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003Speed round, ninety seconds: you draw three cards from a well-shuffled 52-card deck without replacement. What is the probability that all three are of different suits?Quant tradingProp trading firms
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Closest answer, fast.
Show the worked solution
About 39.8%. The first card can be anything. The second must come from one of the three other suits: 39 of the 51 cards left. The third must avoid both suits already seen: 26 of the 50 left. Multiply: 39/51 x 26/50 = 1,014/2,550, just under 40%. Drawing with replacement would give 3/4 x 1/2 = 37.5%.
Why does the first card cost nothing?
Picture three guests arriving at a party with four dress colours in the wardrobe, and you want no two to match. The first guest cannot clash with anyone. A condition about the cards differing only bites from the second card on, so the first card contributes a factor of 1 and you start counting from card two. Candidates who write 13/52 for the first card have fixed a particular suit and then have to multiply by the number of suit orders to recover, which is where the slips happen.
The first card is free, the second must avoid one used suit with 39 of 51 cards still good, and the third must avoid two with 26 of 50 still good, so three different suits happen with probability 39.8%. How do you check it a second way in the time?
Count unordered hands. Choose which three suits appear, 4 ways, then one card from each, 13 cubed, and divide by all three-card hands, 52 choose 3. That is 4 x 2,197 = 8,788 over 22,100, which is the same 0.3976. In a speed round you do not have time for both, but knowing the counting route exists lets you sanity-check the product: 0.765 x 0.52 is a little under 0.40.
The relationship39/51 cards of a new suit among those left after one draw 26/50 cards of a third suit after two draws \binom{52}{3} the number of possible three-card hands What it says in wordsSequential dodging and direct counting give the same 39.8%.What is a speed round actually testing?
Thirty questions in forty-five minutes cannot all be worked in full. The skill being tested is choosing the shortest correct route and estimating the product well enough to pick from the options. Here, 39/51 is about 0.76 and 26/50 is 0.52; 0.76 x 0.52 is about 0.40, which eliminates every other option before you finish the exact fraction.
Where candidates lose it
The fast wrong answer is 37.5%, from treating the draws as if cards go back in the deck. It feels close enough, and in a multiple choice round it sits right next to the correct option on purpose.
The other slip is starting with 13/52 for the first card, which silently fixes that card's suit. You then need to multiply by 4 for the suit choice, and under time pressure most people forget.
What the interviewer asks next
- What is the probability that four cards are all of different suits?
- What is the probability that three cards share a suit?
- Draw until you have seen all four suits. What is the expected number of cards?
015You roll three fair dice. Is a total of 9 or a total of 10 more likely, given that each can be written as exactly six unordered combinations of three faces?Quant tradingProp trading firms
Try it first
Which total is more likely?
Show the worked solution
10 is more likely: 27 ways out of 216 against 25. The equally likely outcomes are the 216 ordered rolls, not the unordered combinations. A combination of three different faces covers six ordered rolls, one with a pair covers three, and a triple covers one. Total 9 includes 3 + 3 + 3, which counts once, and one fewer all-different combination, so it loses two orderings to 10.
Why are combinations the wrong thing to count?
Think of dealing two cards and asking whether a pair of kings or a king with a queen is more likely. There is one combination of each, but king and queen can arrive in either order while two kings are just two kings. Probability comes from counting outcomes that are equally likely, and with dice those are the ordered rolls: first die, second die, third die, 6 x 6 x 6 = 216 of them. Unordered combinations bundle different numbers of those outcomes, so counting combinations gives the wrong weights.
Totals of 9 and 10 each have six unordered combinations, but weighting each combination by its number of orderings gives 27 of 216 rolls for 10 against 25 for 9, mostly because 9 includes the triple 3, 3, 3, which happens only one way. How do you count the orderings fast?
Classify each combination by its repeats. Three different faces give 3! = 6 orders, a pair gives 3 orders, one for each position of the odd die, and a triple gives 1. For 9: 1 2 6, 1 3 5 and 2 3 4 are all different, 18; 1 4 4 and 2 2 5 are pairs, 6; 3 3 3 is a triple, 1; total 25. For 10: 1 3 6, 1 4 5 and 2 3 5 give 18; 2 2 6, 2 4 4 and 3 3 4 give 9; total 27. So 10 comes up 12.5% of the time and 9 only 11.6%.
The relationship6, 3, 1 the orderings of an all-different, a pair and a triple combination 216 the ordered outcomes of three dice What it says in wordsWeight each combination by its orderings and 10 beats 9 by two rolls in 216.Is there a shortcut that avoids listing?
Yes: symmetry. Replacing each face x by 7 - x maps a total of t to 21 - t, so the distribution of three dice is symmetric about 10.5, and 10 and 11 are the two most likely totals, each 27/216. Anything further from 10.5, including 9, must be less likely or equal; a quick count confirms it is 25. Historically this is the question gamblers put to Galileo, who answered it by counting ordered outcomes, which is still the method.
Where candidates lose it
The trap is the question's own framing: six combinations each invites the answer that the totals are equally likely. The interviewer wants you to reject the framing, not accept it.
The second loss is listing all 216 rolls, or writing out every ordering. Classifying by repeats, six, three or one, gets both totals in under a minute.
What the interviewer asks next
- What is the most likely total with four dice, and its probability?
- What is P(total is 9) with two dice, and why does 9 behave differently?
- How many ordered outcomes of three dice sum to 7?
067Five traders drop their business cards in a bowl and each draws one at random. What is the probability that nobody draws their own card, and what does it approach as the number of traders grows?Quant tradingQuant research
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As the number of traders grows very large, the chance nobody gets their own card...
Show the worked solution
44/120, about 36.7%, and it tends to 1/e, about 36.8%. Count the orderings with no fixed point by inclusion and exclusion: 5! times (1 - 1 + 1/2! - 1/3! + 1/4! - 1/5!) = 44 of the 120 ways. The bracket is the start of the series for e^(-1), so the answer barely moves with the number of traders: it is within about 0.001 of 1/e at five people and closer still after that.
Why does the answer not go to 0 or to 1?
Think of a secret gift exchange at the office. With more people, each person is less likely to draw their own name, but there are more people who could. Each trader matches with chance 1/n and there are n traders, so the expected number of matches is exactly 1 at every size, and the chance of zero matches settles rather than vanishing. That is why the answer converges to a constant. It is also why the question is asked: it checks whether you can count an event defined by an absence, which needs inclusion and exclusion.
The chance that nobody draws their own card swings above and below 1/e for small groups and is 44/120 = 0.3667 for five traders, against 1/e = 0.3679, so the number of traders barely matters once there are more than four. How does inclusion and exclusion give 44?
Count the orderings where at least one trader gets their own card, then subtract. Fix any one trader: 4! orderings each, five traders, 120 in total, but that counts orderings with two fixed traders twice. Alternately subtract and add the counts with one, two, three, four and five traders fixed, and you get 120 - 120 + 60 - 20 + 5 - 1 = 44 orderings with no match. Dividing by 120 gives the probability as 1 - 1 + 1/2 - 1/6 + 1/24 - 1/120, and each term is the e^(-1) series truncated.
The relationshipD_n the number of orderings of n items with no item in its own place, a derangement count (-1)^k / k! the inclusion and exclusion term for k traders fixed What it says in wordsThe chance of no match is the alternating series for e to the minus 1, cut off after n terms.How do you check 44 another way?
Use the recursion. Trader 1 takes some other trader's card, say trader j's, in n - 1 ways. Either j takes trader 1's card back, leaving n - 2 traders to derange, or j does not, which is the same as deranging n - 1 traders. So D(n) = (n - 1)(D(n - 1) + D(n - 2)), and from D(1) = 0 and D(2) = 1 you get 2, 9 and then 4 x (9 + 2) = 44. The number of matches is close to Poisson with mean 1 for any decent n, so exactly one match has about the same chance as none: for five traders it is 3/8 = 45/120.
Where candidates lose it
The trap is answering (4/5)^5, about 33%, by treating each trader's miss as independent. The draws are without replacement, so the events are linked; the right answer is close but not equal, and the reasoning is wrong.
The second loss is getting 44/120 and stopping, when the follow-up about large groups is the point. Say the series is the start of e^(-1), so the answer is about 37% for any number of traders.
What the interviewer asks next
- What is the expected number of traders who draw their own card?
- What is the probability that exactly one trader draws their own card?
- With 100 traders, roughly what is the chance that at least two draw their own card?