What Is The Probability That You Draw 2 Cards And They Sum 10?

Playing Blackjack

You are playing blackjack at the Turning Stone Casino. They are using a single standard deck of 52 playing cards, and have already played out the hands of all the players before you. The cards that you run into on the table are:

J ix            A 5 7 8            x 2 Grand            6 6          ? 9
hand 1      hand two            hand 3             you        dealer

The dealer has 1 card turned down that you lot cannot run across, denoted "?".

The rules of blackjack are: all cards count their confront value; J, Q, and G count as 10; aces A count as 1 or 11 at the players pick. You desire to draw cards and score as shut as possible to 21 without going over.

a) It is your plough to draw or agree. If you draw a card, what is the probability that you will become over 21?

b) What is the probability that you will end up with exactly 21?

c) You actually depict a six. How likely was that?

d) After you draw 6 in part (c), you have a total of 18. What is the probability that the dealer already has a total greater than 18?

Solution :

a) We volition go over 21 if we draw a card of value 10. There are 16 such cards in a standard 52 bill of fare deck: four Ks, iv Qs, 4 Js, 4 10s (henceforth we will denote all these cards equally "10"s). Of the 13 cards on the table, in that location are iii cards that nosotros know are 10s, plus the unknown menu "?" of the dealer which might or might not be a 10.

The probability we volition depict a 10 depends on whether "?" is a 10 or non. If "?" is a x, and then in that location are 4 10s on the table, and then 12 10s left in the deck. Since the deck contains 39 remaining cards, nosotros have:

prob to depict a 10 if "?" is a x = 12/39

If "?" is not a 10, then there are only 3 10s on the table, and then xiii 10s left in the deck. Since the deck has 39 cards remaining, nosotros have:

prob to draw a ten if "?" is non a 10 = 13/39

Now nosotros need to effigy out the probability that "?" is a 10 or non. Aside from "?" itself, in that location are 12 cards on the table, iii of which are 10s. "?" clearly cannot exist i of these 12 cards. It must be 1 of the 40 other cards, thirteen of which are 10s. Hence:

prob "?" is a 10 = 13/40

The probability "?" is not a 10 is just ane-(13/40), since "?" either is or is not a 10, and so the probability for these two mutually exclusive events must sum to unity. Hence:

prob "?" is not a 10 = 27/40

At present nosotros need to combine the above pieces to become our answer. Nosotros can divide up all outcomes in which nosotros practice draw a x into two mutually exclusive categories: (ane) "?" is a 10, and (2) "?" is not a ten. The total probability is therefore the sum of probabilities for the outcome to be in each category, i.due east.

prob draw a x = (prob to draw a ten if "?" is a 10) x (prob "?" is a x)

                         + (prob to draw a 10 if "?" is not a x) x (prob "?" is non a 10)

Notation, at present that we did the calculation the long way, nosotros meet that the respond is the same was we would take gotten if we had completely ignored the card "?", i.e. causeless it was not on the table. In that example the probability to draw a x is only the xiii 10s not face up on the table, divided by the forty cards not face on the table. This works out correctly because, since we practise not know annihilation virtually the value of "?", it plays a role identical to any of the 39 cards remining in the deck. This way of thinking is a flake subtle, then if you don't see it, the long fashion of doing the calculation as done to a higher place is the proof that it really is so.

b) For us to end up exactly with 21 we need to draw a 9. There are 4 9s in the deck, two of which are confront on the table. We can once again practice the problem two ways. The long way as above is:

prob depict a ix = (prob to draw a ix if "?" is a 9) ten (prob "?" is a 9)

                       + (prob to draw a 9 if "?" is not a 9) x (prob "?" is not a 9)

The brusk style is to ignore "?". At that place are then two 9s left in forty cards, so the probability to draw a nine is ii/twoscore = 0.05, the same answer as above.

c) There ar 4 6s in the deck, two of which are confront on the table. To get the probability we depict a six, nosotros can employ the "brusque" method. There are ii 6s left in xl cards. The probability we drew a 6 is therefore

ii/40 = 0.05

d) We have drawn a vi. There are at present 13 cards face up on the table plus "?". The probability that the dealer already has a total greater than 18 is the probability that "?" has value 10, or is an Ace (which tin can count every bit 11 giving the dealer 20). There are 20 cards in the deck which are either 10 or A. Four of these cards are face upward on the table, so sixteen remain in the 39 cards that are non face on the table. "?" is equally likely to be whatever of these 39 cards. The probability the dealer already has greater than 18 is therefore:

16/39 = 0.41

Source: http://www.pas.rochester.edu/~stte/phy104-F00/HW/blackjack.html

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