Suppose you’re dealt two cards face-down. You peak at one, and see an ace.
Being the superstitious type, you shuffle them, so you no longer know which card is which. You look again, and see an ace.
What are the chances both cards are aces?
If you look at the first ace and then look at the second ace and they’re different suits, then the chances are 100% that the other card is an ace (or you’ve got a crooked deck).
I am guessing 3/51. That was the probability that the other card you didn’t look at before shuffling was an Ace to begin with.
Despite being a reasonable mathematician, I’m generally hopeless at probability, so I’m going to post the obvious answer in the expectation of being proved wrong. My guess is that the chance is 27/51. Reasoning: 50% of the time, the ace you see will be different from the first one. Then you are 100% certain you have two aces. The other 50% of the time, you will be seeing the first ace again. Then your chance of having two aces amounts to the chance of the second card being an ace, which is 3/51. 3/51 * 1/2 = 3/102. 3/102 + 1/2 = 3/102 + 51/102 = 54/102 = 27/51. I haven’t explained the assumptions surrounding these calculations, I hope they’re obvious. If my answer is wrong, I suppose the error lies in making these assumptions.
But you have more information now than you did.
Here’s my wild stab. Either you looked at the same ace twice or you looked at two different aces. Chances are 50% either way.
If you looked at the same ace twice, there’s a 3/51 chance that the other card, unseen, is an ace.
If you looked at two different aces, then there’s a probability of 1 that you have two aces.
Working out the total probabilities, I get ((3/51) * (1/2)) + 1/2
(3/102) + (51/102)
54 / 102
27 / 51
or 52.94%
Anyone else wanna take a whack at correcting this? 
ETA: Hi Dead Cat. Looks like we followed similar reasoning and got the same numbers.
It seems to me that you’re forgetting that the OP may be seeing a second (different) ace after shuffling. If this happens, the probability of having two aces is obviously 100%.
Otto, that occured to me, too. I guess we are meant to assume that the OP identifies that the card is an ace each time, but not what suit it is. Or there are four aces of the same suit in the deck instead of the usual four aces, and otherwise the deck is as normal. On this assumption, I stand by my answer (for now).
Yeah, you guys are right. I was forgetting about the possibility of it being the same ace.
I think chrisk is right.
50% of the time you have a 100% chance of knowing you have two Aces (different card after shuffling)
50% of the time you have a 3/51 chance of knowing (same ace picked up after shuffing)
Mind if I hijack a bit?
We have a seven card stud player who bitches every time you give him a deuce. I am math challenged and can’t figure out the chances he would get a deuce, with 7 players, being dealt out three cards to start, there are 21 cards on the table.
It seems to me the odds of someone NOT getting a deuce in 7 cards is smaller than the chance of them getting one.
There are three options,
-
You look at the other card the second time (50%) and thus you definitely have two aces (100%). Total:50%
-
You look at the same card the second time (50%) and the remaining card is one of the other aces (3/55). Total:3/110
-
You look at the same card the second time (50%) and the remaining card is not one of other aces (52/55). Total:52/110
Thus the odds of you getting two aces is 3/110 + 55/110 = 58/110 or 29/55 or ~52.7%
Where do you guys get that there is a 3/51 chance that the other card is an ace when you see that same card twice?
I have a new analysis, based on following through the experiment from the beginning and eliminating possibility cases that do not match up with observations:
Start with all possible two-card deals from the deck:
2652 possibilities. (52 * 51)
2256 have no aces. (48 * 47)
12 have two aces. (4 * 3)
384 have one ace. (by subtraction.)
Turn over the left spot and get an ace
2448 possibilities are eliminated.
204 possibilities remain
0 have no aces (all eliminated)
12 have two aces (none eliminated)
192 have one ace (by subtraction, or 4 * 48)
shuffle the cards, either swapping left and right spots or not
all possibilities are thus twinned, thus giving us 408
24 have two aces
384 have one ace
Note that at this point, ‘ace of spades dealt to the left, ace of hearts dealt to the right, swapped’ is a seperate possibility than ‘ace of hearts dealt to the left, ace of spades dealt to the right, not swapped’ even though the cards in each position after the shuffle are the same. But a different ace has been exposed in each case.
Now, we eliminate all possibilties where there is not an ace in the right spot. (picked at random, could be the left again without affecting the possibilities.) None of the 24 two-ace possibilities are eliminated, but half of the 384 one-ace possibilities should be.
So this gives us 24/(24 + 384) or 6/101 of having two aces.
I think that this is more sound than the earlier analysis, because the fact that we saw two aces makes it much likelier than 50% that we actually saw the same card twice.
ETA: And this is not much higher than dgrdfd’s guess, since 3/51 is 6/102
BIG math error - I forgot to actually divide 384 in half. :smack:
So that should be 24/(24 + 192) or 1/9
For the sake of the problem we’ll have to assume you’re forgetful as well as superstitious - iow, you don’t remember the suit of the first ace.
There are 52 cards in a deck, one of which you’ve already seen, which leaves 51. Of those 51, 3 are aces.
Where are you getting 55?
Deck with Knights in it as well as the other 13 ranks? Or four jokers??
(scamper)
Stupidity.
Chris, I think you’re right, though I don’t quite understand how you got there.
For reference, here’s how I did it:
You’ve got either AA or Ax, and you’re either looking at the same card twice, or you’re not. Chances of AA = 3/51. Chances of Ax = 48/51. So if you diagram, it ought to look like this:
…same card…different card
AA… 1.5/51…1.5/51
Ax…24/51…24/51
Now we know we didn’t see an “x” so we can eliminate the lower right box. That leaves 27 possibilites (51 minus 24, or 1.5+1.5+24).
So 3/27 (=1/9) that we have AA, and 24/27 (=8/9) that we don’t.
btw, you got it much faster than I did.
This is why I could never understand conditional probability questions like this one; it seems so intuitively obvious that the chance of seeing the same card twice is 50%. But I suppose the question really is: ‘given that we have seen an ace (the second time), what is the probability that it is the same ace?’. Is that right? Perhaps someone can explain it all a bit more clearly (and verify that chrisk’s second (third) attempt is correct)?
ETA: LinusK’s solution helps :).
Is it just me, or have you all read the question wrong? I read it as:
- Hold’em Player is dealt 2 cards from a standard deck of cards
- Player looks at one card; sees that it’s an ace(he does not note the suit)
- Player shuffles his 2 card hand so that he doesn’t know which card is which
- Player looks at one card; it’s an ace.
Given all that, what is the probability that the player was dealt Pocket Rockets?
It’s 50%, right? Because he has a 50% chance of seeing the card he saw the first time.
I think that’s it exactly. Try this:
Suppose you go to the doctor, and he tells you one out of a thousand people have a particular disease. He gives you a test, and the test is 99.9 percent accurate.
The test comes back positive.
What are the chances you have the disease?
Here, incidentally, is how it first occurred to me to do it:
Let B be the event that both cards are aces, and let D be the event that we draw an ace from the two before the swap followed by drawing an ace from the two after the swap. We want to calculate P(B | D), which, by Bayes’ Theorem, is equal to P(D | B) * P(B)/P(D). As it happens, P(D | B) is obviously 1 (if both cards are aces, we can’t help but draw aces). Furthermore, P(B) is, of course, 4/52 * 3/51. And, P(D) is 4/52 * (1/2 * 1 + 1/2 * 3/51) [the probability that the first one is an ace, plus the conditional probability that the second is an ace given that (split into the two cases, same ace or different ace)]. Plugging this all back in, we get the answer P(B | D) = 1/9. Ta-da!