Question

A poker deck consists of cards ranked 2,3,4,5,6,7,8,9,10,J,Q,K,A, (13 different ranks) each in four suits, for a total of 52 distinct cards. A pinochle deck consists of two copies each of cards ranked 9,10,J,Q,K,A, each in four suits, for a total of 48 cards (24 pairs of identical cards).(1) What is the probability that a five-card poker hand drawn from a pinochle deck contains no duplicate cards?(2) Suppose I select a deck of cards at random, and that deck has a 50% chance of being a pinochle deck and a 50% chance of being a poker deck. I then deal a random five-card poker hand and give it to you. You look at the hand and determine that it could have come either from a poker deck or a pinochle deck. In other words, the hand you are dealt contains no duplicates and it only has cards ranked 9 or higher. Use Bayes’ Theorem to determine the probability that I drew from a pinochle deck.

Answer 1

Answer:

1) The probability to be given a Poker hand from a Pinochle deck is 0.7943

2) The probability that a hand that could be either a Poker hand or a Pinochle hand was given from a Pinochle deck is 0,3675

Step-by-step explanation:

Letc denote some events:

X = The hand i receive is from a Pinochle deck.

Y = The hand i receive is from a Poker deck

Z = The hand i receive is a Poker hand (from any deck)

Note that Y is the complementary event of X, therefore P(Y) = 1-P(X). Also, if Y is true so is Z, thus

P(Z | Y) = 1

we want to know P(Z | X) in order to resolve item 1)

The first card i receive from the five card hand can be anything, the second one can be anything different from the first card, and there is one copy of that card remaining in the deck, thus, 46 cards from the deck works, from a total of 47. After the second card was given, if it was different from the first card, we will have 2 cards that are copies of the 2 cards i receive, thus, there are, from the 46 remaining cases, only 44 are favourable. For the fourth card we will have 3 cards we dont want to draw, therefore we have 42 cases favourable over 45, and for the last card from a total of 44 cards, only 40 can help us.

With all this information, we have that the probability that a hand dealt from a Pinochle deck is a Poker hand is

P(Z | X) = 46/47* 44/46 * 42/45 * 40/44 = 112/141 = 0.7943

2) Since rach deck has a 1/2 probability of being selected, then

P(X) = P(Y) = 1/2.

In order to compute the probability that i draw a hand that could be either kind from a Pinochle deck, i will calculate first the probability that i draw a 'Pinochle hand' from a poker deck, in other words, if we call W the event that i drew a Pinochle hand, then we want P(W | Y)

Since a Poker deck has 52 cards and only 24 works for us, then P(W | Y) = 24/52 = 6/13 = 0.4615.

Note that P(W | X) = 1

Now, lets calculate what the problem wants, and that is P(X | W∩Z) (The probability to have drawn from a Pinochle deck a hand that could be either Poker or Pinochle. For that we use the Bayes Theorem

$P(X | W \cap Z) = \frac{P(W \cap Z |X) * P(X)}{P(W \cap Z |X) * P(X) + P(W \cap Z |Y) * P(Y)} = \frac{P(Z | X) * 1/2}{P(Z |X)*1/2+P(W|Y)*1/2} = \frac{6/26}{6/26+56/141} = \frac{423}{1151} = 0.3675$

[ Note that if X is given, then W always holds, thats why p(W∩Z|X) = P(Z|X). In a similar way P(W∩Z|Y) = P(W|Y), because given Y, Z always happens (a hand given from a Poker deck is always a poker hand). ]

We conclude that the probability that a hand that can be from any deck is in fact given from a Pinochle deck is 0.3675.