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        riddles >> medium >> Clock Solitaire
        
(Message started by: Icarus on Jan 30^th, 2008, 7:59pm)

Title: Clock Solitaire
Post by Icarus on Jan 30^th, 2008, 7:59pm

When I was kid, I would occasionally play a game called "Clock Solitaire". You would deal out the entire deck into 13 facedown piles of 4 cards in a clock arrangement with the 13th pile in the middle. Each pile represented a card value, Ace-Queen on the outside, kings in the middle. You then turn over the top card of the king pile, and put it face up on the bottom of its corresponding pile (so if it was a 7, you would put it under the 7s pile), and repeat with the top card from that pile. Play continues until you put the 4th king in the kings pile. There will no longer be an upside down card on the pile, so you cannot keep going. If at this point all the cards have been turned up, you win. If any of the piles still has face down cards, you lose.

What is the probability of winning a hand of clock solitaire?

Title: Re: Clock Solitaire
Post by SMQ on Jan 31^st, 2008, 5:36am

I'll run some simulations ;) but I think it's just [hide]one in thirteen. The rules create permutation of the deck, and you win iff the final card in that permutation is a king. So long as the permutation is uniformly random there should be an equal chance of any card being the final card.[/hide]

[edit]supported by simulation: [hide]76,935,662/1,000,000,000 = 1 in 12.9979[/hide][/edit]

--SMQ

Title: Re: Clock Solitaire
Post by rmsgrey on Jan 31^st, 2008, 7:29am

There's an "easier" variant whereby you win if, after the fourth king has been played, all the cards are in the correct pile, regardless of whether they're face up or face down, which would take more work to find the winning probability for.

Title: Re: Clock Solitaire
Post by SMQ on Jan 31^st, 2008, 8:03am

Wow, much easier! By simulation: [hide]20,280,913/100,000,000 = 1 in 4.93074 -- more than 2.5 times as likely to win[/hide].

--SMQ

Title: Re: Clock Solitaire
Post by rmsgrey on Feb 1^st, 2008, 11:44am

Well, you close to double your chances just looking at the times where the last king is the penultimate card, and another a bit less than half as many when it's the antepenultimate card, and so on.

Title: Re: Clock Solitaire
Post by Hippo on Feb 1^st, 2008, 2:54pm

on 01/31/08 at 05:36:28, SMQ wrote:

I think it's just [hide]one in thirteen. The rules create permutation of the deck, and you win iff the final card in that permutation is a king. So long as the permutation is uniformly random there should be an equal chance of any card being the final card.[/hide]
--SMQ

I haven't understand your argument at first, I will try to say it by my words. [hide]Change the game slightly ... delay dealing cards until you should look at the card. So you need to turn the top card of Kings pile. Deal the card and turn it ... put it on corresponding pile, deal second card and turn it ... actualy there is no reason to put the card to the pile before turning it ... so deal cards looking on them to the corresponding piles. Stop when 4th King is dealt. This is the same game with another timing. Yes you win if the last card of the pack is King. So the same game would be to look to the random card only (1/13 prob). But it would probably be less funny.[/hide]

With the other rules the situation is more complicated: [hide]Let p_k1,k2,k3,k4 be the probability there are k=k1+2k2+3k3+4k4 cards after last king ki card values are i-times among them. The winning position has probability \sum_k1,k2,k3,k4 p_k1,k2,k3,k4*2^k2*(3!)^k3*(4!)^k4/k!.
As Grimbal mentioned for k=0,1 p_k,0,0,0=1/13.
\sum_k fixed p_k1,k2,k3,k4=1/13. p_k,0,0,0=13!/(13-k)!/13^(k+1) ... in the sum
this accounts by kC13/13^(k+1) ...
[/hide]

Title: Re: Clock Solitaire
Post by Icarus on Feb 2^nd, 2008, 9:58am

Clearly, I was overthinking this, or else I would have put it in Easy.

Here is a way of looking at the game to see how it represents a permutation of the deck, as SMQ said. Make two changes to the rules:
1. Instead of putting the cards you turn over in the appropriate pile, simply place them in a discard pile. You still take the next card from the pile indicated by the current card. This does not affect the game play any. It just changes the final disposition of the cards.
2. Institute a "Finish even if winning is impossible" rule. Whenever the game requires you to take a card from a pile that is already removed, take a card from the highest numbered of the remaining piles instead. The game is still won only if the final card is the 4th king (in which case, this new rule would not have any effect, so winning games play exactly as before).

With these two rules, the game becomes simply a fixed means of rearranging your deck, with the discard pile being the final permutation. Further, any potential discard pile can be backed out to find an initial arrangement of the cards that results in it. So the permutation method is an endomorphism of decks. In particular, this means that all arrangements of the discard pile are equally likely (since all arrangements of the original deck are). Since you win if and only if the top card of the discard pile is a king, the probability of winning is 1/13.

This analysis also provides another way of looking rmsgrey's variant: A discard pile represents a winning hand if and only all cards after the 4th king are arranged in decreasing order.

Title: Re: Clock Solitaire
Post by Icarus on Feb 2^nd, 2008, 6:00pm

Astrix apparently decided not to point out after all that he had posted this before. But I think credit should be given where it is due. Besides, there are points in the earlier thread not brought up here:

http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_easy;action=display;num=1067108039