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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?Counting and combinatoricsWarm upQuant tradingProp trading firms
Try it first
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 relationship
$P = \frac{39}{51}\cdot\frac{26}{50} = \frac{4\cdot 13^3}{\binom{52}{3}} = \frac{8788}{22100} \approx 0.398$
39/51 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?

39/51	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

Try it first

Why does the first card cost nothing?

How do you check it a second way in the time?

What is a speed round actually testing?

Where candidates lose it

What the interviewer asks next