Okay, I’m feeling cocky. Having won three games in a row, I’m gonna start a new game but automatically dealing 10 cards from the stock regardless of the initial deal. Again we will look at the same questions as last time.
- How many cards are we guaranteed to turn over even if all face-down cards happen to contain faces of random politicians? (and the rules say we can’t do anything with them).
- What are the chances of drawing a good card (assuming each rank occurs with probability 1/13 and politicians occur with probability zero).
This is much harder than the previous example. As I mentioned earlier, one of the non-trivial aspects of Spider Solitaire is exposed cards are not necessarily in descending sequence (now you can see why).
To answer the first question, I recommend the beginner should break the problem down by considering each column in isolation. Assuming the columns are numbered 1-10 from left to right is it possible to turn over a card in column 1, even if we were willing to trash the rest of the tableau in every way possible? Then repeat the same question for column 2, column 3 etc. The answers for each column are summarised below:
- Clearly impossible: to shift the Deuce we need another Three, but the only other Three is covered by a king. Note that if the 2-3 were in-suit then yes, it could be done.
- Clearly impossible: the Jack can be shifted in two moves, but again there is no way to shift the Deuce
- This column contains a king. Don’t. Even. Think. About. It.
- Yes, even my Dad can do this.
- No. There is no spare Ten to shift the Nine.
- Yes, but only because we are lucky enough to have a suited connector.
So our initial analysis shows that we can achieve a turnover in six out of 10 columns. But we haven’t established that it is possible to guarantee all six turnovers simultaneously. For instance, there could be supply/demand issues (remember in the earlier “non-cocky” example we had three sixes and only one seven) or there could be complex timing issues. But at least we know that the answer to Question 1 is at most 6.
Unfortunately there is no easy way to answer this question and we have to visualise (or experiment with 85,78,68,79). The beginner may find the latter helpful at first, but I strongly recommend he she or ze try to discard this dangerous habit as soon as possible.
You should find that 5 turnovers are possible. Further experimentation might reveal the reason why we can’t get all 6 turnovers: three of the “good” columns contain Sixes and we only have two Sevens. Therefore we can only obtain turnovers in two out of the three columns, (but at least we get to choose which ones). It is pure coincidence in both examples, we had too many Sixes and not enough Sevens.
The diagram below shows the game state after obtaining 5 turnovers. I have replaced the newly exposed cards with random politicians so the reader can easily verify that (i) No further turnovers are possible without knowledge of these cards (ii) This position can in fact be reached from the original position. (In the actual game, the new cards gave me 2 extra turnovers).
Whew! That was a complex analysis: hopefully I have inserted more than enough corny jokes to prevent the reader from being totally confused and/or bored and I haven’t even attempted to answer the second question, so that will be for the next post. If you understood all of this perfectly and are really bored, perhaps you could try writing your own rap song for this position 😊
That’s it for now, happy spidering 😊