What is the player expectation in this simple card game?

Simple game. 4 cards are dealt face down from a deck of 52. Aces are always high. Player turns over card of his choice. He then guesses if the next card he turns over is higher or lower. If successful, he guesses whether the next card he turns over will be higher or lower than the previous card he just turned over, and if he guesses right again, he guesses one last time if the last card left will be higher or lower than the 3rd card he turned over. All ties are losses. The player must guess correctly the hi-lo question 3 times to win the game.

Assuming he plays correctly, what is the expected % of times the player should win?

I don't know the answer, so I'll be impressed if someone can figure this out accurately via formula.

Quote: shakhtar

Simple game. 4 cards are dealt face down from a deck of 52. Aces are always high. Player turns over card of his choice. He then guesses if the next card he turns over is higher or lower. If successful, he guesses whether the next card he turns over will be higher or lower than the previous card he just turned over, and if he guesses right again, he guesses one last time if the last card left will be higher or lower than the 3rd card he turned over. All ties are losses. The player must guess correctly the hi-lo question 3 times to win the game.

Assuming he plays correctly, what is the expected % of times the player should win?

I don't know the answer, so I'll be impressed if someone can figure this out accurately via formula.

Card Sharks is a reasonably close game to what you describe, if I am understanding your description correctly:

https://wizardofodds.com/games/card-sharks/

I don't know if there's a formula, as you have to take into account what the earlier card(s) are, especially if the current card is an 8. (If the first card is a 3 and the second is an 8, there are 24 cards remaining that are higher than 8, but only 23 that are lower.) It sounds more like a brute force (go through every deal of 4 cards) problem.

If I counted these right, I get 35.0023% (2274240 out of 6497400).

Quote: Mission146

Card Sharks is a reasonably close game to what you describe, if I am understanding your description correctly:

https://wizardofodds.com/games/card-sharks/

I'm not trying to be mean, but that is nothing like what I described.

Quote: ThatDonGuy

I don't know if there's a formula, as you have to take into account what the earlier card(s) are, especially if the current card is an 8. (If the first card is a 3 and the second is an 8, there are 24 cards remaining that are higher than 8, but only 23 that are lower.) It sounds more like a brute force (go through every deal of 4 cards) problem.

If I counted these right, I get 35.0023% (2274240 out of 6497400).

If that's true, it's a little higher than I thought. Thanks for the help. Be curious if anyone else would get that figure and I'd consider it confirmed.

Quote: ThatDonGuy

I don't know if there's a formula, as you have to take into account what the earlier card(s) are, especially if the current card is an 8. (If the first card is a 3 and the second is an 8, there are 24 cards remaining that are higher than 8, but only 23 that are lower.) It sounds more like a brute force (go through every deal of 4 cards) problem.

If I counted these right, I get 35.0023% (2274240 out of 6497400).

Quote: shakhtar

If that's true, it's a little higher than I thought. Thanks for the help. Be curious if anyone else would get that figure and I'd consider it confirmed.

That's what I get.