Ten cards, numbered one to ten are shuffled and placed in a

Ten cards, numbered 1 to 10, are shuffled and placed in a pile face down. A player is then asked to turn over the top card.

If the top card shows a "1", he or she wins and the game is over. If the top card is not a "1", but is instead a number we will call "k", then the card is placed in the k-th position in the deck, counting from the top.

The player does this twice more, turning over at most 3 cards with the same rules. What is the probability that the player will win the game?

please respond asap.. thanks

2 is pulled on 1st pull; 2 is inserted in 2nd place

7 is pulled on 2nd pull; 7 is inserted in 7th place

So game is over and player loses at this point, right?
Because he/she knows next card is a 2...

Looked at that again; quite simple:

Probability of getting a "1" in 3 turns = 3/10 :
that would be the answer if the cards turned were not put back in...

The ONLY time putting cards back in affects the outcome is when the
1st card is the "2" AND the 3rd card is the "1"; the "1" is no longer available,
as putting the "2" back in will push it down one spot.

The "2" will be 1st card 1/10 of the time; now there's 9 cards left; the "1" has
a 1/9 chance of being in each spot, so a 1/9 chance of being the 3rd card;
(1/10) * (1/9) = 1/90;
so the 3/10 chance is reduced by 1/90: 3/10 - 1/90 = 13/45 (or .288888.....).

So probability that player wins is 13/45. (Hope someone checks me!)

Another look: 3/10 - 8!/10! = 13/45