This would be zero. I used a different method but also arrived at 1%. Contribute to the database and get 1 month free* Full online access! yes one or two of the previous explanations to the snake problem are correct. I believe the answer to the snake one is no. Some i saw mentioned were expected value and game theory... any others? Again, on your second roll, your EV is 3.5. Nope. That is, the expected hitting time to "6-6" = expected hitting time to "6" + expected hitting time to "6-6" from "6". blue, red, yellow. I had one of the hardest interview probability questions and it was just expectation however it was near impossible to structure the answer. Sales and Trading: Interview Guide written by Hedge Fund Trader, http://en.wikipedia.org/wiki/Fermat%27s_Last_Theorem, Certified Prop Trading Professional - Trader, Certified Equity Research Professional - Vice President. P(X=2) = (5/6)(1/6)=.139 basically, as the team continues losing (and assuming that the .600 success rate does not change) should I bet heavier in favor for a win? and if he did, no one here would understand the proof anyway. In the long run, the average value of two dice rolls using regular dice is 7. dude, just add the expected value of two independent variables and you get 7. no need for all that calculations. sum all of those and theres youre total expected value and therefore how much you'd be willing to pay. i.e., suppose you found a coin that flips with 1/6th probability of heads and 5/6th probability of tails. It depends if the cards are replaced or not...if not you need to make a path dependent tree. In reply to Prove that no three positive by blastoise, In reply to balbasur wrote: 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. the probability in the interviews is nothing that is technically/theoretically advanced, its really the foundation concepts of probability like Bayes theorem but under a great amount of time pressure and in a brainteaser format. If you are looking for something about the history of risk and probability, this was an interesting read. you're gonna have 36 unique combinations. You can exclude the Girl-Girl outcome. The problem is perfectly symmetrical. One red and one yellow snake meeting will lead to 15 blue, not 14 blue. Then, what is the probability that I have a four of a kind in queen ? Don’t look at the solutions until you try each question! Are you sure thats the correct answer? QED, In reply to What to have some fun? Or are you given a paper and pen and given a couple minutes. I agree with the past results (losing streak) not affecting the next game. P{A doesn't gets 2 consecutive heads} = 1 - P(A gets 2 consecutive heads} = 1 - (1/2)(1/2) = 3/4. In reply to how can I solve this card probability question? Since the coin is chosen randomly, we know that P(U) = P(F) = 0.5. Also, you should be multiplying, not adding. Doesn't matter when A gets his heads as long as B gets it first. 4 3 / 36 P(LLLLW) is really only the probability of a four game losing streak with one win, which is different than the probability of one win. i know of few other ways to get a sampling of some great brainteasers than by asking the fine gentlemen on this site. Instead of having a die, suppose you had a fair coin. yea you are right the red herring threw me off, PROBABILITY brainteasers (Originally Posted: 10/07/2010). solving for a we get 1/36a = 7/6 --> a = 42. He's drawn 13 cards, and already knows the first is a Queen, so the question is what are the odds there are another three Queens in that. ^ Disagree. Wouldn't expected # of rolls for tow 6s in general be 12? You're given a loop of copper wire. eg, if a red snake meet a yellow snake, they both change to blue. **Ops read the question wrong. It doesn't matter. The probability that a couple has two boys is 1/4, two girls is also 1/4 etc. Now the 14 blue and 14 red snakes are meeting leading to all snakes being yellow. Prove that no by Art Vandelay, Haha, we did that in IB HL math (Grade 12). Didn't read rest though. Why? See you on the other side! I don't know why this called "Hardest Probability Questions" .. I will appreciate it very much if anyone can tell me whether there is any probability questions asked during the interview for structured finance entry level positions? You assume that the other 3 of the same cards are somewhere within the 12 remaining cards. 4 ways to make Y=6 Why is it not 36? if you roll two 6's right off the bat, that counts as 2 rolls)". P(X=1) = 1/6=.167 Since saying "the odds he will score at least one" is the same as saying "1 minus the odds that he won't score any", we use the "no goals" scenario as our basis. (Originally Posted: 10/05/2012). The expectation operator is a linear one. 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 odds that you get two straight heads on your 3rd roll is (1/4)(1/2)... 1/2 is the odds that first roll is tails The problem was missing information, as the expected value depends on the strategy the user uses.

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