Odds on player, dealer BJs

[shadroch]

Can someone give me the exact odds of both you and a dealer getting BJ in a SD game, off a fresh deck.
More importantly, can someone show me which formula they used to get the answer. Thanks.

[Guynoire]

Easiest way of doing it

Probability of both you and dealer getting blackjack =
(probabllity of you having blackjack) *(probablity of dealer getting blackjack given you have blackjack)

Probability of your blackjack: There are two ways to get blakjack Ace 10 or 10 Ace; 16 10's 4 aces in a 52 card deck

= (4/52*16/51 + 16/52*4/51) = 32 / 663 =.0482

Probability dealer having blackjack given you have blackjack, same idea only a 10 and ace are missing in a 50 card deck.

=(3/50*15/49 +15/50 *3/49)= 9/ 245

Probabilty of both you and dealer having blackjack

= 32/663 *9/245 =.00177 or .177%

[rollem411]

There are basic methods for counting problems such as these. The first is permutations and the second combinations. All card games use combinations because the order of the cards does not matter...ie. a 10,5 is the same value as 5,10.

As for the formula, nCr, simply means your choosing r from n elements. I'm assuming you know about factorials and the symbol ! that denotes it.

A direct formula of nCr looks like:

(n!) / (r!)(n-r)!

This equation can be used to solve any combination probability problems.

I'll do the example.

You already know there are 4Aces and 16Tens and a total of 52 cards.

You can break your question down into a 3 step problem.

1) How many total possible combinations of hands are there.
---Using nCr: a total of 52 cards are being used and you are choosing 2 from those 52. In other words, nCr = 52C2. Plug in n and r in the equation and you will get 1326 possible combinations.

2) The probability that you will receive a blackjack.
--Using the equation nCr: you have 4 Aces and you are choosing 1 from those 4. nCr = 4C1.
-- You also need to choose 1 Ten from the 16 in the deck. nCr = 16C1..again plug in.

Then multiply these 2 answers from step 2 and divide by Answer in step 1.

3) You now find the probability of the dealer having blackjack ALSO.
= 3C1 x 15C1 / 50C2.

Finally multiply the answer you get at the end of step 2 by the answer from step 3.

.048 x .0367 = .0017 is your probability that you and the dealer will both have a BJ.

[rollem411]

I spent forever typing that GUY! Damn you

Also, your way seems easy now, however when you reach tougher problems, the

use of the equation will be of great help. I used to solve these problems the

exact same way as you had just done.

[Guynoire]

Yeah there's more than one way to do it, I just find the way I did it more intuitive. If you want to calculate anything more complicated like poker or bridge hands the combination method is the way to go.

[callipygian]

Originally Posted by Guynoire

Yeah there's more than one way to do it, I just find the way I did it more intuitive.

Actually, it's the same method, you just simplified the math in your head.

Starting with rollem411's equation ...

P = [C(4,1)*C(16,1)/C(52,2)]*[C(3,1)*C(15,1)/C(50,2)]
P = [4*16/(52*51/2)]*[3*15/(50*49/2)]
P = [4/52*16/51*2]*[3/50*15/49*2]
P = [4/52*16/51+4/51*16/52]*[3/50*15/49+3/49*15/50]

... you end up with your equation.

[shadroch]

Okay, thanks for the formula. It's pretty much as I figured it was.
Let me take it a step further, because this is what I'm trying to ascertain.
In Laughlin, you can bet $1 bonus bet on the first hand of a new deck( SD). A BJ pays you 10-1,I believe. If both you and the dealer hit, it pays a progressive, which ranges from $50 to an occasinal $200plus. Is there a dollar figure at which this bet is worth making? Whats throwing me off is the smaller payoffs for only a player BJ. I'm not sure how to factor that payoff in.

[sagefr0g]

ok guys i been trying to learn some of this stuff and could use a bit of an explaination of something i'm not understanding.
like i think i understand at least intuitively the reasoning that comes up with the equation:
nPn = n(n-1)(n-2)(n-3)... 3x2x1
and am ok with
nPn = n!

where i'm getting lost is when the author talks about;
if you have n objects, but want to make ordered arrays each consisting of r objects (r being less than n), then he calls that nPr or "the number of permutations of n things taken r at a time"
so ok i'm fine up to this point but then he goes on to say (here i get lost)
nPr = n(n-1)(n-2)(n-3)... (n-r +1)
he calls the right side of the equation essentily r factors.
and this is part of what i don't understand, i don't get the logic or justification of taking the nPn equation and taking out the 3x2x1 part and replacing it with (n-r +1)
then he goes on to say for the two equations:
nPn = n!
and
nPr = n!/(n-r)!
that we should be easily able to satisfy ourselves that (n-r)! in the lower part of the last of the last written fraction just kills off, by cancellaion, the unwanted tail of n! so as to make it properly stop with the r'th factor (n-r + 1)
and here i'm really lost as far as getting the logic

[Sonny]

Originally Posted by shadroch

Okay, thanks for the formula. It's pretty much as I figured it was.
Let me take it a step further, because this is what I'm trying to ascertain.
In Laughlin, you can bet $1 bonus bet on the first hand of a new deck( SD). A BJ pays you 10-1,I believe. If both you and the dealer hit, it pays a progressive, which ranges from $50 to an occasinal $200plus. Is there a dollar figure at which this bet is worth making? Whats throwing me off is the smaller payoffs for only a player BJ. I'm not sure how to factor that payoff in.

I sent you a PM.

-Sonny-

[Sonny]

Originally Posted by sagefr0g

so ok i'm fine up to this point but then he goes on to say (here i get lost)
nPr = n(n-1)(n-2)(n-3)... (n-r +1)
he calls the right side of the equation essentily r factors.
and this is part of what i don't understand, i don't get the logic or justification of taking the nPn equation and taking out the 3x2x1 part and replacing it with (n-r +1)

It’s just a way of writing the formula so that you can replace the n and r with any numbers you want to use. The (n-r+1) just tells you where to stop. You keep multiplying until you have done it r times. The reason we factor out the 4*3*2*1 is because we are only looking at the first 4 positions. We can ignore the results of the last 4 positions. The example below should make things a bit clearer.

Originally Posted by sagefr0g

then he goes on to say for the two equations:
nPn = n!
and
nPr = n!/(n-r)!
that we should be easily able to satisfy ourselves that (n-r)! in the lower part of the last of the last written fraction just kills off, by cancellaion, the unwanted tail of n! so as to make it properly stop with the r'th factor (n-r + 1)
and here i'm really lost as far as getting the logic

It might help if you write out the fraction to see how everything cancels out. Using the same numbers from above:

n = 8
r = 4
nPr = n!/(n-r)!

nPr = 8*7*6*5*4*3*2*1
              4*3*2*1
nPr = 8*7*6*5
nPr = n*(n-1)*(n-2)*(n-3)

The bottom numbers cancel out the top ones in order to eliminate the unwanted end position results. As an example, let’s look at a horse race with 7 horses. How many different ways can the race end assuming there are no ties? Well, any of the 7 horses could finish first. That leaves 6 horses that could finish second, 5 horses that could finish third, 4 horses that could finish fourth, etc. The total number of possibilities is:

n = 7 horses
nPn = n!
nPn = 7*6*5*4*3*2*1

But what if we are only concerned with the horses that finish in the money? If we only look at the first, second and third horses then we still get 7 horses for first, 6 for second and 5 for third, and then we’re done. We can ignore the horses that come in fourth, fifth and the rest of the positions. Using the formula:

n = 7 horses
r = 3 positions
nPr = n!/(n-r)!

nPr = 7*6*5*4*3*2*1
            4*3*2*1
nPr = 7*6*5

-Sonny-