1
$\begingroup$

Just had quite a discussion with my co-workers on this one.

If we had a pile of four cards. A King of spades, King of Hearts, Ace of Spades, Ace of Hearts, I pick up two cards and look at them. I then tell you, truthfully, that AT LEAST ONE of the cards I have is an Ace. (Q1) What is the probability of my other card being the other ace? (Q2) If I told you that the card I had was an ace of spades, what is the probability of the other card being the ace of hearts?

Now, I'm quite certain that in either case, the possibility that the other card is 1/3. But my co-worker swears it is 1/5. Here is his logic:

If we permute on the possibilities (ignoring order), there can be 6 total possibilities:
Ah, As
Kh, Ks
Kh, Ah
Kh, As
Ks, Ah
Ks, As

If one of the cards is an Ace, then that eliminates the (Kh, Ks) pair. (Q1) Thus there is only a 1/5 chance that the other card is an ace. (Q2) If I knew the card was an ace of spades, then that eliminates 3 possibilities, leaving 3 possibilities, thus is 1/3 chance the other card is the ace of hearts.

My logic: This is incorrect because permuting on the cards' suit (Spades, Hearts), is not relevant to the question asked - doesnt matter how you arrange the suits, the number of possible win conditions do not change since the question only asks if the cards VALUE (Ace, king) are the same.

Help? Can anyone offer an understandable explanation?

Edit:

I also created a small C# program to test a few examples of how I interpret the excercise. I'm getting fairly solid 1/3. If anyone cares to double check my implementations and point out any possible errors, that would be great. It can be found here (Runnable in browser) - https://repl.it/BsJ9/17

Thanks!

$\endgroup$
  • $\begingroup$ Sorry for the incomplete comment - I don't have the time to go into full detail. But your coworker is right. This relates to what is known as the Monty Hall problem. $\endgroup$ – Eric S. Mar 2 '16 at 14:38
  • $\begingroup$ Do you only speak in scenario (1) if you have an ace or two, and in scenario (2) if you have the Ace of Spades, and otherwise you are silent? If so, your friend is correct $\endgroup$ – Henry Mar 2 '16 at 14:38
  • $\begingroup$ @henry In this case, we are just assuming that of the two cards drawn, in this particular instance, (1) I have one or two aces or (2) I DO have an ace of spades. After stating that so far, these are true, what is the probability of the other card being an ace. $\endgroup$ – Jace Mar 2 '16 at 14:46
  • $\begingroup$ "If we had a pile of two cards": is that "two" supposed to be "four"? $\endgroup$ – user940 Mar 2 '16 at 16:15
  • $\begingroup$ @ByronSchmuland Yes! Edited. Thanks $\endgroup$ – Jace Mar 2 '16 at 16:18
1
$\begingroup$

I'm pretty sure that your coworker is right in the first case, and you in he second. Take a look at this.

If we label the cards A B C D, for King of Spades, King of Hearts, Ace of Spades, Ace of Hearts, then we have the following combinations.

$AB\quad BA\\AC\quad CA\\AD\quad DA\\BC\quad CB\\BD\quad DB\\CD\quad DC\\$

This gives us 10 combinations in which you have at least 1 ace : $AC, CA, AD, DA, BC, CB, BD, DB, DC, CD$

Of these we have 2 with 2 aces: $CD, DC$ so you have a probability of $\frac{2}{10}$ or $\frac{1}{5}$.

If we go with your second case - that we have the ace of spades $(C)$, then we have 6 of our combinations: $AC, CA, BC, CB, CD, DC$, of which 2 contain the other ace: $CD, DC$, which gives a probability of $\frac{2}{6}$ or $\frac{1}{3}$

Hope this helps!

$\endgroup$
  • $\begingroup$ If there are 10 combinations in which you have at least 1 ace, and the ace the person is referring to is EITHER C or D. Assuming it is C, this leaves: AC, CA, BC, CB, DC, CD, two combinations (DC, CD) are winning. that is 2/6 or 1/3 winning combinations. Assuming the card they are referring to is D, then the same thing happens (I won't type it again). The card they refer to is not C AND D, it is C OR D and so the number of combinations drop from 10 to 6. Correct, or? $\endgroup$ – Jace Mar 2 '16 at 14:55
  • $\begingroup$ No I don't believe so. The information given tells us that we have an ace, so we assume nothing else about the problem. This gives us $P(no ace) = 0$ and $P(ace) = 1$. What we can do at this point is take all of the combinations in which there is an ace (of which there are 10). All of these are equally likely, so there is a $\frac{1}{10}$ chance of each of them occurring (since $P(ace) = 1$). Since 2 of these 10 are a "Success", that gives us our probability. What you are doing is assuming that we have one ace or the other (like the second problem), but this information isn't given. $\endgroup$ – J. Bush Mar 2 '16 at 15:06
  • $\begingroup$ I believe the flaw in your logic is that you are assuming that you know which ace you have. What the problem states is that you know that you have C OR D, and so you must take all combinations that include either C or D (or both). The case where you know that you have either the ace of spades or the ace of hearts is how you describe it though. $\endgroup$ – J. Bush Mar 2 '16 at 15:09
  • $\begingroup$ I see what you're coming from. I know that we don't know which ace we have, but we MUST have one or the other (or at least, the ace the person is referring to is one of them). Even though the ace isn't specified, we do know that the ace they are referring to IS either the ace of spades or the ace of hearts. Does this really not factor into the probability? $\endgroup$ – Jace Mar 2 '16 at 15:13
  • $\begingroup$ I could see how this might make sense in quantum entanglement, like, the card they refer to when they say "I have an ace" IS literally both Hearts and Spades until revealed, but in this case, given we know one of the aces are in the hand, our options would be either: AC, CA, BC, CB, DC, CD, -OR- AD, DA, BD, DB, CD, DC. No? $\endgroup$ – Jace Mar 2 '16 at 15:18
1
$\begingroup$

The first case is simple enough; there are six different card combinations, and by informing me that you have at least one Ace you have removed one possibility (namely, two Kings). Thus the probability is $\frac15$.

The second case is considerably more delicate and actually requires more assumptions. Specifically, how did you choose which card to tell me about? Did you:

  1. just tell me the first card you picked up (or equivalently, randomly select a card to report?)
  2. always plan to report the Ace of Spades if you picked it up?
  3. randomly select a card to report, but always report an Ace if you have one?

The probability we are looking for is $$P(A|B)=\frac{P(A\cap B)}{P(B)}=\frac{P(A)}{P(B)}P(B\,|\,A)$$ where $A$ is the event "you have two aces" and $B$ is the event "you told me you have the Ace of Spades". Notice that depending on your selection rule above, the probability of $B$ changes and may be distinct from $P(C)$, where $C$ is the event "you have the Ace of Spades".

If you report by rule 1 above, then we have $12$ possible card draws (where order now counts). By telling me your first card was the Ace of Spades, you have eliminated all but $3$ card draws, so the answer is $\frac13$.

If you report by rule 2, then $B=C\supset A$ and so $$P(A\,|\,B)=\frac{P(A)}{P(C)}=\frac{\frac16}{\frac36}=\frac13.$$ However if you report by rule 3, then $P(B\,|\,A)=\frac12$ and $P(B)=\frac5{12}$ (out of the $6$ combinations, you will definitely report the Ace of Spades for $2$ of them and have a $50\%$ chance for $1$ of them). Thus $$P(A\,|\,B)=\frac{\frac16}{\frac5{12}}\cdot\frac12=\frac15,$$ corresponding to the intuition that it shouldn't matter which Ace you report (which makes sense, since this was the option where you reported at random).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.