Help with interview questions
Hello, these are two questions asked on interview for trading firms, any help would be appreciated:)
1.) If the probability of of seeing at least 1 car in an hour is 544/625, what's the probability of seeing no cars in a 15 minute interval?
This question doesn't seem like a pretty hard one, but the problem for me is that we don't know a thing about distribution of this random variable.
2.) Two choices: A: 26 black, 26 Red B: random 26 card from the deck of 52 cards. Take the first two cards, if same color, you win $1, otherwise you lose $1. Which deck is best for you if you are playing? Why?
Considering Bayes's formula, I would know how to calculate probability of getting two cards of the same color, but only in theory. Is there a smarter way to solve this, but with a bit of precision?
1) Let the probability of seeing no cars in 15mins be p. Assuming independence, the probability of seeing no cars in 60mins is p^4. You are given 1-p^4 = 544/625 ( P(no cars in 1hr) = 1 - P(at least 1 car in 1hr) ). Hence, p = 3/5.
2) Consider choice A: Suppose you draw a red (black) card first. The number of remaining reds (blacks) in the deck is 25. The probability of drawing another red (black) card is then 25/51.
Consider choice B: You don't know how many reds and blacks you have drawn. Therefore, you have zero information about what happens to the number of reds/blacks in your remaining cards. i.e. the first and second draws are independent. Hence, you can deduce that the probability of getting a card of the same color is 1/2.
As 1/2 > 25/51, we should go with choice B.
You should be able to to do these in your head.
As 1/2 > 25/51, we should go with choice B
I think the second question is a very simply gambling question, and you are overlooking the variation in the second deck. In case A, you know the distribution, in case B, you are taking a random sample from that distribution. This means that while the expected number of red cards is 13, and the expected number of black cards is 13, there are many other possible outcomes, (e.g. they could all be red, all black, and in either case this means you will have 100% chance of winning $1).
In regards to the claim that the probability of drawing a pair from a deck of 4 cards is the same as drawing a pair of cards from a deck of 26 or 52, I think that this is false. Clearly, the probability of having two of the same color in the deck of 26 red and 26 black is going to be 25/51(49%), which is greater than expected 13 red and 13 black =12/25 (48%).
Now think about the distribution of deck B, which could be 13-13, 12-14, 11-15 etc. Would you take a guaranteed 49% odds? or would you take an expected 46%, with a variance that could improve your odds significantly.
Think about the law of large numbers, this question is more about risk tolerance than probability, though if you clearly miss basic probability concepts, it won't matter anyways.
I was stupid with first question, thank you very much for your answer.
But, for question two, using the same logic could we just deduct 25/51 > 12/25 ?, and choose the set A. I thought of this in that way, but was not sure because of randomness of set B.
I would choose A, intuitively the expected payoffs from the random distribution will not converge to anything greater than 49% (based on the symmetry of the problem).
e.g. probability of having an even split 13-13 yields a 12/25 chance, a 14R-12B split gives you 13/25(14/26) if you chose red first, 11/25(12/26) if black first. a 15R-11B split gives you 14/25*(15/26) if red first, 10/25 (11/26) if black first. At this point you're starting to venture away from the mean, so the intuition that A will be a better choice is confirmed.
Thats just how I would think about it intuitively, sorry for the edits, had some typos.
1000000 trials mean 52: 0.490 mean 26: 0.490 var 52: 0.250 var 26: 0.250
intuitive explanation: say you have a deck of 52 cards, randomly shuffled. in that case, you could get a randomly shuffled deck of 26 cards by throwing away the back 26 cards of the deck. the first two cards are the only ones you look at, so your 52 and 26 card decks are the same. you end up looking at the same two cards. throwing away the front 26 or some arbitrary indexing of the 52 cards is no more random than tossing out the back 26.
incidentally, 25/51 = 0.490
this is an interesting approach so i'm going to think about the problem with your method. i think the part you missed is that as you diverge from the 13R 13B distribution your winrate increases, rather than decreases
the probability of getting a 26-selection with R reds and B blacks is (26 choose R) * (26 choose B) / (52 choose 26). the probability of winning is [(R choose 2) + (B choose 2)] / (26 choose 2)
so then for R = 0 to 26 you'd sum the product of the above terms.
B: 0 R: 26 prob_distr: 0.000 distr_winrate: 1.000 cum_winrate: 0.000 B: 1 R: 25 prob_distr: 0.000 distr_winrate: 0.923 cum_winrate: 0.000 B: 2 R: 24 prob_distr: 0.000 distr_winrate: 0.852 cum_winrate: 0.000 B: 3 R: 23 prob_distr: 0.000 distr_winrate: 0.788 cum_winrate: 0.000 B: 4 R: 22 prob_distr: 0.000 distr_winrate: 0.729 cum_winrate: 0.000 B: 5 R: 21 prob_distr: 0.000 distr_winrate: 0.677 cum_winrate: 0.000 B: 6 R: 20 prob_distr: 0.000 distr_winrate: 0.631 cum_winrate: 0.000 B: 7 R: 19 prob_distr: 0.001 distr_winrate: 0.591 cum_winrate: 0.001 B: 8 R: 18 prob_distr: 0.005 distr_winrate: 0.557 cum_winrate: 0.003 B: 9 R: 17 prob_distr: 0.020 distr_winrate: 0.529 cum_winrate: 0.014 B: 10 R: 16 prob_distr: 0.057 distr_winrate: 0.508 cum_winrate: 0.043 B: 11 R: 15 prob_distr: 0.120 distr_winrate: 0.492 cum_winrate: 0.102 B: 12 R: 14 prob_distr: 0.188 distr_winrate: 0.483 cum_winrate: 0.193 B: 13 R: 13 prob_distr: 0.218 distr_winrate: 0.480 cum_winrate: 0.297 B: 14 R: 12 prob_distr: 0.188 distr_winrate: 0.483 cum_winrate: 0.388 B: 15 R: 11 prob_distr: 0.120 distr_winrate: 0.492 cum_winrate: 0.448 B: 16 R: 10 prob_distr: 0.057 distr_winrate: 0.508 cum_winrate: 0.476 B: 17 R: 9 prob_distr: 0.020 distr_winrate: 0.529 cum_winrate: 0.487 B: 18 R: 8 prob_distr: 0.005 distr_winrate: 0.557 cum_winrate: 0.490 B: 19 R: 7 prob_distr: 0.001 distr_winrate: 0.591 cum_winrate: 0.490 B: 20 R: 6 prob_distr: 0.000 distr_winrate: 0.631 cum_winrate: 0.490 B: 21 R: 5 prob_distr: 0.000 distr_winrate: 0.677 cum_winrate: 0.490 B: 22 R: 4 prob_distr: 0.000 distr_winrate: 0.729 cum_winrate: 0.490 B: 23 R: 3 prob_distr: 0.000 distr_winrate: 0.788 cum_winrate: 0.490 B: 24 R: 2 prob_distr: 0.000 distr_winrate: 0.852 cum_winrate: 0.490 B: 25 R: 1 prob_distr: 0.000 distr_winrate: 0.923 cum_winrate: 0.490 B: 26 R: 0 prob_distr: 0.000 distr_winrate: 1.000 cum_winrate: 0.490
code here: http://pastebin.ca/2476664
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