Anonymous
Anonymous asked in Science & MathematicsMathematics · 6 years ago

# Statistics probability question? (max points)?

You are to pick 4 cards randomly from a condensed deck of cards that contains four suits - ♡,♢,♣,♠ - and the following denomations: Ace, 2, 3, 4, 5, 6, 7, 8, 9, and 10. There are no face-cards in this deck.

What is the probability that you will get

Part (a) a three-of-a-kind?

Part (b) What is the probability that all four cards are of the same suite? For example, all four cards are ♡s?

Part (c) What is the probability you get one-pair?

Part (d) What is the probability you get two aces and two 10s?

Part (e) What is the probability you get two aces and two ♡s?

Relevance

To start with, we will need to know the total number of hands (of any kind) that we can create.

With 40 cards, if we are choosing 4 cards there are "40 choose 4" ways to do that. The "n choose k" formula can be denoted several ways. I'm going to use the notation of C(n,k), where:

C(n,k) = n! / ((n-k)! k!)

C(40,4) = 40 x 39 x 38 x 37 / (4 x 3 x 2 x 1)

= 91390 ways

PART A:

To construct a hand that has 3 of a kind, first pick the rank (A through 10).

C(10,1) = 10 / 1 = 10 ways

From the 4 cards in that rank, pick 3 suits:

C(4,3) = 4 x 3 x 2 / (3 x 2 x 1) = 4 ways

Finally, pick 1 other card from the remaining 36 cards:

C(36,1) = 36 / 1 = 36 ways

Ways to form a 3-of-a-kind:

10 x 4 x 36

= 1440 ways

P(three of a kind) = 1440 / 91390

= 144/9139

≈ 1.6%

PART B:

To construct a flush (all the same suit), first pick the suit:

C(4,1) = 4 / 1 = 4 ways

Now from the 10 cards in that suit, pick 4 cards:

C(10,4) = 10 x 9 x 8 x 7 / (4 x 3 x 2 x 1) = 210 ways

Ways to form a hand of all the same suit (flush):

4 x 210

= 840 ways

P(flush) = 840 / 91390

= 84/9139

≈ 0.92%

PART C:

To construct a hand with one pair, first pick the rank of the pair (A through 10):

C(10,1) = 10 ways

Pick the two suits for the pair:

C(4,2) = 4 x 3 / (2 x 1) = 6 ways

From the 9 remaining ranks, pick 2 cards that aren't part of a pair:

C(9,2) = 9 x 8 / (2 x 1) = 36 ways

Pick the suit for each card:

C(4,1) x C(4,1) = 16 ways

Ways to form one-pair:

10 x 6 x 36 x 16

= 34560 ways

P(one pair) = 34560 / 91390

= 3456/9139

≈ 37.8%

PART D:

Pick the suits for the two aces:

C(4,2) = 4 x 3 / (2 x 1) = 6 ways

Pick the suits for the two tens:

C(4,2) = 4 x 3 / (2 x 1) = 6 ways

Ways to form two aces and two tens:

6 x 6

= 36 ways

P(pair of aces, pair of tens) = 36 / 91390

= 18/45695

≈ 0.04%

PART E:

This one isn't fully clear. Are we to assume the aces can't be one of the hearts?

If so, we have *three* suits to pick for the aces.

C(3,2) = 3 x 2 / (2 x 1) = 3 ways

And we have *nine* cards in the hearts (not counting the ace) that we can pick two cards from.

C(9,2) = 9 x 8 / (2 x 1) = 36 ways

Ways to form two aces and two hearts:

3 x 36

= 108 ways

P(two aces and two hearts) = 108 / 91390

= 54/45695

≈ 0.12%

• Puzzling
Lv 7
6 years agoReport

I suppose they want to include the case of the ace of hearts. If so, we can also have 1 ace non-hearts (3 ways), 1 ace of heart (1 way), 1 heart (9 ways) and then a non-ace, non-heart (27 ways). So that would be an additional 729 ways.
(729 + 108) --> 837/91390 ≈ 0.92%

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• total 40 cards. picking any card between A and T is 4/40 0.1(10% of chance)

A: divide total amount of possible AAAx(4*3*2*36) combinations by totalcombinations(40x39x38x37) . Around 0.0004 so its around 0.045%

B:Divide amount of all possible XsXsXsXs combinations(10x9x8x7) by total amount of hands. 0.23%

Not gonna count anymore:

C: All possible amount of PPxx combinations(40*3*36*35) by total amount of combos around 7%

D: (4*3*4*3) divide by total combos

E: (3*2*9*8) divide by total combos(if third suited ace doesnt count into the combos)

Im not sure if it works,

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