Clock game math

Just curious what the odds of winning the Clock game are. Ive also heard this called Beat The Kings

Its a game i have played since i was young. Its single player like solitaire

A shuffled 52 card standard deck is dealt in its entirety in a counter clockwise shape (hence the name) with 4 cards to each of 13 positions. Each position corresponds to a different card value. All cards are face down

A. . 2. . 3. .4

Q. . . . . . . . 5
. . . . . K
J. . . . . . . . .6

10. . 9. . 8. .7

EDIT: had to add dots to keep formatting

Begin by turning over the top card in the king column. Place that face up on the bottom of the corresponding value pile

Ex if first card is 4 then it goes top right stack face up under the face down cards in the 4 column stack. U then turn over the top card on that stack

Lets say u now have Jack. Place that face up bottom of the Jack position. Keep going until...

All four kings are face up in center (you lose)
Or
Every card is face up.

Basically the only way you can win (beat the king) is if thru sheer luck the final card in the sequence is a king

Suits have no meaning or value. 3 kings showing instills fear in you but the only real decision is made when the fourth king appears

As a non-mathematician my gut tells me either 13:1 or 52:1. But maybe its more complicated than that

Good luck

I'm pretty sure the answer is 1/13.

The game is pretty much equivalent to this:
Take the top card of a deck, and place it face down. Take a particular card based on the card you just drew, pull it out of the deck, and place it face down on top of the card you placed face down earlier. Keep doing this; you win if the last card is a King.
Each of the 52! orderings of the 52 cards is equally likely, so the probability of winning is the probability that the last card is a King.


At first, I thought you were talking about Clock Game on The Price is Right, in which case, the best strategy is:
Start at 512
If the price is higher, say 784 (i.e. 512 + 256); if lower, 256 (i.e. 512 - 256)
Each subsequent jump is half of the previous one.
You will be correct in 10 or fewer guesses; 15 seconds is plenty of time.
Note that there is an "unwritten rule" that no prices on Clock Game are higher than $999; presumably, this is because it becomes harder to say the higher numbers. The few times they have offered cars in that game, the player only had to guess the last three digits.

Doesn't sound like a fun game since there's no skill or choice involved.

Wikipedia says the game is called "Clock Patience" and has 1/13 chance of winning.

Quote: RS

Doesn't sound like a fun game since there's no skill or choice involved.

Wikipedia says the game is called "Clock Patience" and has 1/13 chance of winning.

If you only find skill games fun then yea this is not for you

I find it fun in a how lucky can i get mode. It killa some time without costing anything

but but but...

If you have a situation where the last card in a stack is the same card as that stack.... ie: a 4 in the "4" stack.... you will never get to it. So I"m not believing the 1/13 conclusion. i think it's something less.

Quote: rsactuary

but but but...

If you have a situation where the last card in a stack is the same card as that stack.... ie: a 4 in the "4" stack.... you will never get to it. So I"m not believing the 1/13 conclusion. i think it's something less.

That's what I was thinking before, too. Then again, if all four K's are in the K pile, you immediately lose.

So who knows? You either win or you lose. It's 50/50.

Quote: RS

That's what I was thinking before, too. Then again, if all four K's are in the K pile, you immediately lose.

So who knows? You either win or you lose. It's 50/50.

Its definitely not 50/50

Winning at this game is very infrequent. Maybe its not more than 13:1 but most definitely not like flipping a quarter

Quote: rsactuary

but but but...

If you have a situation where the last card in a stack is the same card as that stack.... ie: a 4 in the "4" stack.... you will never get to it. So I"m not believing the 1/13 conclusion. i think it's something less.

Thats not correct. If the fourth king is the last card u locate all cards are found regardless of their location

In order for all 4 kings to be in the kings column at the start one king would have to be the 13th 26th 39th and 52nd card dealt after a random shuffle

Although i have seen kings turn up in the kings column i hav never seen all four. I have been playing for thirty years off and on. The odds against it are pretty high

It's 1 in 270,725 games you'll get all kings in the kings pile.

Quote: RS

It's 1 in 270,725 games you'll get all kings in the kings pile.

I definitely havent played that many times even in 30 years lol

Even though I trust the citations cited by the webpage cited in the wikipedia article,
https://en.wikipedia.org/wiki/Clock_Patience
http://mathworld.wolfram.com/ClockSolitaire.html

I wrote a quick program to simulate the game.

I get a win rate of about 1 in 13.