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The problem asks us to find the probability of a full house in a well-shuffled deck of $52$ cards.

The solution in the textbook states in the first line "All of the $52\choose 5$ possible hands are equally likely by symmetry, so the naive definition [of Probability] is applicable."

Everything after this involves counting, which I understand. However, I do not see why all outcomes are equally likely. Here is why.

Say we chose a card with rank $7$. The number of $7$'s left are now less than that of other ranks. This must mean the probability of choosing a different rank must be more than that of choosing another $7$. This means that the probability of an outcome (a hand) with ranks $(2, 3, 4, 5, 6)$ must be more than that of a hand with ranks $(2, 2, 2, 3, 3)$.

Kindly explain why the naive definition works here and why we treat every hand (all $52\choose 5$ hands) to be an equally likely outcome.

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    $\begingroup$ By "hand" they mean one single, specific hand. (ace of spades, ace of hearts, eight of hearts, ten of clubs, five of hearts) is a hand. But $(2,3,4,5,6)$ is a set of multiple hands, since the suits factor into what a hand is. $\endgroup$ Sep 9, 2021 at 6:43
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    $\begingroup$ There are ${4\choose 1}^5=1024$ hands with ranks $2,3,4,5,6$ but only ${4 \choose 3}{4 \choose 2}=24$ with ranks $2,2,2,3,3$. Each individual hand is equally likely and this makes the first pattern more likely than the second $\endgroup$
    – Henry
    Sep 9, 2021 at 6:45
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    $\begingroup$ The problem in the textbook might be clarified by instead asking to find the probability of being dealt a full house; i.e.: a single deal of 5 card stud with no card replacements after the initial deal. In this case, there is no "choosing" a particular card; just a random deal of 5 cards out of 52. $\endgroup$
    – spuck
    Sep 9, 2021 at 16:58
  • $\begingroup$ It's true that if the first card dealt is a 7 (a 4/52 chance), the chance of the next card being a seven has dropped to 3/52. But if the first card is a 7 of hearts, there is zero impact on the chance of the next card being one of the other sevens. $\endgroup$
    – CCTO
    Sep 9, 2021 at 17:16
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    $\begingroup$ Another thing to take into account when you're battling with your intuition is that when they say all hands are equally likely is that you should think of a hand as one event not that you look at one card and then reconsider what the rest of the cards will be. $\endgroup$ Sep 9, 2021 at 19:21

4 Answers 4

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Your examples of (2,3,4,5,6) and (2,2,2,3,3) are not hands for the purpose of the question.

(2♦,3♣,4♥,5♠,6♦) and (2♦,2♣,2♠,3♥,3♣) are hands, and are equally likely because the chance of pulling each of the cards involved is equally likely, up to the symmetry of rearranging the order of cards in the hand.

The idea that it's more likely a card will be junk than help to form a useful hand is a good intuition for why some hands are more or less likely, and can form the basis of a different method of calculating the probability of a given hand or class of hand, but is irrelevant in this case.

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A full house means you have $2$ of kind $+$ $3$ of kind. So we need to find the number of possible hands that satisfy that. How? Well we know there are $\binom{52}{5} = 2,598,960$ possible hands.

Think about it, a "full house" means you're choosing two ranks: could be $2$'s and $3$'s, $2$'s and $4$'s, $2$'s and $5$'s, etc. So let's look at one possible example: you get a full house: $3$x$2$'s and $2$x$3$'s. How many hands have this? There are $\binom{4}{3} = 4$ ways to get $3$ two's and $\binom{4}{2} = 6$ ways to get $2$ three's: there are $24$ ways to get this hand (out of ~$2.5$ million hands).

Would it change if, instead, I said we should look at $2$x$3$'s and $3$x$2$'s? So the number of full houses with $2$'s and $3$'s would be $2\cdot24 = 48$ ($24$ ways to get three $2$'s and two $3$'s and $24$ ways to get two $2$'s and three $3$'s). Now how many ways can we choose two ranks? $\binom{13}{2} = 78$. So there are $78\cdot 48 = 3,744$ ways to get a full house.

This result is indeed what is derived everywhere else: Probability of getting a full house

But note, I'm using combinations, not permutations like (at least my reference): so I'm saying (and obviously--they're all mathematically equivalent):

\begin{align*} p(\text{full house}) = \frac{\left(\binom{4}{3}\binom{4}{2}\cdot\binom{2}{1}\right)\binom{13}{2}}{\binom{52}{5}} \end{align*}

Addressing Question about Drawing First Card

I would expect that the probability of a full house doesn't change after your first card. So if we know the first card, then there are now $\binom{51}{4} = 249,900$ ways of choosing the next four. How many ways can we get a full house? Well there are two now:

  1. Get two more of your first card and two-pair of another
  2. Get one more of your first card and three-of-a-kind of another

There are $\binom{3}{2} = 3$ ways of getting two more of your first card and $\binom{12}{1}\binom{4}{2} = 72$ ways of getting two pair from another rank. So $3\cdot72 = 216$ ways of doing this.

There is $\binom{3}{1} = 3$ ways of getting one more of your first card and $\binom{12}{1}\binom{4}{3} = 48$ ways of getting three of a kind of a different rank: $144$ ways.

So there are a total of $360$ ways to get a full house after your first card is dealt. Out of how many? $\binom{51}{4} = 249,900$:

$$ \frac{360}{249900} \approx 0.001\ 441 $$

Again, you will see this corresponds to the probability of getting a full house. In other words: the probability of getting a full house is independent of the first card drawn (now--second card! no).

Second card

Intuition: you are far more likely to get a full house if you draw two-of-kind. Let's try to test that:

Case 1: You draw two-of-a-kind

Two ways to get a full house:

  1. One more of your card and 2-of-a-kind, $\binom{2}{1}$ and $\binom{12}{2} = 66$, leading to $132$ ways
  2. Draw 3-of-a-kind: $\binom{12}{1}\binom{4}{3} = 48$

$180$ possible ways to get a full house.

Case 2: You get two different ranks

Still two ways to get a Full House: but they're obviously symmetric. The probability is draw one more of one, then two more of the other:

$$ 2\cdot\binom{3}{2}\binom{3}{1} = 18 $$

This is all divided by $\binom{50}{3} = 19,600$. So the probability of getting a full house given two of a kind is $\frac{180}{19600} \approx 0.009 184\ 2$ (meaning your chances have increased about 10x) and the probability of a full house given two different ranks is $\frac{18}{19,600} \approx 0.000\ 918$ (meaning your chances have decreased).

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I'm going to add a second answer (highly inappropriate--except I think in this case). My answer and others are making an assumption that in a sense disregards your question. You asked "why do we treat all hands as equally likely" and the answers you received were basically "well if you treat all hands equally likely, here's what happens." (at least that describes my answer, maybe not others).

In truth the problem is far more complicated. We do have to make an assumption: the deck deals out each card with equal probability and without replacement. If we have two players, in five card draw, you get what you get (easiest to analyze), the deck deals a total of $2*5 = 10$ cards. Therefore we have $\binom{52}{10} = 15820024220$ (~$15.8$ billion combinations). For each of those combinations there are $\binom{10}{5} = 252$ possible combinations of sets of "2-hands".

So now we need to find how many of these combinations contain a full house and what's the probability that the full house ends up all on the same side (reminder they could both have full houses!).

I really don't know how to do this computation. I would start with one full house and none of the two ranks. We can figure that out. We know there are ~$3$k ways to get a full house, now we choose 5 cards from the $52 - 8 = 44$, to get that we can have a single full house hand with no other possibility of $\binom{44}{5}*\#(\text{full house hands}) \approx 4066013952$ (~$4.1$ billion)!

We know that there are $252$ ways to "permute" this and that only one (two actually) puts the full house into one hand. So if we assume that a single full house is dealt within $10$ cards, the probability that "you" get it is $\frac{1}{2}\frac{2}{252}$.

But remember--the probability of you even getting into this situation is ~$\frac{4}{15} \approx \frac{1}{3}$, so you still only have a ~$\frac{1}{1000}$ chance of getting a full house (in this specific scenario). Your chances increase as we add more possibilities (but obviously not by much because the chances of having a "10-hand" that has multiple full houses is small comparatively).

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I think your error is in the following assumption... "Say we chose a card with rank 7. The number of 7's left are now less than that of other ranks. This must mean the probability of choosing a different rank must be more than that of choosing another 7."

The chance of any specific card in a shuffled deck occupying any specific place in the deck is equal across the whole deck.

Looking at cards does not change the initial distribution determined by the original deck shuffling. The chance that the second card is a 7 is still 1 in 52.

Shuffling is the event sparking the random distribution...not observing the cards.

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    $\begingroup$ well, yeah, in a shuffled deck, the second card from top is one of the four 7s at a probability of 4/52 (same as the probability any other single card being one of the four 7s). But, taking into account new information from peeking at the top card, we can and have to recalculate. In particular, it's obvious that the probability of the top card being a 7 is no longer 4/52, but either 1/1 or 0/1. We know what it is. And we also know how many 7s there are left in the 51 remaining cards, so with that information the probability of the second card being a 7 is now either 4/51 or 3/51. $\endgroup$
    – ilkkachu
    Sep 9, 2021 at 19:33

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