1) Cards are drawn (without replacement) from a deck (of 52

Question

1) Cards are drawn (without replacement) from a deck (of 52 cards) 
until an ace is obtained. 
If this procedure is repeated often, what is the average number of draws (required to draw the first ace) ?

jcgoodchild · Answer

There are 4 aces in a deck of 52 cards. The probability of obtaining an ace at any point is therefore 1/13. So if this is done into infinity, I would imagine the average number of cards before getting an ACE to be 13.

Yurik79 · Answer

1/13 remains I guess

remgeo · Answer

1 out of every 13 is an ace. so 13 draws required to get an ace. 
Ans=13

thenine · Answer

I can't get to the right answer but i do not think it is 13. 

The reason I do not think the answer is 13 is that the cards are not replaced after each pull.

If the cards where replaced it would be easy. With each pull the odds would be 1/13 because there are 4 aces and 52 cards.

However if the cards are not replaced the odds change each time.

The odds of pulling an ace on the first pull are 4/52 or 1/13.

On the second pull however the ads are 4/51, on the thrid 4/50. 

So since the cards are not being replaced the oddss are getting better with each pull, thus reducing the average. I can't come up with the math to acutally solve the problem however. My guess would be like 1/10

Beyond700 · Answer

Assuming that arithmetic mean of the possible draws need to be calculated. 

Set of possible draws will be,
 {4/52, 4/51, 4/50,.....4/4} total no. of elements can be 49, 49th event will be 4/4 when there is a certain pick of an ace. 

Therefore average - Sum of elements / no. of elements

 = 4 / 49 * (1/52 + 1/51 + 1/50+...+1/4).

Not sure how to resolve this further..

ccax · Answer

Well, I'm not sure if this is correct, but let's give it a try:

There are 4 positions between 1 and 52 inclusive for the
4 aces. That results in 52*51*50*49 = 6497400 different
possibilities for the game.

Now we need to add the number of draws for each of
the possible games and divide this by 6497400 to get
the average number of draws.

Since this isn't trivial, let's start small with a deck with
4 cards and 4 aces. The answer is 1 of course, because
in every possible game there's an ace in first position.

With 5 cards, there are 120 different games. We only need
to have the number of games where there isn't an ace in
first position. Since there are 24 combinations for the aces
at positions 2,3,4 and 5, the average number of draws is
(96*1 + 24*2) / 120 = 6/5.

With 6 cards there are 360 different games. There are 48
combinations where the first ace is in 2. position (BAAAAB)
and 48 positions where the first ace is in 3. position (BBAAAA).
The average number of draws is
((48*2 + 48*3 + (360 - 48 - 48))/360 = 7/5.

With 7 cards there are 840 different games/combinations.....
The average number of draws is 8/5.

So (if I calculated right which I'm not sure about) the
average number of draws for the first ace in a game
with a deck of X cards including 4 aces seems to be
(X+1)/5.

So for the deck with 52 cards, the average number of
draws should be 53/5 = 10.6.

If anybody finds a fault in the argumentation above,
please let me know.

1) Cards are drawn (without replacement) from a deck (of 52

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