Game: Guess the wrong card!

Basically.... I have a normal deck of cards (52 cards total, 13 suites and 4 ranks per suite). You have to make a guess of what the next card will NOT be. ie: You guess '6', if the card is A,2,3,4,5,7,8,9,T,J,Q,K you win and advance to the next card. If that card is a 6, you lose. You gotta get through all 52 cards to win.

I thought it was rather interesting.

By random guessing, I believe the chance of success is (12/13)^52 = 0.01557293512 or ~ 1.55%.

What is the best strategy to play this game assuming you can keep track of each played card? Well, I guess that's easy -- guess the card that has been played the most, and once a card has been played 4 times, you keep guessing that card. Rather -- if you use this strategy [I believe that would be the best strategy? unless there's another better one?], what is the chance of being successful?

It's an interesting concept.

Yeah, if one rank gets exhausted early, picking it for the rest of the game is a no-brainer.

For that reason, I'd include an electronic tracking display, and a rule that you can't pick an exhausted rank. And, obviously, you only make a maximum of 51 guesses.


How many correct guesses are needed before you start to win? What's the paytable look like?


Cool concept but I don't see it working in a casino. Maybe online...

It's not gambling related. Just curious as to the rate of success (of picking all 52 cards wrong) if you use a tracking-strategy.

This is an old form old solitaire. Flop one over at a time. Start at ace go to king then repeat. Match a rank, Don't shuffle but take the exposed cards and place them at the bottom of the deck repeat until you either get an exact repeater (Kh, Kh) or you get all the way through the deck without matching.

Assuming you cant pick an exhausted suit, keep track until there are two suits with only one card and alternate each choice. Your odds will be great at first although they will approach fifty fifty asou get towards the last two cards

Assuming perfect strategy (choose Ace for the first card; if the first card is not an Ace, choose that card, since there are only three of them left in the deck, and keep choosing it until another card (including Ace) has appeared twice, then keep guessing that card until any card has appeared three times, then keep guessing that card until any card has appeared four times, then keep guessing that card as it cannot come up any more), a simulation of 100 million runs shows you win OOPS - 27% of the time.

I show the probability of winning for a perfect counter to be 27%.

To correct something on the show, the probability of winning by guessing sequentially is not (12/13)^52 = 1.56% but 1.65%. It is hard to explain why going sequentially is better than guessing randomly. One way would be to think of the simple case of a 2-card deck with just a king and queen. In such a deck if you guess randomly your odds of winning are .5^2=25%. Going sequentially it is 50%.

Quote: Wizard

I show the probability of winning for a perfect counter to be 27%.

Well, sure, if you take the number of wins and divide by the number of trials like you're supposed to, and not the other way around like I did...

I was going to post this in its own thread, but it's germane to this topic. Mike Sexton talking about Stu Ungar and card counting.

https://youtu.be/JGbm9GMkWvs

Here is a simple strategy -- always pick the last rank drawn (except on the first card where you can pick anything). That has a probability of success of 4.20%. Compare that to the 1.65% of going in sequence.