Probability question regarding blackjack.?

Question

Given a well-mixed deck,what is the probability that neither dealer nor you are dealt blackjack?

- A two-card hand of 21 (an ace plus a ten-value card) is called a "blackjack".

Answer 1

It looks like its easiest to calculate the probability of the complementary event, that you or the dealer DOES get a blackjack, and then subtract from 1.

Let a = ace, and b = K,Q,J,10, and x = anything else.
In one deck, there are 4 a, 16 b, and 32 x.

There are 6 kinds of hand possible:
ab or ba (blackjack), aa, bb, xx, ax or xa, bx or xb

The ways to get at least one blackjack are

(your hand) -- (dealer's hand)
(ab or ba) -- (anything)
(aa) -- (ab or ba)
(bb) -- (ab or ba)
(xx) -- (ab or ba)
(ax or xa) -- (ab or ba)
(bx or xb) -- (ab or ba)

It is fairly straightforward to calculate the probability of each of these ways to get at least one blackjack.
For example, for (aa) -- (ab or ba): the probability of (you) = (aa) is (4/52)*(3/51).
The probability of (dealer) = (ab or ba) is (2/50)*(16/49)+(16/50)*(2/49) = (2/50)*(16/49)*2.
So the probability of this combination is (4/52)*(3/51)*(2/50)*(16/49)*2,
which is (4*3*2*16*2)/(52*51*50*49)

After you calculate the probabilities of all the ways of getting at least one blackjack, add them together to get the total.

The answer is 615680/(52*51*50*49) = 0.094757903

Therefore the probability that NEITHER gets a blackjack is 0.905242097

Answer 2

there is not any such element because of the fact the regulation of averages. All on line casino video games are weighted in favour of the on line casino. Take an ordinary case - roulette - purple or black guess pays out at evens - this means that the of purple or black hazard could desire to be 0.5. yet wait, what's that? this is the fairway 0. So the possibility of purple coming up is slightly decrease than 0.5 (this is 18/37 = 0.486). That distinction ability that interior the long-term you are able to never win. All activities in on line casino video games are hazard ones. All payouts are at slightly decrease than they could desire to be if payouts have been desperate entirely via the probabitity of the form. this means that the on line casino will constantly win interior the long-term, and that the punter never can.

Answer 3

prob of a bj = (1/13)*(4/13)*2
prob of no bj = 1 - [(1/13)*(4/13)*2]
prob of no bj for you and no bj for dealer =
{1 - [(1/13)*(4/13)*2]} * {1 - [(1/13)*(4/13)*2]}

I hope this helps.
Good luck if you are playing blackjack!