A probability puzzler

Of all 56 possible hands, Alice is betting on 44 of them, winning 12, for a 27.3% win rate.
Bob bets on 10 of them, winning 6, for a 60.0% win rate.

Are you sure about that?

28 possible hands, I think, but other than that this reasoning seems correct.

Wait, correction:

Alice: bets 44, wins 12, 27.3%
Bob: bets 14, wins 6, 42.9%

Oops. I figured Alice’s odds wrong. But the actual odds are worse for her than my mistaken odds.

Here’s the correct figures:

The basic odds against a player are 22-6 (or 3.67:1) against the player. Alice’s question eliminates six losing hands and reduces the odds to 16-6 (2.67:1). Bob’s question eliminates eighteen losing hands and three winning hands and reduces the odds to 4-3 (1.33:1). So Alice is losing because she is both making more bets than Bob and is making her bets are worse odds than Bob.

No, I’m wrong of course. The number of winning hands grows more slowly than the number of losing hands when you include more aces. Bob has three hands he can win with (As Ax) and four he can lose with (As Kx). Alice has only six hands (Ax Ay) she can win with and sixteen she can lose with (Ax Ky).

Good little puzzle. :slight_smile:

It turns out not to matter as long as both players get the same odds.

I really like this approach. I was somewhat frustrated with this problem, because I couldn’t think of a good way to get the answer besides sitting down and doing the calculations. But I think that this will generalize at least enough to get a sense of who has an advantage in other problems.

I think people are overthinking this, unless I am missing something (does the “at least” wording screw Alice somehow? In past puzzles seemingly insignificant little throwaway details like that invariably seem to come back and bite you, hard).

Seems very very simple to me-and not sure why people are doing this AxKx stuff.

Once the presence of at least one ace in your hand is confirmed, 7 unknown cards are left for the final draw: 3 cards let you win, 4 cards let you lose-it’s always 3:4 odds, for both players.

Or (if all of your overanalyzers are correct) would Alice be better off just asking if there is one ace in her hand, none of that “at least” crap?

There are one, possibly two, questions that need to be answered for a meaningful answer to OP: what is the payout on a winning bet? If the payout is larger than 2-2/3 per 1 bet, the 2nd question is: is there a limit to how often we get to play, or a limit on how often we get to bet?

If payout over 2-2/3, and a limit on how often we get to play, but no limit on how often we bet, advantage alice
If payout over 2-2/3, and a limit on number of bets, advantage bob
If payout between 1-1/3 and 2-2/3. Advantage Bob
if payout less than 1-1/3, don’t play. If you must, though, play like Bob.

No, because she doesn’t want to bet on a hand that has one ace. She wants a hand that has two aces.

This logic sounds convincing but it doesn’t work. The problem is not all hands with at least one ace are created equal. A hand with an ace of diamonds and an ace of clubs is not a different hand than a hand with an ace of clubs and an ace of diamonds.

Here are all of the possible hands:

Ac Ad
Ac Ah
Ac As
Ad Ah
Ad As
Ah As

Ac Kc
Ac Kd
Ac Kh
Ac Ks
Ad Kc
Ad Kd
Ad Kh
Ad Ks
Ah Kc
Ah Kd
Ah Kh
Ah Ks
As Kc
As Kd
As Kh
As Ks

Kc Kd
Kc Kh
Kc Ks
Kd Kh
Kd Ks
Kh Ks

If you take the twenty-two hands which have at least one ace, you can see that only six of them are a pair of aces and sixteen of them are an ace and a king.

Intuitively that’s what it seems like it should be, but it’s much easier to visualize if you actually write out all 28 possible hands and assume they each come out once in the first 28 deals.

There are 6 AA hands, 6 KK hands, and 16 AK hands. Neither player bets on any of the KK hands. Of the winning AA hands, Alice bets on all of them, while Bob fails to bet on 50% of them, so at this point Alice is at +6 units while Bob is at +3. But of the 16 AK hands, Alice bets on and loses all of them, while Bob only bets on the 4 of them that contain the Ace of spades, so at the end of that Alice is at -10 while Bob is only at -1.

If she asked if there was exactly one Ace and only bet when the answer was “No,” then she’d when on the 6 AA hands and lose on the 6 KK hands, so she’d come out even.

Hmm. Do I get any points for instantly voting “Bob, because Bob has more information, even though I don’t immediately see why that’s relevant”? No? Ah well, multiple choice, cookie for me. (This, for the record, is why multiple choice is just fine for aptitude and intelligence tests, but horrible at determining whether people actually have any idea what they’re talking about.)

I’ll take a crack at the math now, since my statistics skills are a good eight years out of date. Thanks for the practice, ultrafilter!

Here is how I figure it. There are (8 choose 2) = 28 hands. Of these exactly (4 choose 2) = 6 contain two aces and the same number 2 kings. That leaves 16 consisting of an ace and a king. Alice will bet on 22 of them and win only 6. Of the 28 hands exactly 7 include the spade Ace and 3 of these consist of 2 aces, so Bob wins 3/7 of the time.

This is the same conclusion as several other posters, albeit with less effort. To spice it up, I would say that the payoff is 2 for 1 so that Bob wins steadily and Alice.

I might add that my first impression was that Bob would win since his question was more informative, but I am surprised the disparity is so large.

So what it boils down to is that the gambler has such garbage odds in this game that the more restrictive question that is asked keeps one from betting as much and is therefore the “better” strategy. The best strategy is simply not playing the game.

OK…my intuition simply refuses to believe what the stark probability logic says. My intuition tells me that an ace is an ace is an ace, and it doesn’t matter if it is the ace of spades, clubs, diamonds, hearts, or Base.

So why are the odds different from the following change in the rules:

Each player is given one card, and then asks the same question: Is there an Ace in my hand (Alice), or is the Ace of Spades in my hand (Bob)? At that point it seems like the odds are the same in both cases of drawing a 2nd ace: 3:4, as I tried to argue above.

I simply can’t grasp why Bob’s odds in the original game are the same as in this reimagining, while Alice’s suck all of a sudden.

Right, it’s the same 3:4 odds which are likely a losing bet. Since Bob is asking a more restrictive question, he bets less and therefore loses less.

If James asked “Do you have at least one Ace is your hand that the previous owner of the card scrawled threatening references to the United Nations on it?” then James’ strategy would be the “winner” because he wouldn’t bet at all, and therefore would do better than Alice or Bob.

It’s because you’re not drawing the cards one at a time, you’re dealing them both and throwing out the ones that don’t match the criteria. Alice only throws out KKs, while Bob throws out a few winning hands (AA without a spade) and a lot of losing hands (KK, Ac with any king, Ad with any king and Ah with any king).

First, let me show my work. My answer matches what many people are saying, but I’ve laid out the work a little differently for myself.

Analyzing Alice’s betting strategy, we have these scenarios:
First card is an Ace (4/8), second card is an Ace (3/7) = 12/56 (bet - win)
First card is an Ace (4/8) second card is a King (4/7) = 16/56 (bet - lose)
First card is a King (4/8), second card is an Ace (4/7) = 16/56 (bet - lose)
First card is a King (4/8), second card is a King (3/7) = 12/56 (no bet)

Alice’s question means that she bets on the first three. She wins 12/56 hands, and bets 44/56. Thus, she wins 3/11 times that she bets.

For Bob’s betting strategy, we have more scenarios:
First card is an Ace-Spade (1/8), second card is an Ace-Other (3/7) = 3/56 (bet - win)
First card is an Ace-Spade (1/8), second card is a King (4/7) = 4/56 (bet - lose)
First card is an Ace-Other (3/8), second card is an Ace-Spade (1/7) = 3/56 (bet -win)
First card is an Ace-Other (3/8), second card is an Ace-Other (2/7) = 6/56 (no bet)
First card is an Ace-Other (3/8), second card is a King (4/7) = 12/56 (no bet)
First card is a King (4/8), second card is an Ace-Spade (1/7) = 4/56 (bet - lose)
First card is a King (4/8), second card is an Ace-Other (3/7) = 12/56 (no bet)
First card is a King (4/8), second card is a King (3/7) = 12/56 (no bet)

So Bob wins 6/56 hands and he bets 14/56 hands. Thus, he wins 3/7 times that he bets.

3/7 (43%) is clearly better than 3/11 (27%).

I was also surprised that he came out ahead, and by that much.

Trying to understand it on an intuitive level, I think I’ve figured out the difference. A losing K-A/A-K hand can have any Ace, and there’s only one. So losing hands of this type will only contain the Ace-Spade 1/4 times. No real surprise here, but think about winning A-A hands. These contain two aces and thus have “double” the chance for an Ace-Spades. Specifically, of those 12 possible hands, 4 contain an Ace-Spade, or 1/3.

Thus, Bob’s strategy refuses 3/4s of the losing A-K hands, but only 2/3s of the winning A-A hands. Bob doesn’t come out better just because he rejects more hands in a losing game - Bob comes out better because his strategy preferentially rejects losing hands.