Gayle
> This is an easy one. I should warn you - the game is purely mechanical. You
> have no choice as to the way you play - the shuffle alone determines victory.
> You have a far better chance playing Klondike (Yukon is even better).
I know of two other solitaires with a clock-type layout which are *not*
mechanical. One of them is a two-deck game which gives the brain
quite a workout.
> But, I digress...
> Deal twelve cards in a circle, starting roughly where the one is on an analog
> clock (check local historical sites if you forget what one looks like...) and
> continuing, you guessed it, clockwise (I can imagine trying to explain to my
> grandchildren what that means: "Numbers just come and go, nothing turns...")
> so that the twelfth card is on top. Place the thirteenth card dead center
> in the circle. Repeat this process three more times, dealing on top of the
> cards already placed, so that you have thirteen piles of four cards each.
> Flip over the top card of the center pile. Place this card face up beneath
> the pile whose clock position matches the rank of the card. In other words, a
> trey would go in the rightmost pile (at three o'clock), a seven at seven
> o'clock, and so on; aces go at one o'clock, jacks at eleven, and queens at
> twelve. Kings are the cards you DON'T want to see - they go under the center
> pile. After placing the card, flip over the top card of the pile you just put
> it under and place it where it belongs. Repeat this process until either all
> the cards are placed where they belong (in which case you win) or you cannot
> continue because you placed a card beneath a pile that has no face down cards
> left on top (in which case you lose). Since you started with the King-pile,
> the game will end when you have completed the King-pile, the question being
> whether the other twelve piles were completed first.
> Al and Jeff (the professional "Hoyle" name-stealers) rated the odds of winning
> at about one in one hundred. Good luck.
They were a bit pessimistic. The actual odds are closer to 1 in 15 or 20.
Andy
--
================================================================================
My views are mine alone, and sometimes I change my mind.
================================================================================
This is an easy one. I should warn you - the game is purely mechanical. You
have no choice as to the way you play - the shuffle alone determines victory.
You have a far better chance playing Klondike (Yukon is even better).
But, I digress...
Deal twelve cards in a circle, starting roughly where the one is on an analog
clock (check local historical sites if you forget what one looks like...) and
continuing, you guessed it, clockwise (I can imagine trying to explain to my
grandchildren what that means: "Numbers just come and go, nothing turns...")
so that the twelfth card is on top. Place the thirteenth card dead center
in the circle. Repeat this process three more times, dealing on top of the
cards already placed, so that you have thirteen piles of four cards each.
Flip over the top card of the center pile. Place this card face up beneath
the pile whose clock position matches the rank of the card. In other words, a
trey would go in the rightmost pile (at three o'clock), a seven at seven
o'clock, and so on; aces go at one o'clock, jacks at eleven, and queens at
twelve. Kings are the cards you DON'T want to see - they go under the center
pile. After placing the card, flip over the top card of the pile you just put
it under and place it where it belongs. Repeat this process until either all
the cards are placed where they belong (in which case you win) or you cannot
continue because you placed a card beneath a pile that has no face down cards
left on top (in which case you lose). Since you started with the King-pile,
the game will end when you have completed the King-pile, the question being
whether the other twelve piles were completed first.
Al and Jeff (the professional "Hoyle" name-stealers) rated the odds of winning
at about one in one hundred. Good luck.
Adam R. Wood
the Zotmeister
I don't know how clear that is. I'm not really good at writing directions.
Let me know if you have any questions.
Shari
> Al and Jeff (the professional "Hoyle" name-stealers) rated the odds of
> winning at about one in one hundred. Good luck.
They are wrong here. The probability of winning at Clock Solitaire is
exactly 1/13.
Each time you turn up a card, the probability is equal that it is any
one of the remaining cards. (In fact, you could just play with a
shuffled deck, turning over the next card at each stage, and checking
off one box when a card from one of the piles would be taken.) If you
lose, you can turn up the remaining cards one by one so that you get to
see all 52.
If the 52nd card turned up is a king, you must win, because you haven't
seen the fourth king until everything else is played. Otherwise, you
lose, because there are cards left when the fourth king turns up.
One of Martin Gardner's columns has a more complete analysis; among
other things, he shows that you can tell whether you will win by looking
only at the bottom cards of the ace through queen piles. If there is a
cycle (i.e., seven on the bottom of the five pile, jack on the bottom of
the seven pile, five on the bottom of the jack pile), you can never
remove any card in that cycle, so you will lose. If there is no cycle,
then you will win.
--
David Grabiner, grab...@math.harvard.edu
"We are sorry, but the number you have dialed is imaginary."
"Please rotate your phone 90 degrees and try again."
Disclaimer: I speak for no one and no one speaks for me.
Odds in Klondike depend on what version you're playing, but only the
loosest
rules give you as good a chance as Clock.
Michael Keller, World Game Review, 1747 Little Creek Drive,
Baltimore, MD 21207 <Wgre...@aol.com>
Not needed - it's intuitively obvious. Silly me, quoting Al and Jeff - I
should know better by now...
: Odds in Klondike depend on what version you're playing, but only the
: loosest
: rules give you as good a chance as Clock.
The "official" rules (draw one, no redeals, no takebacks) give odds at about
thirty to one. This number I'm more sure of.
> Al and Jeff (the professional "Hoyle" name-stealers) rated the odds of winning
> at about one in one hundred. Good luck.
Clearly, the odds of winning are 1/13. They just _seem_ like 1/100.
__
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> : Clearly, the odds of winning are 1/13. They just _seem_ like 1/100.
> Yeah, I mentioned that in the other thread. You see, that's what I get for
> quoting "Hoyle" - and I'm a math major, for crying out loud!
> I definitely have to stick to the SEG.
In Al and Geoff's defense, I must point out in the preface to their
solitaire book they admit that many of the odds they give are guesses
and "some may be wide of the mark".
What is the SEG? Does it give accurate odds for all the solitaire games?
Andy
--
(from a fortune cookie:) Anyone who makes a blanket statement is a fool.
: Clearly, the odds of winning are 1/13. They just _seem_ like 1/100.
Yeah, I mentioned that in the other thread. You see, that's what I get for
quoting "Hoyle" - and I'm a math major, for crying out loud!
I definitely have to stick to the SEG.
Adam R. Wood
the Zotmeister
The odds of Clock are exactly 1 in 13 but I don't see how this is
inherently
obvious -- there have been half a dozen mathematical papers written
proving
this 'obvious' fact.
Michael Keller, World Game Review, 1747 Little Creek Drive,
Baltimore, MD 21207-5230, <Wgre...@aol.com>
> The odds of Clock are exactly 1 in 13 but I don't see how this is
> inherently
> obvious -- there have been half a dozen mathematical papers written
> proving
> this 'obvious' fact.
I have similar sentiments--and a theory. The last card turned over
must be a king; in that sense, it seems like the odds would be the
same as that for the card on the bottom of a deck being a king.
But what if the ace pile *starts* with 4 aces in it? None will be
reached. Similar blocks are easy to cook up between piles. Someone
mentioned a proof of the odds by Martin Gardner--if he took on this
problem I don't think it was very "obvious".
Andy
P.S. Does anyone know the odds of winning the counting solitaire,
Hit or Miss? This might be done best empirically, with simulation.
Your theory has it right. I remember this game coming up in a class
taught by Don Knuth. He said he'd done a very complicated analysis of
the game and eventually reduced the answer to 1/13, at which point he
"suspected there might be an easier proof". He used this to introduce
the concept of delayed information.
Consider the following variation on the mechanics for the game: Instead
of dealing all the cards ahead of time, wait until you're about to turn
over a card in one of the 13 piles and only then do you actually deal the
card there. I.e., you start with no cards dealt; you deal one to the
center pile and turn it over. If it's a 7 (say) you deal the next card
to the 7s pile and turn it over, and so forth.
It's not hard to see that this is really the same game, just with the
cards sitting in one place instead of starting out spread across the
table. (If it'll help, you can say that if you lose you then deal the
rest of the cards out to fill out each pile to be 4 cards.) And it's
clear that you win if and only if the 4th King is the last card.
-- Don.
Yes, I agree, quoting odds of 1:13 seems very simplistic. For starters,
here is what you need to win the game:
(1) Last card in the draw pile must be a king, or the last cards in the
middle must be kings.
(2) The bottom card in the ace pile cannot be an ace.
(3) The bottom card in the two pile cannot be a two.
Etc.
(4) You cannot have complementary card pairs, e.g., a two in the bottom
of the ace pile and an ace in the bottom of the two pile.
(5) You cannot have complementary card trios, e.g., a three in the bottom
of the ace pile, and ace in the bottom of the two pile, and a two in the
bottome of the three pile.
Etc., for 4-card, 5-card combinations and so forth.
So, simply having a king at the end doesn't guarantee a win. The odds
HAVE to be worse that 1:13.
Interesting statistical problem. Anyone know for sure what the true
odds are?
Dan E.