# Expected number of cards you should turn before finding an ace

NOTE: I want to check my solution only

Same question here

The question is this:

Shuffle an ordinary deck of 52 playing cards containing four aces. Then turn up the cards from the top until the first ace appears. On the average how many cards are required to be turned before producing the first ace ?

I want to check my solution.It definitely matches with the answers in that other question but I don't understand their solutions(haven't done much study on probability yet). I want to check mine:

We consider all the cards except the aces indistinguishable,and we will consider the aces to be indistinguishable.Now a deck is just a binary string with $48$ $C$ and $4$ $A$ ($A$ stands for aces and $C$ stands for the other cards.Obviously,the number of such strings is $\dbinom{52}{4}$. Now we divide into cases:

1)Number of strings with the first A in the $1$st position:$\dbinom{51}{3}$

2)Number of strings with the first A in the $2$nd position:$\dbinom{50}{3}$

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49) Number of strings with the first A in the $49$th position:$\dbinom{3}{3}$

Since there are $4$ A's,the first $A$ cannot be in the $50$th position.

Note that for case $1$ above,we need to turn $0$ cards before getting an ace.For case $2$,we need to turn $1$ card before getting an ace,.....,in the $49$th case,we need to turn $48$ cards before getting an ace.Therefore,the average of it all is:

$$\dfrac{48\dbinom{3}{3}+47\dbinom{4}{3}+....+0\dbinom{51}{3}}{\dbinom{52}{4}}$$

$$=\dfrac{\dbinom{52}{5}}{\dbinom{52}{4}}$$ $$=\dfrac{48}{5}$$

which is indeed the answer given there.Note that the numerator was computed by repeated application of the hockey stick identity.My questions are :

1)Is my solution correct?

2)Why does assuming indistinguishability still preserve the answer?I have seen this several times,but never really thought about it.

3)If we assumed distinguishability of the cards,we will get a huge expression..How can we compute that?

• The question considers the aces to be the same, so there's no reason to distinguish them. – user21820 Feb 8 '15 at 9:29
• @user21820 when we say a deck,I take it to mean that all the cards are distinguishable.So assuming that's what the question meant,I don't get why making them indistinguishable still preserves our answer. – rah4927 Feb 8 '15 at 9:38
• You don't get it. The question only asks about getting an ace. It doesn't care which ace it is. It could be 52 bottles on a wall, with 4 of them empty, and their other characteristics aren't considered by the question. – user21820 Feb 8 '15 at 9:40
• @barakmanos: Look, the question does not change if we use 52 bottles that are identical except that 4 are empty and the rest are full. The erasure of all other information like size, weight, colour, does not affect the probability desired. If you don't get it, just set up the obvious many-one correspondence between the arrangements of the original objects and the arrangements of 4 ones and 48 zeros, and the number of solutions divided by the number of possibilities remains exactly the same because each simplified arrangement corresponds to exactly $4!48!$ original arrangements with no overlap. – user21820 Feb 9 '15 at 5:03
• @rah4927: You could try a small case of say 3 out of 7 cards instead of 4 out of 52, and explicitly list all the solutions and their corresponding binary strings, and you can see for yourself that the correspondence will be exactly $3!4!$ to $1$, hence the desired ratio $\frac{\#(\text{solutions})}{\#(\text{possibilities})}$ remains the same. – user21820 Feb 9 '15 at 5:07

Imagine you have the following setup:

_A1_A2_A3_A4_

Each ace separated out evenly and we are interested in the pile that's before A1. For a standard deck of cards you have 52 cards - 4 aces = 48 cards left, and $$\frac {48} 5 = 9.6$$ cards for each pile. So basically you would have to turn all 9.6 cards + the A1 card in order to see the first ace. So the answer is $$1 + \frac{48} {5} = 10.6$$

The probability that the seven of clubs turns up before the first ace is $$\frac15.$$ This is because each of the five relevant cards (the four aces and the seven of clubs) has the same one-in-five chance of being first.

Likewise, the probability that the jack of diamonds turns up before the first ace is $$\frac15,$$ and the same goes for the queen of hearts, the four of spades, and every other non-ace card in the deck. Since there are $$48$$ non-ace cards, the average number of (non-ace) cards preceding the first ace is $$48\cdot\frac15.$$

• This is a really nice solution. – saulspatz Oct 27 '18 at 16:34

I propose really basic approach which turns out to give really complicated algebra.

Let $X_i = \{Ace,\ Other\}$ be the card drawn at round $i$. We are interested in the quantity $Pr[X_i = Ace \wedge X_j = Other,\ 1 \leq j < i]$ which is the probability that we get the first ace at round $i$.

First we define for $i \in \mathbb{N}$: $$f(i) = \begin{cases} Pr[X_i = Other | X_j = Other,\ 1 \leq j < i] = \frac{48-i+1}{52-i+1} & \text{if } i > 1 \\ Pr[X_1 = Other] = \frac{48}{52} & \text{if } i = 1 \end{cases}$$ Note that $Pr[X_i = Ace | X_j = Other,\ 1 \leq j < i] = 1-f(i)$

Then using the fact $Pr[A \wedge B] = Pr[A|B]\cdot Pr[B]$ we can compute : \begin{align} &Pr[X_i = Ace \wedge X_j = Other,\ 1 \leq j < i] = \\ &Pr[X_i = Ace | X_j = Other,\ 1 \leq j < i] \cdot Pr[ X_j = Other,\ 1 \leq j < i] =& \\ &Pr[X_i = Ace | X_j = Other,\ 1 \leq j < i] \cdot Pr[ X_{i-1} = Other | X_j = Other, \ 1 \leq j < i-1] \cdot Pr[X_j = Other, \ 1 \leq j < i-1]... \end{align}

Continuing to apply the fact in chain we get that : $$Pr[X_i = Ace \wedge X_j = Other,\ 1 \leq j < i] = (1-f(i)) \cdot \prod_{j=1}^{i-1} f(j) = \\(1-\frac{49-i}{53-i}) \cdot \frac{(i-53)(i-52)(i-51)(i-50)}{52\cdot51\cdot50\cdot49} = -\frac{4(i-52)(i-51)(i-50)}{52\cdot51\cdot50\cdot49}$$ Now taking the expectation and using some Mathematica because of laziness we can get : $$\sum_{i=1}^{52}-i \cdot\frac{4(i-52)(i-51)(i-50)}{52\cdot51\cdot50\cdot49} = 53/5$$ So on average you will get the first ace at card number $10.6$ !