3
$\begingroup$

A card deck with 52 playing cards is well mixed and the cards are indicated successively, until the first ace appears. Is it more likely that the next card is the ace of spades or heart two?

My idea was that the probability of drawing the ace of spades or heart two is the same with $\frac{4}{52}$, but that doesn't seem right. Can anyone help me here?

Thanks in advance.

$\endgroup$
  • 1
    $\begingroup$ There is only one ace of spades in the deck, not four. The probability of drawing an ace of spades from a full and shuffled deck is then $\frac{1}{52}$, not $\frac{4}{52}$. Similarly for the two of hearts. As for the specific scenario you describe, you just drew an ace which could have been the ace of spades itself. It is clearly impossible to draw the ace of spades after having just drawn the ace of spades, so knowing that you just drew an ace should decrease the likelyhood that the next is an ace of spades. $\endgroup$ – JMoravitz Oct 26 '17 at 15:53
  • 1
    $\begingroup$ To more formally approach the problem, consider the related scenario where you take the deck of cards and search for and pull out an ace at random (noting that it will be the ace of spades 1/4 of the time). Take the remaining $51$ cards and shuffle them. Find the probability the card drawn from this smaller deck of $51$ cards is the two of hearts versus the ace of spades. $\endgroup$ – JMoravitz Oct 26 '17 at 15:56
2
$\begingroup$

The probability that the next card following the first ace is the two of hearts is 1/52. Surprisingly, the probability that it is the ace of spades is also 1/52.

To find the probability that the next card is the ace of spades, imagine that we deal out all 52 cards. This can be done in $52!$ ways, all of which we assume are equally likely. We would like to count the arrangements in which the ace of spades immediately follows the first ace dealt. Each ordering of the 52 cards can be produced by first dealing out all the cards except the ace of spades, then inserting the ace of spades into that ordering. There are $51!$ ways to arrange the first group of cards, and there is only one place to insert the ace of spades so it immediately follows the first ace. So there are $51!$ arrangements in which the aces of spades follows the first ace. Therefore the probability that the card following the first ace is the ace of spades is $$\frac{51!}{52!} = \frac{1}{52}$$ By the same argument, simply replacing the ace of spades with the two of hearts above, the probability that the card following the first ace is the two of hearts is also $1/52$.

$\endgroup$
  • 2
    $\begingroup$ I don't think so. For the two of hearts you have to think about the possibility that it's the first card in the deck. For the ace of spades you have to think about inserting it as the first of the aces. If you make these corrections you should get the same answer as @drhab . To see this more clearly, think about it for a three card deck with two aces and the two of hearts. $\endgroup$ – Ethan Bolker Oct 26 '17 at 18:52
  • $\begingroup$ @EthanBolker Playing the game with two of hearts and two aces among which the ace of spades convinced me that I am wrong somewhere, and awkward is correct. This surely is a tricky question. $\endgroup$ – drhab Oct 27 '17 at 9:34
  • $\begingroup$ This question seems tricky, if the condition changes to 'until all the 4 Aces appear', what is your answer ? $\endgroup$ – Frazer Nov 2 '18 at 7:24
  • $\begingroup$ @Frazer I agree, it's a very tricky question. But I don't understand your question exactly--you want to know the probability of what? In any case, if you want to ask a related question, it's usually better to post a new question instead of putting the question in a comment. $\endgroup$ – awkward Nov 2 '18 at 12:58
  • $\begingroup$ @awkward, your answer is wrong, but correct for the "heart two" case. Please take a look at my answer. What I meant is to think about the extreme case that, the probability of the next card after the fourth appeared Ace, of course there is no chance to get another Ace, but still chance to get heart two. But your answer may also lead to be equal. $\endgroup$ – Frazer Nov 4 '18 at 11:29
0
$\begingroup$

This might shed some light.

You can get a well mixed deck of cards like this.

Place a card. Then there are two places for the second card to be placed: at the left of the first card or at the right. Let a coin (or a virtual fair two-sided die) decide about that. Then there $3$ places for the third card: at the left, in the middle or at the right of the two cards that were allready placed. Going on like that we come to a point that $51$ cards are placed and $52$ places equiprobable places are there for the last card. There is only one spot for it such that - if it is placed there - we can say that this last placed card is the next of the first ace. This no matter what card it is: heart two, ace of spades or some other card.

Conclusion: in a well mixed deck of cards every card has probability $\frac1{52}$ to be the next card after the first ace.

$\endgroup$
0
$\begingroup$

It is more likely that the next card is heart two, and the probability of heart two versus ace of spades is 4:3. Let $A$ denotes ace of spades, $B$ denotes heart two, and $a_1, a_2, a_3$ denote ace of other 3 colors. There are $3 \times 3!$ ways to let $A$ be after the first $a_i$:

$\{a_1, A, ...\}$$\{a_2, A, ...\}$$\{a_3, A, ...\}$

and there are $4 \times 3!$ ways to let B after the first $A$ or $a_i$:

$\{a_1, B, ...\}$$\{a_2, B, ...\}$$\{a_3, B, ...\}$$\{A, B, ...\}$

Till now, we can get the versus of probability between the two cards. But for the exact probability, we need more computation. For all the 52 cards, there are $52!$ ways to permute them. For the other 47 cards (exclude $A,B,a_1,a_2,a_3$), there are $47!$ ways to permute them. So the probability of $A$ after the first $a_i$ is

$\frac{47! \times 3 \times 3! \times C}{52!}$

and the probability of $B$ after the first $A$ or $a_i$ is

$\frac{47! \times 4 \times 3! \times C}{52!}$

where $C$ is number of ways to insert the 5 cards (have been permuted) into 48 slots of 47 cards, requiring the first 2 cards to be in the same slot. One solution of $C$ is to consider 1,2,3,4 slots separately.

1 slot: $C_{48}^1 \times C_3^0$, like #####

2 slots: $C_{48}^2 \times C_3^1$, like ##|###, ###|##, ####|#

3 slots: $C_{48}^3 \times C_3^2$, like ##|#|##, ##|##|#, ###|#|#

4 slots: $C_{48}^4 \times C_3^3$, like ##|#|#|#

Sum them all, we get 249900 ways. So the probability of ace of spades next to the first ace is $\frac{3}{208}$, and heart two next to the first ace is $\frac{1}{52}$.

$\endgroup$
  • $\begingroup$ As a way of seeing that it's actually 1/52 for all cards, consider the smaller example of an urn with 2 reds and 2 greens. Pick w/o replacement, the probability that the ball following the first G happens to be Green is 1/2. There's 6 orderings of RRGG and 3 of them have consecutive G's. If you distinguish the G's then 1/4 of the time you get G1 G2 consecutive and 1/4 you get G2 G1 consecutive. Lastly, with the cards, by obvious symmetries your solution says that the there are 48 cards with prob 1/48 and 4 with prob 3/208 of being the first after the first Ace. Doesn't add up to 1. $\endgroup$ – Ned Nov 4 '18 at 12:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.