Accounting for Potential of Discarded Card

[SumofSigma]

Hi,

If one draws a single card from the deck of 52, and sees the face, say a 2, then if asked what is the probability your next draw will make a two of a kind, you would say
(4-3)/51.

Can someone explain how to solve the following scenario?

If one draws a single card from the deck of 52, and sees the card, say a 2, then the dealer discards a single card from the deck to the a pile without showing you the face, if asked what is the probability, how would you solve? How do you account for the potential of that card to have been a 2 which would decrease your odds from 3/50 to 2/50?

 

[HallsofIvy]

In a standard 52 card deck, there are 4 cards of each denomination. After you have drawn a card, there are 51 cards left, 3 of the same denomination as the first card drawn. The probability that you will now draw that same denomination, so get "two of a kind", is 3/51, NOT "(4- 3)/51= 1/51".

If you draw a card, there are 51 cards left, 3 of the same denomination as the first card drawn so the probability the card the dealer discards is also that denomination is 3/51 and the probability it is not is (51- 3)/51= 48/51. In the first case, that the card the dealer discards is the same as the first card drawn. there are 50 cards left in the deck, two of them the same as the first card drawn so the probability of "two of a kind" is 2/50= 1/25. In the second case that the card the dealer discards is not the same as the first card drawn, there are 50 cards left in the deck, three of them the same as the first card drawn so the probability of "two of a kind" is 3/50.

Overall, in this scenario, the probability of "two of a kind" is (3/51)(2/50)+ (48/51)(3/50)= 6/2550+ 144/2550= 150/2550= 15/255= 3/51, exactly the same as in the first scenario!

 

[Steven G]

Think of it this way. Suppose the dealer does not put the top card on the bottom but instead gives you the 2nd card (or the 15th card or randomly picks a card from the deleted deck) will that change anything. The probability that any card position is another two is 3/51=1/17. So putting the top card on the bottom and then giving you the 2nd card does not matter.

What would happen if the deck is shuffled after giving you the two (shuffled with the remaining 51 cards). Does this change the probability of the card you are giving is another two?

 

[SumofSigma]

In a standard 52 card deck, there are 4 cards of each denomination. After you have drawn a card, there are 51 cards left, 3 of the same denomination as the first card drawn. The probability that you will now draw that same denomination, so get "two of a kind", is 3/51, NOT "(4- 3)/51= 1/51".

If you draw a card, there are 51 cards left, 3 of the same denomination as the first card drawn so the probability the card the dealer discards is also that denomination is 3/51 and the probability it is not is (51- 3)/51= 48/51. In the first case, that the card the dealer discards is the same as the first card drawn. there are 50 cards left in the deck, two of them the same as the first card drawn so the probability of "two of a kind" is 2/50= 1/25. In the second case that the card the dealer discards is not the same as the first card drawn, there are 50 cards left in the deck, three of them the same as the first card drawn so the probability of "two of a kind" is 3/50.

Overall, in this scenario, the probability of "two of a kind" is (3/51)(2/50)+ (48/51)(3/50)= 6/2550+ 144/2550= 150/2550= 15/255= 3/51, exactly the same as in the first scenario!

That's quite interesting! So even though cards are removed from the available cards in the deck, nothing changes to how you view the probability?

i.e. I remove 10 cards up front, the probability of getting a 2 on the first draw is a 4/52?

 

[Deleted member 4993]

That's quite interesting! So even though cards are removed from the available cards in the deck, nothing changes to how you view the probability?

i.e. I remove 10 cards up front, the probability of getting a 2 on the first draw is a 4/52?

Why don't you calculate it and find out!!

 

[JeffM]

Here is a different way to think about it.

What is the probability that the first two cards form two of a kind.

[MATH]\dfrac{3}{51} = \dfrac{1}{17}.[/MATH]
In your scenario, we need to think about the first three cards.

What is the probability that neither the second or third cards are of the same kind as the first.

[MATH]\dfrac{48}{51} * \dfrac{47}{50} = \dfrac{16}{17} * \dfrac{47}{50}.[/MATH]
What is the probability that the first two cards are alike but the third different.

[MATH]\dfrac{3}{51} * \dfrac{48}{50} = \dfrac{1}{17} * \dfrac{48}{50} .[/MATH]
What is the probability that the first and third cards are of the same kind but not the second?

[MATH]\dfrac{48}{51} * \dfrac{3}{50} = \dfrac{1}{17} * \dfrac{48}{50}.[/MATH]
What is the probability that the first three cards form three of a kind.

[MATH]\dfrac{3}{51} * \dfrac{2}{50} = \dfrac{1}{17} * \dfrac{2}{50}.[/MATH]
Let's check that

[MATH]\dfrac{16}{17} * \dfrac{47}{50} + \dfrac{1}{17} * \dfrac{48}{50} + \dfrac{1}{17} * \dfrac{48}{50} + \dfrac{1}{17} * \dfrac{2}{50} =[/MATH]
[MATH]\dfrac{1}{17} * \dfrac{16 * 47 + 48 + 48 + 2}{50}[/MATH]
[MATH]\dfrac{1}{17} * \dfrac{752 + 98}{50} = \dfrac{1}{17} * \dfrac{850}{50} = 1 \checkmark [/MATH]
In your scenario, success comes from all three alike or or first and third alike. What is the sum of those two probabilities.

[MATH]\dfrac{1}{17} * \dfrac{2}{50} + \dfrac{1}{17} * \dfrac{48}{50} = \dfrac{1}{17} * \dfrac{2 + 48}{50} = \dfrac{1}{17}.[/MATH]