In a certain variety of poker, known as Draw Loss Poker, each possible

Question

In a certain variety of poker, known as Draw Loss Poker, each possible poker card combination has a raw point value. The following table shows the raw point values for the card combinations themselves.

one pair = 12 points
two pair = 36 points
three of a kind = 60 points
straight = 300 points
flush = 540 points
full house = 720 points
four of a kind = 3,600 points
straight flush = 60,000 points

Explanation of the card combinations. A straight is a sequence of five consecutive valued cards (e.g. 8, 9 10, Jack, Queen) of any suits. A flush is a set of five cards of the same suit. A full house is three-of-a-kind of one value of card and a pair of another value of card (e.g. three 7's and two aces). A straight flush is a sequence of five consecutive valued cards all of the same suit.

If one's final card combination does not include any of these special combinations, the hand is worth zero points.

Sajjad1994 · Accepted Answer

Official Explanation

3. Player G received the following cards on the initial five-card draw: 2 of Clubs, 5 of Diamonds, 8 of Hearts, 8 of Spades, and King of Diamonds. Player G will choose to discard exactly two cards, the 2 of Clubs and 5 of Diamonds. Over all possible hands that could result, what is the range of the possible point values of Player G’s final hand?

Explanation

Explanation

The five original cards in Player G’s hand are: 2 of Clubs, 5 of Diamonds, 8 of Hearts, 8 of Spades, and King of Diamonds.  Player G discards the 2 of Clubs and 5 of Diamonds, leaving two 8’s and a king.  We know that Player G’s reducing factor for this hand will be 1/6.

The highest possible hand would be four-of-a-kind, in the unlikely scenario that G picked up the other two 8’s.  That would result in (3600)*(1/6) = 600 points, the maximum.

The lowest possible hand would be if G picks up garbage and just has the two 8’s.  That would result in (12)*(1/6) = 2 points, the minimum.

The range of any set is the max minus the min.  Range = 600 – 2 = 598.

Answer = (B)

Sajjad1994 · Answer

Official Explanation

1. Player A initially draws three-of-a-kind and chooses not to discard at all on the discard round. Player B initially draws three-of-a-kind, discards the remaining two cards. For the following final card combinations of Player B, tell whether Player A outscores Player B.

Explanation

Player A draw three-of-a-kind (worth 60 points), and does not discard, so A’s hand is worth 60 points.   In all three cases, B discards two cards, for a reducing factor of 1/6.

(a) B’s final hand = three-of-a-kind, which is (60)*(1/6) = 10 points.    A outscores B .

(b) B’s final hand = Full House, which is (720)*(1/6) = 120 points.   A does not outscore B .

(c) B’s final hand = Four-of-a-kind, which is (3600)*(1/6) = 600 points.   A does not outscore B .

It wasn’t relevant in any of these scenarios, but notice the exact wording — if A and B were tied, the answer would be “A does not outscore B."

Sajjad1994 · Answer

Official Explanation

2. Suppose player J draws two pair and chooses not to discard any cards. Suppose player K draws three-of-a-kind and choose not to discard any cards. Which of the following discard choices and final card combination would have a higher point value than player J’s hand but not than player K’s hand?

Explanation

Explanation

Player J, with two pair and no discards, has 36 points.   Player K, with three-of-a-kind and no discards, has 60 points.   We want a combination worth more than 30 points but less than 60 points.

(A)  discard 1 card, three of a kind = (60)*(1/2) = 30.  Less than 36, no good.

(B)  discard 2 cards, flush = (540)*(1/6) = 90.  Higher than 60, no good.

(C)   discard 2 cards, straight = (300)*(1/6) = 50.  This could work.

(D)  discard 3 cards, full house = (720)*(1/10) = 72.  Higher than 60, no good.

(E)  discard 4 cards, full house = (720)*(1/60) = 12.  Less than 36, no good.

(C)   is the only hand & discard combination that is in the required range.

Sajjad1994 · Answer

Official Explanation

4. Suppose player S draws a straight on the first five-card draw and choose not to discard any cards. Player T initially draws the following five cards: 4 of Diamonds, 4 of Hearts, 6 of Hearts, 7 of Hearts, 7 of Clubs. Which of Player T’s discard choices and final results will outscore Player S?

Explanation

Explanation

Player S’s hand is worth 300 points.  That’s fixed.  Now, we have to compare T to that 300 point total.  discard 4 of Diamonds & 7 of Clubs, winds up with a flush

Scenario #1: discard 4 of Diamonds & 7 of Clubs, winds up with a flush.  Flush is worth 540, times the reducing factor of 1/6, for a point value of 540*(1/6) = 90, which is less than S.   Player T does not outscore player S .

Scenario #2: discard 6 of Hearts, winds up with a full house.  Full house is worth 720, times reducing factor of 1/2, for a point value of 720*(1/2) = 360, which beats S.    T outscores S .

Scenario #3: discards 4 of Diamonds & 4 of Hearts & 6 of Hearts, winds up with four-of-a-kind.  Four-of-a-kind is worth 3600, times a reducing factor of (1/10), for a point value of 3600*(1/10) = 360, which beats S.    T outscores S .

Gemmie · Answer

1. Player A initially draws three-of-a-kind and chooses not to discard at all on the discard round. Player B initially draws three-of-a-kind, discards the remaining two cards. For the following final card combinations of Player B, tell whether Player A outscores Player B.

Player A: 3 of a Kind & No discard => 60 pts

Player B

Three-of-a-kind & 2 Discards => Less than 60
Full house & 2 Discards => \(720 * \frac{1}{6} = 120\) ==> More than 60
Four of a Kind => Higher score than Full house ==> More than 60

2. Suppose player J draws two pair and chooses not to discard any cards. Suppose player K draws three-of-a-kind and choose not to discard any cards. Which of the following discard choices and final card combination would have a higher point value than player J’s hand but not than player K’s hand?

J: 2 pair & No discard => 36 pts
K: 3 of a kind & No discard => 60 pts

Straight & 2 Discards => \(300 * \frac{1}{6} = 50\)
36 < 50 < 60

Answer: Straight & 2 Discards

3. Player G received the following cards on the initial five-card draw: 2 of Clubs, 5 of Diamonds, 8 of Hearts, 8 of Spades, and King of Diamonds. Player G will choose to discard exactly two cards, the 2 of Clubs and 5 of Diamonds. Over all possible hands that could result, what is the range of the possible point values of Player G’s final hand?

G orginally has

2 C => to be discarded
5 D => to be discarded
8 H
8 S
K D

Lowest score: G will have 1 paird & 2 Discards => \(12 * \frac{1}{6} = 2\)

Highest score: (2+ Range) * 6 = score of a combination
Try to plug in each answer choice => Range = 598 gives us 3600 (4 of a kind)

Answer: 598

4. Suppose player S draws a straight on the first five-card draw and choose not to discard any cards. Player T initially draws the following five cards: 4 of Diamonds, 4 of Hearts, 6 of Hearts, 7 of Hearts, 7 of Clubs. Which of Player T’s discard choices and final results will outscore Player S?

S: Straight & No discard => 300
T originally has

4 D
4 H
6 H
7 H
7 C

Flush & 2 Discards: \(540 * \frac{1}{6} = 90\) ==> Less than 300
Full house & 1 Discard: \(720 * \frac{1}{2} = 360\) ==> More than 300
4 of a kind & 3 Discard: \(3600 * \frac{1}{60} = 360\) ==> More than 300

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In a certain variety of poker, known as Draw Loss Poker, each possible

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Player A outscores Player B	Player A does not outscores Player B
		Three-of-a-kind
		Full House
		Four-of-a-kind

	discard 1 card, three of a kind
	discard 2 cards, flush
	discard 2 cards, straight
	discard 3 cards, full house
	discard 4 cards, full house

Player T outscores player S	Player T does not outscores player S
		discard 4 of Diamonds & 7 of Clubs, winds up with a flush
		discard 6 of Hearts, winds up with a full house
		discards 4 of Diamonds & 4 of Hearts & 6 of Hearts, winds up with four-of-a-kind

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In a certain variety of poker, known as Draw Loss Poker, each possible

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