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.
Best Answer
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:
$$\frac{1}{6}$$
As an aside, your answer is correct., however is rather tedious to calculate directly without a calculator.