And also, as one of the above posters mentioned, since the condition for each of the 4 suits is identical, the probability of the 4th card being each suit has to be equal, which is a much quicker way to get the answer. There is no change or set of changes you can do to make 2 colors seperated by 3. by FIASCO. what is the expected value? In reply to probability question by sdumb. shreve's is the best stuff out there - actually explains most of the math instead of merely laying out the underlying assumptions. //www.wallstreetoasis.com/forums/question-with-card-probability-brainteas…, if not, you need to take into account the first 3 cards, Miscer the probabilities might match so you could be right, the three will spit at 1/4 as well, 16! so, with that being said ... what are your favorite brain teaser, and/or the ones you have personally had in interviews ? There. Anyone know how to do it? http://www.quantnet.com/forum/showthread.php?t=1152, Also, you probably heard about the "Heard on the Street". EE and Oranhotan are right. That leaves Boy-Boy, Boy-Girl and in one of these the kid is a girl, so the probability is 1/2. Maybe see if your library has a copy before buying, though. how can I solve this card probability question? Well I don't think for b it is meant you keep the higher of the two times two, simply you get to keep the higher of the two, so I would think the value of the 2nd roll has to be either the probability of rolling higher than expected return from first roll, or >3.50 (i.e. P(X=5) = (5/6)^4(1/6)=.080 I assume it is a kind of Joke .. Financial Modeling Training assuming a strategy where you only chose to roll the die again if the outcome of the first die was below the expected value of rolling a fair die (3.5), you get 4.25 because you still have (1/6)(4 + 5 + 6) but you also add (1/12)(1 + 2 + 3 + 4 + 5 + 6). By inspection , only a difference between any two snake numbers that is divisible by 3 will work . Or is it just innate skill? let's say A keep tossing a fair coin, until he get 2 consecutive heads, define X to be the number of tosses for this process; B keep tossing another fair coin, until he get 3 consecutive heads, define Y to be the number of the tosses for this process. What’s the probability that our 3 rolls are strictly in increasing order? it was the same thing except you dont get the higher of the 2, if you roll the 2nd time you get the payout of the 2nd roll, not the greater of the two, but i'll give yours a shot. interview probability will be simple EV type stuff that just gets marginally harder to make sure you can think analytically/prove you can do some sort of math. The snakes cannot all become the same colour simply because of the relative difference between sankes gaiven being divisible by two. Broken down: Book to learn how to answer tough probability questions? "You have a friend who has two children. So you get (1/2)(48.5/95)(49/94) = ~0.133, slightly greater than (1/2)^3, which is intuitive because these are non-independent in a way where each previous step positively impacts the probability of the subsequent step? if you roll two 6's right off the bat, that counts as 2 rolls)". This covers all cases because this defines when X>Y, and it becomes an easy problem. WSO depends on everyone being able to pitch in when they know something. Probability questions for S&T interview (Originally Posted: 04/01/2008). Intuitively, I think it would be: leaving out the first queen, you now have 51 cards to choose from to form the remaining 12 cards in your hand. in this one we have first part denotes hitting tails right away so back to a, second part is hitting Heads and then hitting Tails, so again back to square one, finally last part is hitting 2 heads in a row on the first two flips. Given that you have drawn {1, 2, 3, 4}, the probability that you have drawn them in exactly that order is 1/4! Given the hint, the P( max | x - x_n_trials | > d ) d ) (Boole's inequality) (Originally Posted: 09/22/2010), Not sure whether I remember the question correctly, but here it goes -. Nope. Certified Hedge Fund Professional - Investment Analyst, 7 tips for picking up women and landing the job you want, Want to be a Trader? You have a deck of 97 cards and I will pay you $10 if I draw 4 cards and they are in ascending order (not necessarily consecutive order) and you pay me $1 if they are not. “...all truth passes through three stages. In reply to THE MOST difficult probability question by mass_marines. of course its basic, but you gave two answers, depending on how you interpret it. "Oh - the ladies ever tell you that you look like a fucking optical illusion?". Conditional expected value: (2/6)2 + (1/6)(3+4+5+6) = 3.6667, First roll outcome: 1 Game theory was so long ago.... Well according to Murphy's law, there is a 100% chance that you will get a probability question, given how worried you seem. sums to .668. I don't know why this called "Hardest Probability Questions" .. You can draw 4 cards in 4! for example, a lot of pricing books (ie - Hull) will simply say 'and here we solve the black scholes PDE' - but they don't always explain how they derive or solve it, or really what it means. c'mon what're the odds you get probability questions? What is the expected number of rolls needed to roll two 6's in a row? you need to explain more specifically what you are trying to do. 4 ways to make Y=6 You have a friend who has two children. Conditional expected value: (4/6)4 + (1/6)(5+6) = 4.5, First roll outcome: 3 Probability Question (Originally Posted: 01/25/2008), a player has a 30% probability to score in a game. I agree with the past results (losing streak) not affecting the next game. Unlock with your email and get bonus: 6 financial modeling lessons free ($199 value). at the end of each combo, you take the meax of the first and 2nd roll. Don’t look at the solutions until you try each question! Notice that, in order to get to "6-6", you have to first get to "6". I had two other clever questions pop up in an interview recently, both of which made you go through long-winded calculations whose answer was most of the work of the problem, but not the brief final step from there that they wanted. Wouldn't it just be (3/51)(2/50)(1/49) [The probability of doing it with the 3 cards] * 12 C 3 [# of ways to arrange that in 12 cards], so (3/51)(2/51)(1/49)* 12!/(3!9!) basically you have to relate the confidence interval (involving standard deviation of the sample mean, which is a summation of bernoullis (i.e. I'll double-check and get back to you. Rolling a "5" has a probability of 1/6. That is how I would break it down. (Originally Posted: 10/05/2012). If I asked you about diamonds in particular, you'd still tell me 10/13. If I asked about spades, you'd still tell me 10/13. 6 * 1/6 = 6/6, Adding them together gives: if you want, but you don't have to, you can roll the die again and get paid what the second roll shows instead of the first. Anyone figure out number 1 yet? i know of few other ways to get a sampling of some great brainteasers than by asking the fine gentlemen on this site. 7 6 / 36 Furthermore, only 1 of those 24 ways will yield cards in ascending order. But I'm not sure you can say the same with a baseball game, so this is the crux: Is the HISTORICAL winning percentage necessarily equal to the probability they will win a certain game? So the question really is what's the probability that there will be 3 more of the first card in the remaining 12 face down. Having never taken probability, I didn't get it. Unless there is a way to get it so there are 15 of each, which i can't figure out. Or have a quick look into wikipedia: http://en.wikipedia.org/wiki/Fermat%27s_Last_Theorem, In reply to balbasur wrote: Normally the question is, what's the expected value of two rolls assuming you can roll again after the first roll, and only your last roll counts? The probability that a couple has two boys is 1/4, two girls is also 1/4 etc.

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