# Cool brainteasers list

another I got recently, it’s in some books though. you have 64 teams numbered from 1 to 64 in order of rank, assume that the lower-numbered team will always beat the higher numbered team. Randomly select pairs of teams and the winners will advance to the next round. what is the probability that team 1 will play 2 in the final round?

P* = P / (2^200), where P is my original guess

DoubleDip Wrote: ------------------------------------------------------- > i had something like that on a recent interview, > but it was pr(getting more than 60 heads) from 100 > flips of a fair coin. Not at all same, approach changes when you are talking about consecutive occurrence (6 in a row) and pattern occurrence (4H1T) VS cumulative occurrence

The Movie Line There is a long line of people waiting to get into a movie. One by one, each person enters and states their birthday (Month and Day only, no year) to the ticket checker. The first person to pass through the door and state a birthday that has ALREADY been stated wins the big prize. What is the optimal place to stand in line?

^64th or 65th I think.

sublimity Wrote: ------------------------------------------------------- > P* = P / (2^200), where P is my original guess actually, i shouldn’t divide by 2^200. the solution seemed so clear to me for arbitrary k that i didn’t bother to check for very low k, thus giving rise to that stupid mistake back to thinking…

Sanka’s Mom Wrote: ------------------------------------------------------- > An oldy but goody. > > The monte hall three doors problem. > > Suppose you’re on a game show, and you’re given > the choice of three doors: Behind one door is a > car; behind the others, goats. You pick a door, > say No. 1, and the host, who knows what’s behind > the doors, opens another door, say No. 3, which > has a goat. He then says to you, “Do you want to > switch your pick door No. 2?” Is it to your > advantage to switch your choice? Yes.

another question: you have two urns containing a mix of red and blue balls. Urn 1 contains 30% red balls, 70% blue balls and Urn 2 contains 70% red balls, 30% blue balls Select an urn at random and draw 12 balls from it without replacement. Out of 12 balls chosen from the urn, 8 are red and 4 are blue. What is probability that balls were drawn from Urn 2?

DoubleDip Wrote: ------------------------------------------------------- > another question: you have two urns containing a > mix of red and blue balls. > > Urn 1 contains 30% red balls, 70% blue balls and > Urn 2 contains 70% red balls, 30% blue balls > > Select an urn at random and draw 12 balls from it > without replacement. > > Out of 12 balls chosen from the urn, 8 are red and > 4 are blue. What is probability that balls were > drawn from Urn 2? somebody’s been reading the L3 cfai books. unfortunately, i couldn’t figure out how they got the answer, so I ignored it and moved on. can anyone shed some light on this?

I figured out how to solve the movie queue question, but I probably made a math error somewhere - my answer is 20. My solution is to express P(n) in a formula, where P(n) is the probability of winning the prize when you are in position n. You will get: P(n) = {365*364*363…*(365-n+2)(n-1)}/365^n Then, you need to find the smallest n such that P(n) > P(n+1). That is: {365*364*363…*(365-n+2)(n-1)}/365^n > {365*364*363…*(365-n+2)(365 -n +1)(n)}/365^(n+1) This reduces to: (n-1)/365^n > (365-n+1)*n/365^(n+1) Solve the resulting quadratic equation to get the answer.

dspapo: yes, it’s in there but also in the CFA webinar I’m watching now on behavioral finance. Use Bayes Rule. P(A|B) = P(B|A)*P(A)/P(B) Let A = event that the balls were drawn from Urn 2, and let B be the event that 8 red balls were drawn from 12. Background info: The prior probability that (balls were drawn from Urn 2) = 1/2

Hello Mister Walrus Wrote: ------------------------------------------------------- > I figured out how to solve the movie queue > question, but I probably made a math error > somewhere - my answer is 20. > > My solution is to express P(n) in a formula, where > P(n) is the probability of winning the prize when > you are in position n. You will get: > > P(n) = {365*364*363…*(365-n+2)(n-1)}/365^n > > Then, you need to find the smallest n such that > P(n) > P(n+1). That is: > > {365*364*363…*(365-n+2)(n-1)}/365^n > > {365*364*363…*(365-n+2)(365 -n > +1)(n)}/365^(n+1) > > This reduces to: > > (n-1)/365^n > (365-n+1)*n/365^(n+1) > > Solve the resulting quadratic equation to get the > answer. You got it.

Wait, that was correct? Wow, lucky shot there.

Hello Mister Walrus Wrote: ------------------------------------------------------- > Wait, that was correct? Wow, lucky shot there. yeah, 20 is correct.

K, I think I’ve got it…might still be some slight errors I’m not aware. I didn’t consult any references… > Mobius Striptease Wrote: > ------------------------------------------------------ > what is the probability that you will get a > sequence of least 6 consecutive heads or tails > after flipping a fair coing 200 times Hmmm, let’s take the base case of “6”. There are are 200-5 slots for this to occur, each with a probability of (1/2)^6. So for, we expect this to occur the following TIMES: k = 6, (200-5)/2^6 [UPDATE: This is the expected number of TIMES, not PROBABILITY for this to occur. To get probability, you need to divide by 200.] Since you introduced the term “at least”, that means we will have to consider everything from k = 6 to k = 200. For k = 7, (200-6)/2^7 … For general k: define f(k) = (200 - (k -1))/(2^k) Add these up, we have N(k= 6, 7, …, 200) = SUM(k = 6, 7, …, 200) f(k) So probability = N(k = 6, 7, …, 200)/200 ------------------- Also, we could say that this probability is 1 - P(5 or less). P(5) = (200-4)/( 200 * 2^5) = 0.030625 P(4) = (200-3)/( 200 * 2^4) = 0.0615625 P(3) = (200-2)/( 200 * 2^3) = 0.12375 P(2) = (200-1)/( 200 * 2^2) = 0.24875 P(1) = (200-0)/( 200 * 2^1) = 0.5 ---------- P(5 or less)= P(1) + P(2) + P(3) + P(4) + P(5) = 0.9646875 1 - P(5 or less) = 1 - 0.9646875 = 0.0353125 ---------- So, after all was said and done, the probability of getting 6 or more heads or tails after tossing a coin 200 times is: p= 0.0353125 Please verify, if someone knows the solution. Thanks.

80% for 6 out of 200.

^ do you know if that’s the answer for sure? the problem is 6 or more consecutive for 200 tosses.

king_kong Wrote: ------------------------------------------------------- > Sanka’s Mom Wrote: > -------------------------------------------------- > ----- > > An oldy but goody. > > > > The monte hall three doors problem. > > > > Suppose you’re on a game show, and you’re given > > the choice of three doors: Behind one door is a > > car; behind the others, goats. You pick a door, > > say No. 1, and the host, who knows what’s > behind > > the doors, opens another door, say No. 3, which > > has a goat. He then says to you, “Do you want > to > > switch your pick door No. 2?” Is it to your > > advantage to switch your choice? > > > Yes. Correct. Originally, you had a .33 chance of getting the right door (.67 chance of the right door being in one of the two other doors). If they remove one of the other doors, the original probability does not change. So there is still a 33% chance you picked the right door, and a 67% chance it is in the only other option.

I hate probability questions.

sublimity Wrote: ------------------------------------------------------- >> > p= 0.0353125 > > Please verify, if someone knows the solution. > Thanks. intuitively speaking, do you find your answer reasonable? in other words, if someone asked you to generate a random sequence of 200 coin flips by just coming up with it (rather than actually flipping the coin and recording it), then your sequence is not very likely to include large (>5) blocks/sequences of heads or tails. and if someone shows you a sequence of 200 coin flips that has 6 or more consecutive heads/tails, you are likely to conclude that this might be a non-random event, i.e. the coin is possibly rigged?