# A probability puzzler **URL:** \ **Category:** In My Humble Opinion\ **Created:** [July 19, 2014, 6:38am UTC](https://boards.straightdope.com/t/a-probability-puzzler/693518 "2014-07-19T06:38:50Z")\ **Posts on this page:** 20\ **Page:** 2
**Author:** ![Munch](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/munch/32/5281_2.png) [@Munch](https://boards.straightdope.com/u/Munch)\ **Post date:** [July 19, 2014, 6:29pm UTC](https://boards.straightdope.com/t/a-probability-puzzler/693518/21 "2014-07-19T18:29:17Z")
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. ---
**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [July 19, 2014, 6:33pm UTC](https://boards.straightdope.com/t/a-probability-puzzler/693518/22 "2014-07-19T18:33:40Z")
> [@Giraffe](#): > > The game is actually “is the second card a king or an ace?” Are you sure about that? ---
**Author:** ![Larry\_Borgia](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/larry_borgia/32/156_2.png) [@Larry\_Borgia](https://boards.straightdope.com/u/Larry_Borgia)\ **Post date:** [July 19, 2014, 6:40pm UTC](https://boards.straightdope.com/t/a-probability-puzzler/693518/23 "2014-07-19T18:40:41Z")
> [@Munch](#): > > 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. 28 possible hands, I think, but other than that this reasoning seems correct. ---
**Author:** ![Munch](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/munch/32/5281_2.png) [@Munch](https://boards.straightdope.com/u/Munch)\ **Post date:** [July 19, 2014, 6:41pm UTC](https://boards.straightdope.com/t/a-probability-puzzler/693518/24 "2014-07-19T18:41:27Z")
Wait, correction: Alice: bets 44, wins 12, 27.3% Bob: bets 14, wins 6, 42.9% ---
**Author:** ![Little\_Nemo](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/little_nemo/32/3120_2.png) [@Little\_Nemo](https://boards.straightdope.com/u/Little_Nemo)\ **Post date:** [July 19, 2014, 7:05pm UTC](https://boards.straightdope.com/t/a-probability-puzzler/693518/25 "2014-07-19T19:05:03Z")
Oops. I figured Alice’s odds wrong. But the actual odds are worse for her than my mistaken odds. Here’s the correct figures: > [@](#): > > I voted Alice but that was just off the top of my head. > > Then I tried to figure out the actual odds and now I think Bob might have a better strategy. > > If I figured correctly (and it is 3 am) then there are 28 possible hands. 6 will win and 22 will lose. > > A yes answer to Alice’s question _ **will eliminate 6 losing hands** _. So her odds are improved to 6 winning hands to _ **16 losing hands** _ - _ **2.67:1** _ against her. > > A yes answer to Bob’s question will eliminate 3 winning hands and 18 losing hands. So his odds are improved to 3 winning hands to 4 losing hands - 1.33:1 against him. > > Bob’s question did a much better job of pushing the odds in his favor. The cost was that Bob will be placing a lot fewer bets. Alice will drop out with a no _ **6/28** _ or _ **about 21%** _ of the time. Bob will drop out with a no 21/28 or 75% of the time. > [@](#): > > If my figures are correct, let’s assume an average run of 280 hands. Assume all bets are a dollar. > > Alice will pass _ **60 times** _ and make _ **220 bets** _. She’ll win 60 times and _ **lose 160 times** _. She loses an _ **average of about 36 cents** _ a hand. > > Bob will pass 210 times and make 70 bets. He’ll win 30 times and lose 40 times. He loses an average of 3.5 cents a hand. ---
**Author:** ![Little\_Nemo](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/little_nemo/32/3120_2.png) [@Little\_Nemo](https://boards.straightdope.com/u/Little_Nemo)\ **Post date:** [July 19, 2014, 7:13pm UTC](https://boards.straightdope.com/t/a-probability-puzzler/693518/26 "2014-07-19T19:13:46Z")
> [@VarlosZ](#): > > Why will Bob win a higher percentage of the bets he makes? > > I may be missing something obvious, but it seems like Bob does better (or less bad), but only because he bets less often. > [@Giraffe](#): > > This is the correct answer. The game is actually “is the second card a king or an ace?”, and because of the setup we know the hole cards contain more kings than aces. So both players are making a bet with a negative expectation, the only difference in the odds between any ace vs. ace of spades is in the decision of whether to play or not. Alice will lose faster (in terms of wall clock time) but both players will lose at the same rate as a function of the number of bets they make. 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. ---
**Author:** ![Giraffe](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/giraffe/32/129_2.png) [@Giraffe](https://boards.straightdope.com/u/Giraffe)\ **Post date:** [July 19, 2014, 11:38pm UTC](https://boards.straightdope.com/t/a-probability-puzzler/693518/27 "2014-07-19T23:38:34Z")
> [@ultrafilter](#): > > Are you sure about that? 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. 🙂 ---
**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [July 20, 2014, 12:54am UTC](https://boards.straightdope.com/t/a-probability-puzzler/693518/28 "2014-07-20T00:54:48Z")
> [@Robot\_Arm](#): > > I haven’t confirmed your numbers, but it sounds like you’re on the right track, except that the OP doesn’t say what odds the player is getting on his (or her) wager. If it’s even money, then the best strategy is one that plays fewer hands. But if the bet pays 2-to-1 (or anything else, really) then we’d have to figure that into the calculation. It turns out not to matter as long as both players get the same odds. > [@Cheesesteak](#): > > Definitely Bob. Lets add Chris, Daniel and Evan each using Bob’s method but on the other suits. What you find is that they bet on all the hands Alice bets on, but double up their bets on winning hands never doubling up on losing hands. As a group they do better than Alice, and are identical to each other. 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. ---
**Author:** ![John\_DiFool](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/john_difool/32/19543_2.png) [@John\_DiFool](https://boards.straightdope.com/u/John_DiFool)\ **Post date:** [July 20, 2014, 1:21am UTC](https://boards.straightdope.com/t/a-probability-puzzler/693518/29 "2014-07-20T01:21:44Z")
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? ---
**Author:** ![Isosleepy](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/isosleepy/32/479_2.png) [@Isosleepy](https://boards.straightdope.com/u/Isosleepy)\ **Post date:** [July 20, 2014, 1:38am UTC](https://boards.straightdope.com/t/a-probability-puzzler/693518/30 "2014-07-20T01:38:21Z")
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. ---
**Author:** ![Little\_Nemo](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/little_nemo/32/3120_2.png) [@Little\_Nemo](https://boards.straightdope.com/u/Little_Nemo)\ **Post date:** [July 20, 2014, 2:21am UTC](https://boards.straightdope.com/t/a-probability-puzzler/693518/31 "2014-07-20T02:21:58Z")
> [@John\_DiFool](#): > > 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? No, because she doesn’t want to bet on a hand that has one ace. She wants a hand that has two aces. ---
**Author:** ![Little\_Nemo](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/little_nemo/32/3120_2.png) [@Little\_Nemo](https://boards.straightdope.com/u/Little_Nemo)\ **Post date:** [July 20, 2014, 2:29am UTC](https://boards.straightdope.com/t/a-probability-puzzler/693518/32 "2014-07-20T02:29:46Z")
> [@John\_DiFool](#): > > 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. 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. ---
**Author:** ![VarlosZ](https://avatars.discourse-cdn.com/v4/letter/v/d78d45/32.png) [@VarlosZ](https://boards.straightdope.com/u/VarlosZ)\ **Post date:** [July 20, 2014, 2:34am UTC](https://boards.straightdope.com/t/a-probability-puzzler/693518/33 "2014-07-20T02:34:18Z")
> [@John\_DiFool](#): > > 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. 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. > [@](#): > > 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? 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. ---
**Author:** ![Roland\_Orzabal](https://avatars.discourse-cdn.com/v4/letter/r/c2a13f/32.png) [@Roland\_Orzabal](https://boards.straightdope.com/u/Roland_Orzabal)\ **Post date:** [July 20, 2014, 3:16am UTC](https://boards.straightdope.com/t/a-probability-puzzler/693518/34 "2014-07-20T03:16:54Z")
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**! ---
**Author:** ![Hari\_Seldon](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/hari_seldon/32/5173_2.png) [@Hari\_Seldon](https://boards.straightdope.com/u/Hari_Seldon)\ **Post date:** [July 20, 2014, 3:09pm UTC](https://boards.straightdope.com/t/a-probability-puzzler/693518/35 "2014-07-20T15:09:38Z")
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. ---
**Author:** ![UltraVires](https://avatars.discourse-cdn.com/v4/letter/u/ecccb3/32.png) [@UltraVires](https://boards.straightdope.com/u/UltraVires)\ **Post date:** [July 20, 2014, 4:33pm UTC](https://boards.straightdope.com/t/a-probability-puzzler/693518/36 "2014-07-20T16:33:08Z")
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. ---
**Author:** ![John\_DiFool](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/john_difool/32/19543_2.png) [@John\_DiFool](https://boards.straightdope.com/u/John_DiFool)\ **Post date:** [July 20, 2014, 5:17pm UTC](https://boards.straightdope.com/t/a-probability-puzzler/693518/37 "2014-07-20T17:17:32Z")
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. ---
**Author:** ![UltraVires](https://avatars.discourse-cdn.com/v4/letter/u/ecccb3/32.png) [@UltraVires](https://boards.straightdope.com/u/UltraVires)\ **Post date:** [July 20, 2014, 5:24pm UTC](https://boards.straightdope.com/t/a-probability-puzzler/693518/38 "2014-07-20T17:24:32Z")
> [@John\_DiFool](#): > > 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. ---
**Author:** ![Giraffe](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/giraffe/32/129_2.png) [@Giraffe](https://boards.straightdope.com/u/Giraffe)\ **Post date:** [July 20, 2014, 5:27pm UTC](https://boards.straightdope.com/t/a-probability-puzzler/693518/39 "2014-07-20T17:27:17Z")
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). ---
**Author:** ![dracoi](https://avatars.discourse-cdn.com/v4/letter/d/90db22/32.png) [@dracoi](https://boards.straightdope.com/u/dracoi)\ **Post date:** [July 20, 2014, 5:44pm UTC](https://boards.straightdope.com/t/a-probability-puzzler/693518/40 "2014-07-20T17:44:45Z")
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. [Previous page](https://boards.straightdope.com/t/a-probability-puzzler/693518.md?page=1) [Next page](https://boards.straightdope.com/t/a-probability-puzzler/693518.md?page=3)