Thread: probability of a deck ofcards

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Originally Posted by badboychow

You draw 2 cards from a deck of 52 cards and the return as follow:

Result
How much you win
Blackjack or Double Aces
$50

Total is more than 16
$10

None of the above
- $5

(Assume the face value of an ace is 11)

Blackjack or double aces

[(16/52) x (4/51)] + [(4/52) x (3/51)]
*is this correct?

I'm not sure of the rest. Anybody help pls

Blackjack or Double Aces.

first card is a "10" second an ace (16/52)x(4/51)

first card is an ace second card an ace or "10" (4/52)x(19/51)

so probability is: (16/52)x(4/51) + (4/52)x(19/51)

RonL

thanks but how about the probability where total is more than 16 and none of the above...i have no idea...

Let's see if we can pick it apart.

The probability of drawing two cards from a deck and their sum is > 16.

There's a trick here because of Jack, Queen, King also count as 10.

We can have Ace-Ace and Ace-not Ace. To sum over 16 we must have A6, A7, A8, A9, A10, AJ, AQ, AK, or AA.

For 10, J, Q, K we have

For 9, we have

For 8, 7, and 6 we have

Add them up and get

I could easily be over or under counting.

thanks thanks....now i playing cards...seeing how many probability....haha

Originally Posted by badboychow

You draw 2 cards from a deck of 52 cards and the return as follow:
Result
How much you win
Blackjack or Double Aces
$50
Total is more than 16
$10

total more than 16 but not blackjack or double ace:

1st card A 2nd card 6, 7, 8, 9 (4/52)x(4/51)

1st card "10" 2nd card 7, 8, 9, "10" (4/52)x(27/51)

1st card 9 2nd card 8, 9, "10", A (4/52)x(27/51)

1st card 8 2nd card 9, "10", A (4/52)x(24/51)

1st card 7 2nd card "10" A (4/52)x(20/51)

1st card 6 2nd card A (4/52)x(5/51)

so probability is:

(4/52)x(4/51) + (4/52)x(27/51) + (4/52)x(27/51) + (4/52)x(24/51) + (4/52)x(20/51) + (4/52)x(5/51)

RonL

Hello, badboychow!

You draw 2 cards from a deck of 52 cards.

. .

(Assume the face value of an Ace is 11)

Find the expected value.

There are: .

Double Aces: .There are ways.
Blackjack: .There are ways to get an Ace.
. . . . . and ways to get a card worth 10.
. . So there are: . Blackjacks.

Hence, there are: . 70 ways to win $50.


Sum greater than 16

We can have an Ace (4 choices) and any of 6, 7, 8, or 9 (16 choices).
. . There are: . ways.

We can have a 7 (4 ways) and any "10" (16 ways).
. . There are: . ways.

We can have an 8 (4 ways) and a 9 or a "10" (20 ways).
. . There are" . ways.

We can have two 9's. .There are: . ways.

We can have a 9 (4 ways) and a "10" (16 ways).
. . There are: . ways.

We can have two "10's". .There are: . ways.

Hence, there are: . 398 ways to win $10.


Then there are: . 858 ways to lose $5.


The expected value is: .

. .


Therefore, we can expect to win an average of about $2.41 per game.

thank you so much....i think i know how to solve it....