In your game, the gambler only has information about the first card. In the original game, the gambler gets information about both cards. These are different games, so there’s no reason to think that they’d have the same odds.
I get it now-I think…
Okay, I think I’ve got the winning question.
As per the OP’s rules, you’re not allowed to ask a question if the answer would guarantee you a winning bet. So you can’t ask “Do my cards contain one or more kings?” or “Do my cards contain two aces?”
But “Do my cards contain a king of clubs, a king of diamonds, and/or a king of hearts?” is an acceptable question. It doesn’t produce a sure thing bet but it does weigh the odds in your favor. You ask this question and then bet whenever the answer is no.
A no answer to this question eliminates eighteen of the possible losing hands and none of the six possible winning hands. So of the ten remaining possible hands, there are six winning hands and only four losing hands. The odds are now 3-2 in your favor.
What if you add a joker to the mix --a card that can be an ace but is not an ace?
Bayes’ Theorem FTW.
The question is essentially, given these two questions:
- given at least 1 Ace, what is the probability of two Aces?
- given the Ace of spades, what is the probability of two Aces?
Bayes’ theorem says P(A|B) = P(B|A)P(B)/P(A), where the notation P(A|B) indicates the probability of A, given B.
For Alice’s strategy, we need to calculate P(>0A|2A), P(>0A) and P(2A).
P(>0A|2A) = 1, since if both cards are Aces, at least one card is an ace.
P(>01) = 1 - P(0A) = 1 - (4/8)*(3/7) = 1 - 12/56 = 11/14.
P(2A) = 4C2 / 8C2 = 6/28 = 3/14. (4C2 means “4 choose 2”).
Using Bayes theorem, P(2A|>0A) = (3/14)/(11/14) = 3/11.
For Bob’s strategy, given that we know one card is the ace of spades, there are 3 ways the other card could be an ace (the 3 other suits) out of 7 cards, so
P(2A|A spades) = 3/7. (You can also work this out using Bayes’s theorem, of course).
Since Bob’s strategy has a higher probability of yielding a winning bet, his strategy is better. But really, neither of them should play the game at all.
Sorry, that should have read “P(A|B) = P(B|A)P(A)/P(B)”
That clearly isn’t true, as it is the absolute value of the money that matters, so as the odds get higher alice’s strategy gets better.
IE, assume it pays 1 million to one:
Based on previous poster’s work:
Alice wins: 12/56 hands
Bob wins: 6/56 hands
So every 56 hands, Alice wins just shy of $12 million, while Bob wins just shy of $6 million. Because the initial stake bet is trivial vs the payoff, clearly the best strategy is to bet every hand that has any possibility of wining, if the odds are high enough. Thus the odds definitely do matter.
I got that question once on a combinatorial probability exam during my actuarial degree.
Alice’s strategy is only better when the game has a positive expectation for a player who asks no questions (and even then, it has to be sufficiently better). There’s a standard assumption that that’s not the case, so I’m not considering it.
I don’t think that’s correct.
Let’s say the payoff is 3-1 (a person gets three dollars when he wins and loses a dollar when he loses). Let’s assume there are 280 hands played and assume average results.
Zeke (the guy who never asks any questions and bets every hand) will play all 280 hands. He will win 60 hands for a gain of $180 and lose 220 hands for a loss of $220.
Alice plays her strategy. She passes on 60 hands and bets on 220 hands. She wins 60 times for a gain of $180 and loses 160 times for a loss of $160.
Bob, meanwhile, plays his strategy. He passes 210 times and bets on 70 hands. He’ll win 30 times for a gain of $90 and loses 40 times for a loss of $40.
So Alice does do better than Zeke but not as good as Bob at 3-1 odds. The tipping point where Alice’s strategy equals Bob’s is a 4-1 payoff. At that payoff, both Alice and Bob win an average of $80 during a 280 hand run. At payoffs better than 4-1, Alice’s strategy is better because of her higher betting rate.
Zeke’s strategy will never beat Alice’s because Alice never misses a winning hand. But she does skip some losing hands than Zeke plays. So Alice will always win as much as Zeke and lose less than Zeke.
But it is possible to have a better strategy than Alice’s to take advantage of high payoffs. You want a strategy that covers all the winning hands while minimizing the losing hands as much as possible.
So we have Carla. Carla’s question is “Does my hand contain an ace of clubs, an ace of diamonds, and/or an ace of hearts?” This is an allowable question because its answer doesn’t guarantee a win. However a yes answer to this question will include all six possible winning hands while eliminating ten of the twenty-two possible losing hands.
In an average 280 hand run with a 4-1 payoff, Carla passes on 100 hands and bets on 180 hands. She wins 60 hands for a gain of $240 and loses 120 hands for a loss of $120.
Didn’t Dr. Strangelove provide the best possible question in post #9? His question, “are there either two aces, or a king-spades/king-hearts pair?”, eliminates all but one of the losing combinations, while not guaranteeing a winning bet.
Of course, the OP doesn’t say that the question has to be solely about the cards, so part of the question could be about anything the OP knows that you don’t, e.g. “is it true that either there are two aces, or your social security number is <some random valid number>?” If he answers “yes”, you still don’t know for sure that they’re both aces, but they almost certainly are.
Huh. I read the question, did a quick and dirty calculation and voted “neither is better” because they are both losing strategies. I think I over thunk.
Yep. Or “What would your answer to ‘are there two aces?’ be if you lied to me 0.0001% of the time?” Add as many zeroes as necessary.
You’re right. I had missed his post.
Well, I ran a sim of both strategies-Alice won 27% of the time, Bob 45%…
Bingo.
This is a GREAT question. Unfortunately (or fortunately, depending on your POV) it is open-ended.
Hari Seldon’s analysis is correct and complete iff we assume the odds offered are 2:1.
But the question cannot be answered independently of the odds.
Having just done the maths on the back of a envelope I found that for odds of P : 1 Alice has the better strategy for P > 4 (the stratégies are equally good for P = 4)
I’ll be back with the maths after dinner (or you can work it out yourselves in the meantime).
Probability that Alice wins = 3/11
Probability that Bob wins = 3/7
Proportion of games played by Alice = 22/28
Proportion of games played by Bob = 7/28
(détails of the calculations required are given by Hari Seldon and Little Nemo (who also offered the 4:1 equilibrium long before me) above).
Noting the odds as P:1 the expected gains for Alice and Bob are:
EA =(3/11 P - 8/11) x 22/28 + 0 x 6/28
EB =(3/7 P - 4/7) x 7/28 + 0 x 21/28
simplifying:
EA = (6P - 16)/28
EB = (3P - 4)/28
So the game is worth playing:
for Alice iff P > 8/3 and
for Bob iff P > 4/3
Solving EA > EB gives:
6P - 16 > 3P - 4
3P > 12
P > 4
i.e. if P > 4 Alice has the superior strategy.
This too is based on an assumption about constraints or their absence. For instance, under either strategy, the house will be bankrupted at the same dollar amount. If the payoff is such that either approach will win over the long run, then either approach will eventually accomplish that end. Maybe one will get there faster or slower, but even that’s dependent on another constraint: a ceiling on any individual bet, or limited ability of a player to borrow money to back this sure thing.
Now a maximum bet, combined with a limited number of betting opportunities, would make relevant the speed at which one wins, assuming also that those limits would keep at least one approach from bankrupting the house.
If the house has infinite resources, and there are no limits on bets and the game keeps going forever, then if both approaches win in the long run, then either of them will make one wealthy to the tune of countably infinite wealth. (Since money comes in rational-number quantities.) And of course all countably infinite quantities are equivalent.