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In a game of poker you are dealt $5$ cards at random from a standard deck of $52$. A standard deck of $52$ contains $13$ different denominations of cards (Ace, 2, 3, ..., 10, jack, queen, king), each in four different suits

(a) What is the probability of being dealt a hand where all $5$ cards are different denominations?

(b) What is the probability of being dealt a hand that includes at least one Ace?

My attempts:

(a) Let D = the event where a hand is dealt such that all $5$ cards are different denominations.

There are ($\binom{13}{5}$ ways to pick the denominations $\times \binom{4}{1}$ ways to pick the suit)/ ($\binom{52}{5}$ total combinations)

(b) Let A = the event where a hand is dealt such that it includes at least one ace.

$P(a) =$ ($\binom{4}{1}$ ways to pick the ace $\times \binom{13}{4}$ ways to pick the other cards $\times \binom{4}{1}$ ways to pick the suit)/ ($\binom{52}{5}$ total combinations)

If someone could walk me through a problem similar to this it'd be really appreciated this stuff is sort of confusing. (poor cs major)

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    $\begingroup$ In (a) you need to pick a suit for each card. $\endgroup$ – Angina Seng Sep 24 '19 at 2:48
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    $\begingroup$ In (b) you neglect the possibility that there may be two or more aces. $\endgroup$ – Angina Seng Sep 24 '19 at 2:49
  • $\begingroup$ @LordSharktheUnknown wait how would I represent that numerically? I thought saying 4 choose one ways to pick the suit would cover that (part a). $\endgroup$ – Mathissohardlmao Sep 24 '19 at 2:54
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(a) What is the probability of being dealt a hand where all 5 cards are different denominations?

We can multiply together the probabilities that have each successive draw satisfy the condition. $$\frac{52}{52}\cdot \frac{48}{51}\cdot\frac{44}{50}\cdot\frac{40}{49}\cdot\frac{36}{48}=\frac{2112}{4165}\approx0.50708$$ On the first draw, you have a probability of $\frac{52}{52}$ of selecting a card that is different from a card in your hand (naturally, as you have no cards in your hand).
On the second draw, you have a probability of $\frac{48}{51}$ of selecting a card that is different from the card you had just drawn (that is, only $\frac3{51}$ match the card in your hand).
Apply for the next three draws.

(b) What is the probability of being dealt a hand that includes at least one Ace?

To get at least one Ace, we must fail the action of getting no Aces. $$1-\frac{48}{52}\cdot\frac{47}{51}\cdot\frac{46}{50}\cdot\frac{45}{49}\cdot\frac{44}{48}=1-\frac{35673}{54145}\approx0.34116$$ Alternatively, we can think of this as the number of ways to choose $5$ from $48$ (non-Aces), out of the number of ways to choose $5$ from $52$ (total cards in deck): $$1-\frac{\binom{48}5}{\binom{52}5}=1-\frac{\frac{48!}{5!\ (48-5)!}}{\frac{52!}{5!\ (52-5)!}}=1-\frac{48!\cdot47!}{43!\cdot52!}=1-\frac{48}{52}\cdot\frac{47}{51}\cdot\frac{46}{50}\cdot\frac{45}{49}\cdot\frac{44}{48}=1-\frac{35673}{54145}\approx0.34116$$ The same calculation! Seems the suits did not have anything to do with the probabilities, after all!

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5 sheep, 5 goats, 5 chickens, each animal type is numbered 1-5

What's the probability of drawing 3 animals with different numbers?

We'll draw one at a time--

1st draw doesnt matter--15 animals

2nd draw can't be the same # as the first--12 choices

3rd draw can't be the same # as the first two--9 choices

Total admissible draws = 15*12*9 divided by 3 factorial (divided by 3! since eg ordered draw (goat #1, chicken #2, sheep #4) and (chicken #2, sheep #4, goat #1) and other reorderings, should count only once); total possible = 15 choose 3; take ratio as you know--that's probability.

The probability a 3-animal draw has at least one goat? I'd recast this as what's the probability we get no goats, then take the complement (1-p)--should be easy to see p is just 10 choose 3 divided by 15 choose 3.

Hope that helps!

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  • $\begingroup$ Thank you. I appreciate this and I'm going to save it lol $\endgroup$ – Mathissohardlmao Sep 24 '19 at 3:18
  • $\begingroup$ Note the correction for the first case--order shouldn't matter, so it's actually (15*12*9)/3!. $\endgroup$ – J Prestone Sep 24 '19 at 5:23
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What is the probability of being dealt a hand where all five cards are of different denominations?

There are $\binom{52}{5}$ ways to select a five-card hand.

There are $\binom{13}{5}$ ways to select five different denominations. For each such denomination, there are $\binom{4}{1}$ ways to choose the suit of the card of that denomination. Therefore, there are $\binom{13}{5}\binom{4}{1}^5$ favorable cases.

Hence, $$\Pr(\text{five different denominations}) = \frac{\dbinom{13}{5}\dbinom{4}{1}^5}{\dbinom{52}{5}}$$ As Lord Shark the Unknown indicated in the comments, you failed to pick a suit for each denomination.

What is the probability of being dealt a hand with at least one ace?

Method 1: Since there are four aces in the deck, such a hand either contains one ace and four other cards, two aces and three other cards, three aces and two other cards, or four aces and one other card. Since there are $4$ aces and $48$ other cards in the deck, $$\Pr(\text{at least one ace}) = \frac{\dbinom{4}{1}\dbinom{48}{4} + \dbinom{4}{2}\dbinom{48}{3} + \dbinom{4}{3}\dbinom{48}{2} + \dbinom{4}{4}\dbinom{48}{1}}{\dbinom{52}{5}}$$

Method 2: We subtract the probability of selecting no aces from $1$. If no aces are selected, we must select five of the other $48$ cards. Hence, $$\Pr(\text{no aces}) = \frac{\dbinom{48}{5}}{\dbinom{52}{5}}$$ Thus, $$\Pr(\text{at least one ace}) = 1 - \frac{\dbinom{48}{5}}{\dbinom{52}{5}}$$

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