I got a puzzle:

The game:

- Uses a standard 52 card deck.
- Only one player at a time.
- Deck is shuffled into a random configuration at the start of each hand (all hands are equally likely to be dealt to the player).
- The player's hand consists of 2 face down cards, that the player doesn't get to see the cards until the end of the hand.

Bob is playing this game, but he brought along his *azure magic wand*. This wand when ran over his 2 face down cards, will lite up if there is 1 or more Queens in the hand.

Alice is also playing this game (at a separate table, with a separate deck of cards). Alice brought along her *ruby magic wand*. This wand lite up if the Queen of Hearts is in the 2 card face down hand.

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You walk into the casino, Bob and Alice have each been dealt their face down hands (2 cards each); each one scans their own hand with their own wand, and both wands lite up.

The question:

**Who is more likely to have a pair of Queens?**

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/e And don't be intimidated of being wrong; essentially everyone gets this one wrong the first time! Oh and I forgot to mention, the magic wands are 100% accurate.

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Well, dang it, Nyar! My brain hurts now...almost as much as it hurt when I took Math 101 ;)

All we know is that Bob and Alice each have two cards, of which at least one is a queen. Each pack had three remaining queens out of 51 cards when the second cards were dealt, so Bob and Alice each have one chance in 17 of getting a second queen. The odds are the same for both.

I’m wondering, though, if there’s any statistical significance for Alice’s queen being a specific suit. I suspect there is.

I'll try to plant a tiny seed of doubt:

Imagine playing the game yourself at home and doing 1000 hands (one at a time, reshuffling of course). And each time you scan each hand with both wands. And you keep track of the number of times each wand went off during the 1000 runs...

Alice’s wand only lights up for one of 52 cards. Bob’s wand would probably light up far more often. If his wand lights up there’s has a one in 13 chance it’s a queen and a one in 26 chance it’s two queens.

But even if Bob's wand lights up, he has no control over the cards dealt. I think the suit of Alice's card is a red herring. Whatever suit the first queen, you still have three out of 51 chances to get a second one.

But I think it has more to do with the wands lighting up than of the cards dealt. Bob’s wand will light up for one or two queens of any suit. Alice’s wand will only light up if she has the queen of hearts. Both wands are lighting up. A wand light is a hint that each player now has a chance of having two queens. But does one wand light indicate a better chance because of the conditions under which it will light up?

I want to say yes but my little pea brain has not led me to why (or which player!) yet. Grrrrrr......where’s that bottle of melatonin? I need to get some sleep.

Both wands lite up, their chances are exactly the same. Bob is likely to hit a pair faster than Alice as he is playing 75% more hands but every hand has exactly the same odds. There are 52 cards in the deck and only two count. If both wands light up, the chance of having a second queen are exactly the same.

he is playing 75% more hands

I think they're both playing exactly the same number of hands. And even if Alice's wand doesn't light up, she still has the same chance of receiving two queens.

@Algebe: (WRONG I MADE AN ASSUMPTION) - No, they are getting the same number of hands. Bob gets to 'play' and turn a second card over to see if it is a queen 75% more of the time. Bob is playing with 4 queens and Alice only one. If Alice's wand does not go off there is no reason for her to turn her cards. (MY ASSUMPTION WAS WRONG - ALL HANDS ARE PLAYED)

OOPS! YOU ARE CORRECT, they are playing the same number of hands as "They both turn over their cards at the end of the hand." The odds are exactly the same in any event. The wands light, each has at least one queen and the same odds of catching a second queen.

Let me try to plant another seed of doubt (against the idea that the probabilities are identical/symmetric):

For each wand; I'll compile of list of the hands that will trigger the wand

ANDhave 2 Queens:-----------------------------------------

Azure wand (Bob's wand, detects presence of 1 or more Queens):

-----------------------------------------

Ruby wand (Alice's wand, detects presence of Queen of Hearts):

-----------------------------------------

It seems when you peek under the hood a little bit, the symmetry that originally looked obvious (it looked obvious to me the first time too!), disappears.

The hands do not matter, Both wands have been triggered. Both have at least one queen. The odds of hitting a second queen are exactly the same. They each have a hit on their wands. Each have at least one queen, Each have 3 queens left in the deck. The odds do not change.

This is the ole gambler's ploy. "I have not had a pair of aces in a long time so I am due." Odds do not work that way. The probability of hitting two aces is exactly the same no matter how many hands you play. Odds of receiving a specific pocket pair: 0.45% or 220 to 1.

But the question was about two players, each playing one hand. They both have one queen, and three queens remain in the deck, so the chances are the same (3 in 51, or 1 in 17) IMHO.

Even if they play multiple hands, assuming that the previous cards are returned and the deck is shuffled, the odds revert to the same as in the first hand.

But in one hand with both wands lit, the odds of the second card being another queen are equal IMHO.

@Algebe, “

But the question was about two players, each playing one hand. They both have one queen, and three queens remain in the deck,”

Not necessarily. Bob’s wand lights up if he has one OR MORE queens. Alice’s wand lights up for only one queen.

@Algebe: That's my conclusion as well. The odds do not change.

Yeh I thought the odds were going to be the same, but didn't write it as I was sure that Nyar would have some trick up his sleeve that would make me look an idiot (which I am by the way).

I have to admit, I am of the same opinion that the odds are the same for both players having a pair of queens. But I somehow get the feeling Nyar has some sort of trick up his sleeve that will leave me looking/feeling rather foolish.... *chuckle*...

We can all fight about the answer once he posts it. I might even discover why I am losing at poker sometimes.

Ooh a puzzle! Time to overthink it!

First assumption is the rules are fair and as presented with no "cheap tricks" kind of like they are playing at a casino single person, and single 52 card deck (dont know of any casinos that would play that, but I think that is beyond the scope of this puzzle.

No "magic" wands it looks like it is a fairly standard chance to get 2 queens. 4 in 52 + 3 in 51. Obviously this would not matter if the cards were face up for face down, this is the odds of getting 2 queens with no special wands, cheating etc.

The azure wand has a special power to detect if there is at least 1 queen in the hand. Where as the ruby can detect if the queen of hearts is one of the two cards. How do these wands affect the odds of having 2 queens when they light up?

We assume within the unspoken rules of this puzzle cannot do simple things like: place the two cards further apart, and use the azure wand to actually verify 2 queens.

Let's look at the odds for both wand light up scenarios within the assumed unspoken rules.

For the azure wand. We know one of the two must be a queen. What are the odds for both being queen? We have 1 in 1 of one of one of the 2 cards being queen, and 3 in 51 chance of 2nd one being queen.

For the ruby wand. We know one of the two must be the queen of hearts and then the other card is still the 3 in 51 chance of being one of the 3 other queens. 1 in 1 for one queen (queen of hearts) 3 in 51 for one of the other queens.

So far it appears to me, even though there is more possible combinations of results for azure, then ruby, that does not affect the odds of 2 queens. While the wands lighting up give information that there is a queen present in different ways, once they both light up, neither wand that provided that information, in no way that additional information affects the odds of the 2nd card dealt before the information was known, being queen or not. The odds are same for both, 3 in 51. This seems to be the easy obvious answer so I immediately doubt it. Would make for a poor puzzle (imho) if this was the answer.

More idle speculation... Azure is just presence of queen(s), ruby, is presence of 1 type of the 4 queens.

Perhaps look at it as, one card as.. 1 in 52 instead of 4 in 52, and.. oh wait i got it..(I think) I will stop here and not ruin it for the rest. And see later if I was right.

Hand-waving explanation:

Let's pretend for a moment that Alice and Bob are advertising their wands as "2 queen detectors".

Here are some facts we can get without much math:

2 queen detectors).So we have two effects: one increasing Alice's accuracy, one decreasing it. We've gone about as far as hand-waving can take us, as we'll need the actual numbers to see which of these two effects is stronger, or if they balance each other.

I'll post the math below, but the punchline is

it is almost twice as likely that Alice has 2 queens (as compared to Bob).---------------------------

Some math:

The probability of getting 2 Queens; is equal to the number of possible hands that contain two queens, divided by the number of possible hands.

As many people have pointed out, when the cards are initially dealt: the number of ways to get 2 queens is 6 for both players. The number of possible hands is 1326 for both players. So before the wands are used, both players have a 6/1326 ≈ 0.5% chance of having 2 queens.

---------------------------

When Alice's wand lights up, we know one of her cards is the Queen of Hearts, and the other card is one of the 51 remaining cards, so after her wand lights up, there are only 51 possible hands she could have.

Also (when Alice's wand lights up), there are only 3 ways for her to get 2 queens ([Q♥, Q♠], [Q♥, Q♦], [Q♥, Q♣]).

So the chance of Alice having 2 Queens is 3/51 ≈ 6%---------------------------

We might notice that when Bob's wand lights up, there are still 6 ways for him to to get 2 queens (his wand catches every way to get two queens).

Calculating how many possible hands exist for Bob (after his wand lights up) is not as simple as it is for Alice, but the result is 198 possible hands.

So the chance of Bob having 2 Queens is 6/198 ≈ 3%--------------------------------------------------

--------------------------------------------------

How can we make sense of this?!? (or, return to hand-waving)

So Alice's accuracy is about 1/2 * 4 ≈ 2 times better than Bob's.Not only did I get this wrong the first time (I assumed it was equal probabilities); I got it wrong the 2nd time when I noticed Alice's wand would miss 1/2 of the two queen hands (I naturally assumed this meant Bob's wand was more accurate at finding 2 queen hands, it isn't).

Well, I have to admit, THAT was very interesting. Gonna have to wait till a little later to study the explanation a little more in depth, but it is a cool brain twister.

That was wonderful, Nyar! Thanks for posting it. Although I’m noticeably a derp because of it, it was fun. :)

@RE: Alice misses about 1/2 the hands that have 2 queens, yet Bob doesn't miss these: this cuts Alice's accuracy (in comparison to Bob) by 1/2.

This was my initial thought when I made the distinction of hands played, vs, hands turned over. As Algebe pointed out, "Both Cards are turned over at the end." Nobody misses anything. In my mind, this throws your statistics out the window. Alice does not miss the hands that have two queens. She just does not detect some of them before hand. The only effect on Alice is that her wand does not light up on some of the two queen hands. Alice's accuracy is not effected at all as she is in no way trying to be accurate. All hands are played. If her wand lights up or not, she will play all two card hands. The false positives appear no longer important.

I was sloppy when I said Alice's (or Bob's) accuracy; I meant the accuracy of the associated wand.

In my opinion, this problem has very little to do with poker. Just consider:

One of the pitfalls built into this is Bob's wand (detects 1 or more queens). In almost all cases in real life (like say poker), we determine what the value of a card is (9, 10, Jack, Queen, etc) by looking at it. In almost all cases in real life we determine what the suit (clubs, hearts, etc) of a card is by looking at it. On in other words: in real life we typically learn the value and the suit of a card at the same time.

So it is very rare in real life for us to be faced with the situation that Bob's wand puts us in: where we know the value (it's a queen) but we don't know the suit. In short, your "poker" intuition probably isn't that helpful here because, while it superficially looks kind of like poker; there are a lot of subtle differences between this question and any realistic poker question I can think of.

Another way to think of it might be:

Imagine both players start with a piece of paper that lists all 1326 possible hands: [A♥, 2♥], [A♥, 3♥], [A♥, 4♥], ...

Then when they test the hands with their wands, based on the result they erase the hands they can't possibly have that were written on their paper. The remaining number of hands listed on each person's paper will be different for each player. Leading to the probabilities being different.

I actually thought of it in terms of imagine the wands are real, using real technology.

The wands were actually simple detectors of a small amount of particular (useful to the situation) radioactive isotope that could not be detected unless they had a detector attuned to the radiation. One could detect only the queen of hearts, where the other could detect presence of any queen, but not differentiate. (The cards has the radioactive isotope inbedded in the correct cards paper before the game started.)

I actually mostly figured it out, (did not get all the details right) about this puzzle, by thinking about a somewhat similar puzzle I think we discussed here before. The 3 doors, prize behind 1 door scenario.

Where at first glance it seems an even 33/33/33, but because the host eliminated a door and know's which doors are which when he eliminated one, we use that "new" information to find the better "odds." (Always switch doors after a door is eliminated.)

Then you won't be surprised to learn that this problem is largely just the Monty Hall (the 3 door problem) problem in disguise. The Monty Hall version is just as counter-intuitive(imo); but much more widely known; so the details were changed slightly to disguise it.

I find that this is a re-occurring motif in many of these brain busters. The problem tells you something that obviously contains some useful information: there is nothing behind door 3, or you have at least 1 Queen, etc. But often there is more information contained in what you were told than what is immediately obvious (it's almost never obvious to me the first time I read them).

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