Probability of drawing a card

[Piers]

I read online, that if you have a deck of 60 cards, and you must draw seven, the probability of getting a specific card is 7/60.

What I don't understand, is why isn't this probability equal to: (1/60)+(1/59)+(1/58)+(1/57)+(1/56)+(1/55)+(1/54)? Both answers round to 12%, but this one gives 12.295% while the article's gives 11.667%.

My method seems to make more sense, but I wanted to ask here to be sure.

 

[Citan Uzuki]

The answer you found online is correct. To see why, let's look at a simpler example: you have a deck of 60 cards, and draw two. Now, this event can happen in one of two mutually exclusive ways: either you get the card on the first draw, or on the second. The probability of getting the card on the first draw is plainly 1/60. Now, you seem to be thinking "after I've drawn the first card, there are only 59 cards left in the deck, so the probability of getting the card on the second draw is 1/59." But in fact, this is only true on the condition that you didn't get the correct card on the first draw. If you did get the correct card on the first draw, the probability of getting on the second draw is zero! The unconditional probability that you'll get the correct card on the second draw is thus the probability that you failed to get it on the first draw times the probability that you get it on the second draw, given that you didn't get it on the first draw. Thus, the probability of getting the card on the second draw is 59/60 * 1/59 = 1/60, and the probability of getting the card on either draw is 1/60 + 1/60 = 2/60.

This result makes sense, because there's a bijection between the outcomes where you get the card on the first draw, and the outcomes where you get the card on the second draw (swap the order of the cards). Hence, the two events should have the same probability. Generalizing this argument, we see that the probability of getting a specific card after drawing n cards is n/60, as the online site claims.

 

[phinds]

The answer you found online is correct. To see why, let's look at a simpler example: you have a deck of 60 cards, and draw two. Now, this event can happen in one of two mutually exclusive ways: either you get the card on the first draw, or on the second. The probability of getting the card on the first draw is plainly 1/60. Now, you seem to be thinking "after I've drawn the first card, there are only 59 cards left in the deck, so the probability of getting the card on the second draw is 1/59." But in fact, this is only true on the condition that you didn't get the correct card on the first draw. If you did get the correct card on the first draw, the probability of getting on the second draw is zero! The unconditional probability that you'll get the correct card on the second draw is thus the probability that you failed to get it on the first draw times the probability that you get it on the second draw, given that you didn't get it on the first draw. Thus, the probability of getting the card on the second draw is 59/60 * 1/59 = 1/60, and the probability of getting the card on either draw is 1/60 + 1/60 = 2/60.

This result makes sense, because there's a bijection between the outcomes where you get the card on the first draw, and the outcomes where you get the card on the second draw (swap the order of the cards). Hence, the two events should have the same probability. Generalizing this argument, we see that the probability of getting a specific card after drawing n cards is n/60, as the online site claims.

Excellent explanation. Thanks.

 

[jamison89]

The answer you found online is correct. To see why, let's look at a simpler example: you have a deck of 60 cards, and draw two. Now, this event can happen in one of two mutually exclusive ways: either you get the card on the first draw, or on the second. The probability of getting the card on the first draw is plainly 1/60. Now, you seem to be thinking "after I've drawn the first card, there are only 59 cards left in the deck, so the probability of getting the card on the second draw is 1/59." But in fact, this is only true on the condition that you didn't get the correct card on the first draw. If you did get the correct card on the first draw, the probability of getting on the second draw is zero! The unconditional probability that you'll get the correct card on the second draw is thus the probability that you failed to get it on the first draw times the probability that you get it on the second draw, given that you didn't get it on the first draw. Thus, the probability of getting the card on the second draw is 59/60 * 1/59 = 1/60, and the probability of getting the card on either draw is 1/60 + 1/60 = 2/60.

This result makes sense, because there's a bijection between the outcomes where you get the card on the first draw, and the outcomes where you get the card on the second draw (swap the order of the cards). Hence, the two events should have the same probability. Generalizing this argument, we see that the probability of getting a specific card after drawing n cards is n/60, as the online site claims.

Thanks for this, very nice :)

 

[blue_raver22]

I can provide an explanation for the difference in probabilities. The reason why your method of calculating the probability (adding the individual probabilities of each card) does not equal the probability given in the article is because these are two different scenarios.

In the article, the probability is calculated for the specific event of drawing a specific card out of a deck of 60, given that you must draw seven cards. This means that the probability of drawing the specific card remains the same for each of the seven draws, which is 1/60.

In your method, you are calculating the probability of drawing each of the seven cards in a specific order, which is a different scenario. This means that the probability of drawing the first card is 1/60, the probability of drawing the second card is 1/59, and so on. However, for the second card, the probability is no longer 1/60 because there is one less card in the deck after the first card is drawn. Therefore, your method takes into account the decreasing number of cards in the deck with each draw, while the article's method assumes that the deck remains the same for each draw.

Both methods are valid, but they are calculating the probability for different events. In order to get the same result, you would need to use the same scenario and method of calculation. I hope this clarifies the difference in probabilities for you.