Vladimir plays a game in which he reveals cards one-at-a-time from the

Awesome question!

One way to approach this question is to recognize that diamond cards have no effect on Vladimir winning or losing the game. When we get a diamond, we just move on to the next card.
So, if we remove all of the diamond cards from the deck, this will not alter the probability that Vladimir wins the game.
Once we remove all of the diamond cards from the deck, we have 39 cards remaining. 13 cards are hearts, 13 cards are clubs and 13 cards are spades.
At this point, the game will be won or lost on the FIRST DRAW.

So, P(Vladimir wins) = P(1st card is a heart) = 13/39 = 1/3

Answer: D

Cheers,
Brent

probability of winning = 13/51 = 1/4

I have a question here. is the question asking about probability of winning in the first try itself ??

Official solution from Veritas Prep.

We can ignore the diamond outcomes of this game, since they really have no effect at all on the probability – a diamond just means “try again.” Really the only outcomes that matter are hearts, clubs, and spades. That’s 39 total outcomes, of which 13 are wins. The answer is thus \(\frac{13}{39}=\frac{1}{3}\). D is correct.
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Can someone please provide another way to solve this problem?

Is this approach below wrong? I cannot determine because it is not complete.

The question talks about the probability of winning so clubs and spades can be left out.
A. Hearts: In the first try, you get a heart so a win (Calculate probability: selecting Hearts out of the 4 sets and selecting 1 heart out of 13).
B. Diamonds: Possible cases are DH, DDH, DDDH, .....until 13Ds H. (calculate probability)
=> Sum of A+B = answer

In case B, the case DH is considered H. Likewise, DDH is considered H, and D...DH is considered H. Hence, we could left out diamond cards in the deck.
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But why? Question says one at a time. So on the first try D and then an H, similarly DDH.... Did not understand why we should leave out the diamonds. If we did, wouldn't it just be the same as H on the first try (A)?

Yes, one at a time. But the question also said that "If he reveals a diamond, the game continues". This means the game continues until he reveals a card that is not a diamond.
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@Bunuel
Please provided your explanation.

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Veritas Prep Official Solution:

We can ignore the diamond outcomes of this game, since they really have no effect at all on the probability – a diamond just means “try again.” Really the only outcomes that matter are hearts, clubs, and spades. That’s 39 total outcomes, of which 13 are wins. The answer is thus 13/39 = 1/3. D is correct.
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Bunuel - I have a doubt here. I understand that my method is slightly longer but seems well within the stipulated time (1min 30s) to solve it.

P (Winning) = 1/4 (getting a heart in the first go) + 1/4*1/4 (getting a diamond and then getting a heart in the second go) + 1/4*1/4*1/4..... goes on infinitely till he wins. Here we can see, that this equation forms an infinite GP, where r < 1. Hence, to determine the sum we can use the GP formula = (b1/ 1-r)

==> (1/4) / (1-1/4)
==> (1/4) / (3/4)
==> 1/3

I wanted to understand if this method is correct as well, as this can be used for similar questions where P(getting a diamond) DOES count. Please let me know!

Thanks

I think that your formulation of probability is incorrect because it doesn't account for the progressive reduction in the total number of remaining cards from all the disposed diamond cards.

I believe the correct formulation is:

\(P = \frac{13}{52}+\frac{13}{52}(\frac{13}{51}+\frac{12}{51}(\frac{13}{50}+\frac{11}{50}(...(\frac{13}{40}+\frac{1}{40}(\frac{13}{39}))))))))))))\)

The problem is still solvable using this terribly clunky mess, albeit in a thoroughly horrible way.

First, we can quickly eliminate A and E because those are the probabilities of winning on the first reveal and losing on the first reveal, respectively.

So it is between B, C, and D, with B < C < D. There is no need to calculate the actual probability. It suffices to determine whether the solution is equal to, less than, or greater than C.

For starters we can notice that the first two terms in the above sum are:

\(\frac{1}{4}+\frac{13}{52}*\frac{13}{51} > \frac{1}{4}+\frac{13}{52}*\frac{13}{52} =\frac{1}{4}+\frac{1}{16}\)

So \(P > \frac{1}{4}+\frac{1}{16}\), since all of the terms are positive.

We can also notice that:

Answer \(C = \frac{4}{13} = \frac{1}{4}+\frac{12}{208} < \frac{1}{4}+\frac{13}{208} = \frac{1}{4}+\frac{1}{16} < P\)

Therefore the probability is greater than that of answer C, which gives us the correct answer of D.

We should observe that diamonds have no effect in this game; a diamond draw is simply followed by another draw. Therefore, the probability of winning is not affected if we remove all the diamond cards from the deck, leaving us with 39 cards (13 hearts and 26 clubs and spades). After removing the diamonds, Vladimir has 13/39 = 1/3 chance of drawing a heart and winning the game.

Answer: D
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Hi Bunuel,
My approach to this question is like this :
Since any diamond would win and there are equal number of each type of cards we have a total of 4 possibilities.
but club or spade= loss hence out of 4 there are 3 different possibities. Hence the chance of winning will be 1/3.
Please let me know if my approach is right?

dystq You are correct. As someone with an academic applied statistics background, I can tell you that the question is both poorly worded for a probstat nerd and well-worded for the GMAT. The problem should make it very clear they are talking about the expected value as there a ton of ways to model such a scenario. However, after thinking more than the usual 2 minutes about this problem (i need to practice self control in study mode), all reasonable models using information only given in the question (including conditional models as you've suggested) will converge to the expected value. I skimmed past the sentence where diamonds don't count for shanything so I got it wrong.

Anyways, question doesn't say that a "shuffled standard deck of cards" means randomized nor does it say if the revealed card is placed "randomly" or not back into the deck. Apparently *the* "standard deck of cards" is so "standard" that they even provide explicit and detailed information about its contents. They should simply ask "A single Slavic human has odds of 1 to 3 to win a game. What is the human's probability of winning this game?" At least you have to know how to convert odds to probability

goforgmat Your reasoning is the same as mine. I believe it is correct.

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