Consider betting on a simple card game. I take a standard deck and remove everything but the aces and the kings. I then draw two cards in the usual manner. If they’re both aces, the gambler wins. Otherwise, it’s a loss. Simple enough?
I’m feeling generous, so I’ll answer a single yes or no question before the gambler places their bet as long as it doesn’t determine whether the hand is a win. It won’t even cost anything.
There are two players of interest here. Alice asks me if at least one of the cards is an ace, and bets whenever I say yes. Bob asks me if the hand has the ace of spades, and also bets whenever I say yes. Neither player is aware of the other or of the answers to the other’s question.
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 10 losing hands. So her odds are improved to 6 winning hands to 12 losing hands - 2:1 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 10/28 or about 38% of the time. Bob will drop out with a no 21/28 or 75% of the time.
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.
There are 56 possible hands. Order matters, so the count is just 8*7.
Of these, 12 are AA, 16 are AK, 16 are KA, and 12 are KK. Alice wins 12/(12+16+16) = 27.3% of the time.
There are 3 hands that go As-A, 3 of A-As, 4 of As-K, and 4 of K-As. Bob wins (3+3)/(3+3+4+4)=42.9% of the time.
[/spoiler]My strategy would be “are there either two aces, or a king-spades/king-hearts pair?” 92.3% chance of winning :).
Assuming the real world, the deck only has 8 cards in it after you’ve removed everything but Kings and Aces.
Asking if both cards are Kings doesn’t tell the person whether it’s a winning hand, and thus allowed by the rules.
Since we draw two cards at a time, there will only be four hands before a reshuffle.
If the dealer ever answers, “Yes”, to the above question, then you know that all the other hands are composed of 4 Aces and 2 Kings.
If you always bet after that (until a reshuffle), then you’ll win on average. If you continue to ask whether both cards are Kings, and (obviously) don’t bet when he answers yes, then you’ll win even more on average.
I think shuffle after each game was assumed, because otherwise the question would be about card counting…Anyway this doesn’t improve on "is there at least one ace ? "
because its the same question… well the information derivable from the answer is the same.
My mistake was that I didn’t interpret the question correctly, at all. I was only counting two scenarios: win and loss (i.e. if Alice’s strategy was right she won money, if Alice’s strategy returned false she lost money).
The long and short is: Alice wins far more games than Bob, but she also loses far more than Bob. Where Bob comes out ahead is when he doesn’t bet. I totally forgot to take into account that “not betting” and “losing” were distinct and went with “Alice wins more than Bob”, which is correct, but irrelevant.
Basically, I was playing two games with different reward functions; not using two betters with different strategies. Alice and Bob always bet, but Alice got paid out when there were two Aces, and Bob got paid out when he got the Ace of Spades and another ace.
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.
Identical odds. Bob, once he bets, has exactly the same chance of winning as Alice, when she bets. The only difference is that Alice bets four times as often as Bob does. Bob is simply making the same bet as Alice, except that Bob passes and Alice bets when the Ace is not a spade.
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.