A new cards game for four people is consisting of 105 cards: two norma

Question

A new cards game for four people is consisting of 105 cards: two normal cards deck with 52 cards each, and one additional card which is Ace of Spades. The game starts when each player, in turn, pulls randomly 1 card and leaves it in his hand. What is the probability that the third player will pull a black Ace?

(A) 1/18
(B) 1/19
(C) 1/20
(D) 1/21
(E) 1/22

(A) 1/18
(B) 1/19
(C) 1/20
(D) 1/21
(E) 1/22

KarishmaB · Accepted Answer

The probability of picking a black ace on the third pick is the same as the probability of picking a black ace on the first pick. Since there are 5 black aces (2 in each pack and 1 extra), the probability of picking a black ace on the first pick will be 5/105 = 1/21. SO probability of picking a black ace on the third pick will also be 1/21

Answer (D)

Think why conceptually - Let's consider the case of probability of picking a black ace on the second pick.
If a black ace is picked on the first pick, then probability of picking a black ace on the second pick reduces.  (\frac{5}{105} * \frac{4}{104}) 
If a non black ace card is picked on the first pick, the probability of picking a black ace card on the second pick increases.  (\frac{100}{105} * \frac{5}{104})

So overall, taking both cases, the impact is that the probability stays the same.

(\frac{5}{105} * \frac{4}{104}) + (\frac{100}{105} * \frac{5}{104})

= \frac{5*104}{105*104}

= 1/21

For every pick, this would be the case. The probability of picking a black ace will stay the same whether it is the second pick or the third or the fourth etc.

va789 · Answer

Bunuel   VeritasKarishma   chetan2u  other experts,

Please help with this question. Thanks!

chetan2u · Answer

vansh789 
 Since all cards are unbiased, the probability will remain the same for each card . 
So picking black Ace in first pick will be the same as picking black A on 10th or last card. 
Total black aces= 2 each on both packs and 1 extra Ace of spade = 5 
Thus, P= \frac{5}{105}=\frac{1}{21}

Proof : Let the third card not be a black Ace. 
So the third card could be anything else so 100 ways, while total way to pick that card is 105 ways 
1-(100/105)=1-(20/21)=1/21

OR

If third card is not a black ace, the 5 aces could be anywhere in the 104 picks so 104C5 ways. 
Total ways = 105C5
P= 1- \frac{104C5}{105C5}=1-\frac{\frac{104*103*102*101*100}{5!}}{\frac{105*104*103*102*101}{5!}}=1-\frac{100}{105}=1-\frac{20}{21}=\frac{1}{21}

va789 · Answer

Thanks  chetan2u  for your reply.

I got the same answer but followed a different (and a relatively long) approach. 
There could be four possible scenarios in which the third card could be a black ace: +++, +-+, -++, --+ (where "+" represents a black ace and "-" represents not a black ace). Calculating the probability of these cases and then adding them up results in a total probability of 1/21.

Could you please validate whether this approach is correct? Thanks!

PyjamaScientist · Answer

Nice answer but I have a small query. While calculating no black ace in third pick, why did  chetan2u  not consider total available cards as 103 since two were taken in the first two picks?
I did this question the way Vansh did, that is, "There could be four possible scenarios in which the third card could be a black ace: +++, +-+, -++, --+ (+ denotes Black ace). And that was longer than the other approach.
Though, a seemingly simple question, Karishma and Chetan's approach stumped me for sure.
 IanStewart  Could you kindly share your two cents on the theory behind this?

IanStewart · Answer

The fact that two cards have been dealt out really doesn't matter, unless we have some information about those two cards. If we know nothing about those two cards, then the question just becomes "what is the probability the third card from the top of the deck is a black ace?", and the answer to that question is the same as the answer to the question "what is the probability the top card is a black ace" or "what is the probability the fifteenth card from the top is a black ace". A different way to think about it: if a magician spreads this deck of cards on a table and asks you to pick one card at random, what is the probability you pick a black ace? It's 5/105. But the random card you pick could be third from the top, so the answer has to be 5/105.

And if you think about how, reducing the denominator by one each time you go deeper into the deck, you'd answer the question "what is the probability, if all the cards are dealt out that the very last card is a black ace?" then you'll see the answer you get won't make sense (your denominator would need to become '1' in that case).

Mind you, everything changes if you have some information about the first two cards dealt. If you knew, for example, that neither of those cards was a black ace, then the answer does indeed become 5/103, but with no info at all about those first two cards, the third card is just as random as every other card, and is just as likely as any other card to be an ace or a deuce or a red card or whatever the question asks about.

TheUnenlightened · Answer

Wait, is a black ace an Ace of spades or is it it a card that appears twice - once in the first and once in the second deck?

IanStewart · Answer

A standard deck of cards has two black aces, the ace of spades and the ace of clubs.

You don't need to know about decks of cards for the GMAT, though.

mmdfl · Answer

you can also compute the probability of each round:

├── 1st round: get black ace. prob = 5/105.
│ ├── 2nd round: get black ace. prob = 4/104
│ │ ├──  3rd round: get black ace. prob = 3/103 
│ │ └── 3rd round: DONT get black ace. prob = ...
│ └── 2nd round: DONT get black ace. prob = 100/104
│ │ ├──  3rd round: get black ace. prob = 4/103 
│ │ └── 3rd round: DONT get black ace. prob = ...
└──1st round: DONT get black ace. prob = 100/105.
 ├── 2nd round: get black ace. prob = 5/104
 │ ├──  3rd round: get black ace. prob = 4/103 
 │ └── 3rd round: DONT get black ace. prob = ...
 └── 2nd round: DONT get black ace. prob = 99/104
 ├──  3rd round: get black ace. prob = 5/103 
 └── 3rd round: DONT get black ace. prob = ...

prob of third player get black ace = (5/105)* +(100/105)*  = 1/21

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A new cards game for four people is consisting of 105 cards: two norma

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