probability question

Game: You pay $1 to play the game.
Set Up: You have a deck of cards. Out of the 52 you choose six cards and turn them face down.

Roll 1 die(1-6)
What ever # shows up on the dice turn the card over. Ex: If I rolled a 2 I have to choose the 2nd card out of the six.
1 roll per game

How to win:
Small prize: if the card is a face card you win $3
Large prize: if the card is a ace you win $5
No prize: if the card is a number card you dont win anything buy you do loose a dollar that you payed to play the game

What is the probability that the card you turn over is an ace out of the six card deal(probabily of winning a large prize)?
What is the probability that the card is a face card out of the six card deal( probabilty of winning a small prize)?
What is the probability that you choose a face card or ace card out of the six card deal(probability of winning)?
What is the probability that you choose a number card out of the six card deal(probability of loosing)?

Please show your work
also if this was an carnival game would you play the game based on your chances of winning, explain why?

One would think that drawing an Ace from the 6 selected would be the same as just drawing an Ace from the deck in the first place, but I am not so sure it works that way. Maybe pka or someone well-versed in probability may get involved. In other words, the probability of selecting an Ace after you roll the die would be the same as just choosing an Ace form the deck: 4/52=1/13.


The probability of having at least one Ace in a 6 card draw is the opposite of the probability of no Aces.

Therefore, the probability of at least one Ace is:

\(\displaystyle \L\\1-(48/52)(47/51)(46/50)(45/49)(44/48)(43/47)=\frac{21508}{54145}\approx{0.39722941}\)

Then, the probability you pick an Ace out of the six is 1/6th

\(\displaystyle \L\\\frac{21508}{54145}\cdot\frac{1}{6}=\frac{10754}{162435}\approx{0.066205}\)

I hope I am correct or at least on to something. You can never tell with these dang counting problems.

The probability of you having one Ace in the 6 cards and then drawing it would be:

\(\displaystyle \L\\\frac{C(4,1)C(48,5)}{C(52,6)}\cdot\frac{1}{6}\approx{0.056071...}\)

The probability of having 2 Acres in the 6 card draw and then choosing it is:

\(\displaystyle \L\\\frac{C(4,2)C(48,4)}{C(52,6)}\cdot\frac{1}{3}=\approx{0.01911533...}\)

And so on.