the three cubic roots of 2+11i:
 * $$r_1=2+i$$
 * $$r_2=-1+\frac{\sqrt 3}2-\frac i 2-i\sqrt 3\approx -0.133975-2.23205i$$
 * $$r_3=-1-\frac{\sqrt 3}2-\frac i 2+i\sqrt 3\approx -1.86603+1.23205i$$

Bo Jacoby (talk) 08:01, 8 December 2015 (UTC).