Probability Question Doubt

[Jay_NIT]

I have tried enough but could not able to figure out how to do it, I know it is conditional probability but need help. Thank you

In blackjack there's a 52-cards deck (4 suits of 13 cards each: 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King, Ace). Digits are count as their face value, Aces count for 11 and pictures count for 10. Player with maximum total value of cards wins, provided that the value doesn't exceed 21. Croupier has two pictures. Player has an ace. What's the probability that player wins as the next card is dealt?

 

[pka]

I have tried enough but could not able to figure out how to do it, I know it is conditional probability but need help.
In blackjack there's a 52-cards deck (4 suits of 13 cards each: 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King, Ace). Digits are count as their face value, Aces count for 11 and pictures count for 10. Player with maximum total value of cards wins, provided that the value doesn't exceed 21. Croupier has two pictures. Player has an ace. What's the probability that player wins as the next card is dealt?

Do I understand the setup correctly? There are 49 cards remaining in the deck. The player must get either face card or a ten to be a clear winner. Getting an ace would exceed the limit. What happens if the player draws a nine?

 

[Jay_NIT]

Yes I think you understand correct, Actually we want player to win against croupier. Actually answer is 2/7 but I don't know how? I tried but not getting correct answer.

 

[Jay_NIT]

Thank you after you clarification I got my solution. Actually I was making mistake in understanding. It is very simple than what I was doing, like total we are remain with 49 cards after substracting 3 cards already with players and now for next time I want 10 points which can be earned in total 14 ways(4 cards of 10 and 10 remaining pictures cards) so it's just 14/49 ==> 2/7.
So inherently we are using conditional probability that total doesn't exceed 21.