Odds in Blackjack

Jan 2007
352
0
Question :

In the game of 'Blackjack' or '21' what are the odds of making a hand: 17,
18, 19, 20 and also 14, using 6 or 8 standard deck of 52 cards?


Example:
10+7; 11(ace)+6; 8+6+3; 9+8; etc.

10+4; 9+5; 8+6; 7+7; 10+2+2; etc.

8+2+6+4; 10+10; 9+5+6; etc.
 

ThePerfectHacker

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Question :

In the game of 'Blackjack' or '21' what are the odds of making a hand: 17,
18, 19, 20 and also 14, using 6 or 8 standard deck of 52 cards?
You are dealt two cards, what is the probability it is a 21?
I will answer this question the other questions are similar.

You need to know a formula. Called # of combinations. You have 3 bullets: .22 cal, .36 cal, .45 cal. You choose 2, how many different way are there?
The list is this
Code:
.22 + .36
.22 + .45
.36 + .45
There are 3 different ways.
To our luck there is a formula to that.
The notation in this case is,
\(\displaystyle _3C_2\)
Meaning choosing 2 objects (this case bullets) from 3 objects.
And we found by listing all the combinations the result was 3. Thus,
\(\displaystyle _3C_2=3\).

Now, there is a formula. If you have \(\displaystyle n\) objects and you choose \(\displaystyle m\) objects. Then the number of combinations is,
\(\displaystyle _nC_m=\frac{n!}{m!(n-m)!}\)
Where the ! excalmation means "factorial".
It is computed as follows:
1!=1
2!=(2)(1)=2
3!=(3)(2)(1)=6
4!=(4)(3)(2)(1)=24
...
See the pattern?

Okay, thus we can compute \(\displaystyle _3C_2\) without a list. By the formula we have,
\(\displaystyle \frac{3!}{2!(3-2)!}=\frac{6}{(2)(1)}=3\).
---------------
Now we can talk about probability.
Probability is the ratio of favorable outcomes to possible outcomes.
The possible outcomes is the number of ways of choosing 2cards from 52 cards.
That is,
\(\displaystyle _{52}C_2=\frac{52!}{2!(50)!}=1326\)
The favorable ways is the number of different ways to get a blackjack.
Code:
10,J,Q,K + A
Meaning we need to choose 1 Ace and 1 card which has value 10. There are 4 aces and since we are taking only one the number of combinations is simply 4. There are 4 different value cards for 10 and 4 suits, in total there are 16 diffferent 10 value cards, we are choosing only 1. The number of combinations is simply 16.
Thus, in total between an ace and a ten value card there are (16)(4)=64 total number of combinations.
Thus, the probability is,
\(\displaystyle \frac{64}{1326}\approx .0482\)
About 1 in 20.
 
Jan 2007
352
0
ok

I wonder if you would be kind enough to provide me with the answers I seek?