Suppose we have a deck of $500$ cards numbered from 1 to 500. If the cards are shuffled randomly and you are asked to pick three cards (without replacement), one at a time, what's the probability of each subsequent card being larger than the previous drawn card?

My solution:

Let $i$ be the second card that is picked, then $i-1$ cards will be less that $i$ and $500 - i$ cards will be greater than $i$. Thus:

P(subsequent card being larger than the previous card) ${\displaystyle = \sum_{i=1}^{500} \frac {(i -1)(500 - i)}{500 \cdot 499 \cdot 498}}$

I'm not sure if my answer is correct.

  • $\begingroup$ Please read this tutorial on how to typeset mathematics on this site. $\endgroup$ – N. F. Taussig Sep 26 '19 at 14:32

Suppose you draw three cards. We don't care what they actually are...

Now... without loss of generality, suppose the cards were $1,2,3$. Recognize that each of the six possible orders that you could have drawn them in were equally likely to occur: $123,132,213,231,312,321$.

Exactly one of those six equally likely scenarios will result in the cards occurring in increasing order.

The probability then:


As an aside, your answer is correct., however is rather tedious to calculate directly without a calculator.


We can use permutations and combinations to solve this problem.

Number of ways of picking cards in increasing order will be 500 choose 3 (unordered)

Total number of ways to pick 3 cards will be 500 permute 3 (ordered) $$P = {500C3\over 500P3}$$ $$P = {500C3\over 500C3}{\times}{1\over 3!}$$ $$P = {1\over 6}$$

  • $\begingroup$ It seems the result is correct but I don't get why this works. Could you elaborate on why this method can solve the problem? $\endgroup$ – Sanghyun Lee Dec 15 '19 at 20:06
  • 1
    $\begingroup$ The total ways in which you could pick $3$ cards is $500P3$. Now when you choose any $3$ non-ordered cards, you can arrange them such that they are in increasing order. Also this arrangement is unique with respect to the $3$ cards chosen. Hence $500C3$ is the number of valid cases. $\endgroup$ – Sam Dec 17 '19 at 14:50

An easier way to think about the problem using combination and permutation is the following:

For any given three random distinct cards drawn from the pile, there is exactly one ordered combination such that it is in an increasing order while there is six possible ordered combinations of three cards. Hence, the probability of that three cards drawn are in an increasing order is 1/6.

We can generalize this problem to any number of cards drawn -- If N cards drawn then the probability of interest is 1/P(N,N) = 1/(N!).


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