# Why is the probability that second card is ace in a deck is $\frac{4}{52}$.

I did a question "We have a deck of card and we take one card we don't know what it is then what's the probability that the second card is an ace" and correct answer is $$4/52$$ why? book says it's because we don't know what the first card was.

Another question that I did was " some guy has a key chain and there are $$n$$ keys on the keychain and he wants to open a door. he uses first key if it doesn't work he discards it then use second one and so on. what's the probability the 3rd key opens the door" answer to this question turns out to be $$1/n$$ Why?

Edit: both questions are without replacement.

• The Ace of spades, say, is equally likely to be in any position in the deck, yes? so the probability that it is in position $2$ is $\frac 1{52}$. Similarly for the other three aces, and $4\times \frac 1{52}=\frac 4{52}$ Similarly for your key. – lulu Sep 24 '19 at 18:06
• If you were asked the probability of the first card being an ace, would it worry you that you don't know the second card? – Angina Seng Sep 24 '19 at 18:07
• If you have $n$ children and today is one of their birthdays, what is the probability that it is your second eldest child's birthday? If you randomly select a bottle of water from a shelf of $n$ identical water bottles, what is the probability that you select the second bottle from the left? – Andrew Chin Sep 24 '19 at 18:09
• In the second problem, you need to assume that (1) exactly one of the keys on the chain opens the door, and (2) each of the keys is equally likely to be the one that opens the door. As the question is stated, the answer could very well be $0$ (e.g. if he doesn't have a key to that door) or $1$ (if he has arranged the keychain in advance so that the third key is the one...). – Robert Israel Sep 24 '19 at 19:27

If the question was "what is the chance the first card is an ace", then you would say $$\frac{4}{52}$$, right? And if the question was "what is the probability the first key is the one that opens the door", you would say $$\frac{1}{n}$$, right?

OK, so now think about this: is the first card any more or less likely to be an ace than any other card in the deck? Is any key on the ring any more or less likely to be the key you're looking for? No! So, those probabilities are all the same.

I think what you're confused about is the following: Suppose the person with the keys tries the first and second key, and they both fail to be the right key. Then, what is the chance that the third key is the right key? Now, it is of course no longer $$\frac{1}{n}$$, but rather $$\frac{1}{n-2}$$. However, that was not the question ... the question was to figure out the chance of the third key in the chain of keys being the right key before we tried any of the keys. And, as such, the situation is exactly the same for every key on the chain: they all have the same chance of being the right key, so they all must have a chance of $$\frac{1}{n}$$ of being the right key. So, don;t confuse these two questions.

OK, one more: suppose you are one of $$n$$ people, and you are told to stand in a line, and then someone will randomly pick one of you from that line to give $$100$$ dollars. Now, if you want to optimize your chances to win the $$100$$ dollars, where should you stand? And the answer is of course: it doesn't matter where you stand! You all have the same chance of getting picked (assuming this is indeed truly random ...)

Well, it's symmetry in a way.

The second card could literally be any card, with equal chance. What's the chance that the $$47$$th card is an ace? Well, like any other rank, it's $$\displaystyle \frac{4}{52}$$.

Without using symmetry, we can do case-by-case analysis.

The probability that the first card is an ace and the second card is an ace is $$\displaystyle \frac{4}{52} \cdot \frac{3}{51}$$.

The chance that the first card is not an ace and the second one is an ace is $$\displaystyle \frac{48}{52} \cdot \frac{4}{51}$$.

Add these two and you get $$\displaystyle \frac{4}{52}$$.

Check that whether the first card was an ace or not (which changes the conditional probabilities that the second card is an ace) still adds up to what theyre saying...

Ace, ace plus not ace, ace: \begin{align} & \frac{4}{52} \cdot \frac{3}{51} + \frac{48}{52} \cdot \frac{4}{51} \\ &= \frac{4}{52} \cdot \left( \frac{3}{51} + \frac{48}{51} \right) \\ &= \frac{4}{52}. \end{align}

• @Leucippus Don't be silly. Of course it's an answer. Not the best answer, but still. – Robert Israel Sep 24 '19 at 19:31

Note that in any of your $$2$$ cases, instead of drawing cards/keys one by one, we can at the begginging just arrange them in a sequence (for cards it's $$(x_1,...,x_{52})$$, where every $$x_i$$ is one of the cards, and for keys it's a sequence $$(y_1,...,y_n)$$, where any $$y_i$$ is one of the keys. Note there are $$4$$ cards that will satisfy you when they'll be at $$x_2$$, whereas there is only $$1$$ key that you want to be $$y_3$$. In how many ways u can arrange such $$52-$$ and $$n-$$ sequence? For cards. Choose one ace from $$4$$ and put in on $$x_2$$, then in any order place everything left on $$51$$ places left ( $$51!$$ ways). Divide everything by $$52!$$ for it is the total amount of ways to place your $$52$$ cards. So we get $$\frac{4 \cdot 51!}{52!} = \frac{4}{52}$$. With keys it's just the same type: Choose your good key to be on place $$y_3$$ and arrange every key left with $$(n-1)!$$ ways. Divide by $$n!$$ ways, to get $$\frac{1}{n}$$

Note that we could choose any of $$x_i$$ and $$y_i$$ and answer will be the same.

It's due to fact, that probability isn't affected by occurence of an event, but only by knowledge we can have due to occurence of that event.