Shuffle a deck of cards; then look at bottom card: probability that it's a King = ?
Clock solitaire or clock patience is a card game played by one person where all 52 cards are used and dealt out in 13 piles (4 cards in each pile) with 12 piles representing the numbers on a clock (1-12) and the 13th pile placed in the middle of the clock.
play starts by turning over a card in the middle, for example if that card was a seven you place it at the corresponding 7 pile in respect to the clock and then turn over another card from that pile, for example a jack then you would place it n the coresponding 11 pile and then take a card from that pile. this continues until you draw a king. then you begin again drawing another card from the pile in the middle.
the game is only won if a king is last card to be turned over.
I have been playing this game since I was pretty young and recently began wondering what the probability of winning the game is and i began to think that the probability is very low. i looked at a number of other websites that claim that the probability of winning is 1/13, where this is the probability of drawing a King last within the sequence of drawing cards.
however, they never factor in tha tthe game cannot be won if at the bottom of a pile is the corresponding number. for example if at the bottom of the pile corresponding to 4 on the clock was a 4 card then game cannot be won.
i hope i've explained the game properly so that you can understand.
Hence my last post: king has 1/13 probability.
You're turning cards over until all 4 kings have been turned (hoping 4th is 52nd card):
so minimun cards turned is 4 (all kings ended up in middle),
maximum is 51 (no need to turn 52nd card: 4th king by default!).
Anyhow, that's the way I understand your explanation of the game.
For example, let's say that under the "12" on the clock, the pile is ordered as follows: A, 2, 3, Q. Notice the queen is on the bottom. Somewhere out in the other piles are the other three queens. Now, let's just say that I end up playing those three queens before having played all four kings. The first queen I play will direct me to the "1" on the clock, the second will direct me to the "2" on the clock, and the third will direct me to the "3" on the clock. Now I will continue playing the clock, but will never get back to the "12" on the clock to pick up the final queen because there are no more queens. It's a guarantee that the kings will all be used up first and therefore, a king will not be last in the play sequence. This still fits into the original 12/13 chance that a king will not be the last card we play in the sequence.
In mathematical terms, there are 52! (or 52P52) ways to arrange the cards on the clock. Each one of those will direct us around the clock differently, granted some of them will end in an equivalent result (with the game stopping because all four kings are used), but the arrangement of the cards being different. Out of those arrangements, there will be 51!x4 that will have the king at the end of the sequence (or you could write 51P51 x 4P1). So the probability of getting the arrangement with the king at the end is (51!x4)/52! which simplifies to 4/52 = 1/13
Hope that helps!