What Is the Probability of Drawing Three Cards in Order from a Deck of Ten?

[sputnik]

So, there is a deck of 10 card and a player picks 3 cards one at a time without replacement. What is the probability that the three cards are selected in sorted (increasing) order?

I am not fully understand this question, for me there are two possibilities:
1. The final three cards are sorted in order. (e.g. 1st = 3, 2nd = 1, 3rd = 2)
2. Or, it is strictly increasing, 1st card < 2nd card < 3rd card.

I do favor the 2nd possibility but i can not convince my self, perhaps anyone can help me.

So, my approach is (for the 2nd possibility) to group those 10 cards to a group of 3 cards (in increasing order). Let's call it A, so

A = {(1,2,3),(2,3,4),(3,4,5),(4,5,6),(5,6,7),(6,7,8),(7,8,9),(8,9,10)}
and S is 10x9x8.

So P(A) = 8/(10x9x8) = 1/90

Is this the right thing to do??

 

[HallsofIvy]

So, there is a deck of 10 card and a player picks 3 cards one at a time without replacement. What is the probability that the three cards are selected in sorted (increasing) order?

I am not fully understand this question, for me there are two possibilities:
1. The final three cards are sorted in order. (e.g. 1st = 3, 2nd = 1, 3rd = 2)

No, that would be "consectutive numbers", not "in sorted order".

2. Or, it is strictly increasing, 1st card < 2nd card < 3rd card.

Yes, that is what they mean.

I do favor the 2nd possibility but i can not convince my self, perhaps anyone can help me.

So, my approach is (for the 2nd possibility) to group those 10 cards to a group of 3 cards (in increasing order). Let's call it A, so

A = {(1,2,3),(2,3,4),(3,4,5),(4,5,6),(5,6,7),(6,7,8),(7,8,9),(8,9,10)}
and S is 10x9x8.

So P(A) = 8/(10x9x8) = 1/90

Is this the right thing to do??

No, you are missing such "orders" as (1, 5, 8), (3, 6, 9), etc.