A closer look at the Hole-In-One game

In the last post I looked at the “hole-in-one” game, where the aim was to get an empty column before being forced to deal another row of 10 cards from the stock. I asked the following questions:

  • What are the chances of winning, assuming perfect play and a “proper” random number generator?
  • How would you go about estimating it?

The first question was intentionally difficult. It is not trivial to find a perfect strategy, even for this simplified game where the player doesn’t have to worry about what happens after 10 cards are added to the tableau. Even I don’t have a definitive answer. Instead I wanna focus on the second question.

If you know anything about research, you might think the win rate can be estimated by playing a large number of games, tallying up your wins and losses and then doing some basic math to arrive at a winning percentage between 0 and 100. Or if you want to spice things up, you might guess a winning percentage before playing and see how accurate your guess was 😊

In this case, I estimated my win rate should be 30%. I played 30 games (to give myself the chance of an exact guess if I happened to win exactly nine games). Alas I won ten games out of 30, one too many – so my estimated win rate should be 33%.

This figure should be taken with a few pinches of NaCl because:

  •  my sample size of 30 games is small. Ergo, 33% could be an overestimate or underestimate.
  • Even the noble Spider GM may not have played the perfect strategy, so the correct win rate has been underestimated.
  • A more subtle point is optimal strategy may change when we only focus on empty columns. In a real game nobody would trash their game in every way possible just for the sake of increasing their miserly chances of getting that hole-in-one from 2% to 3%. In other words, 33% could be an over-estimate of our chances of getting that hole-in-one.

The real point of the exercise is to give the student a feel of how the simplified game works. If you have played 30 (or more) games of hole-in-one you may have come across some interesting decisions. Here is an example:

There are a number of options to consider:

  • Two of Clubs onto the Three of Hearts, turning over a new card in the only column containing only three face-down cards.
  • Three of Hearts onto the Four of Clubs, ensuring the previous option won’t decrease our minimum guaranteed turnovers.
  • Two of Hearts onto the Three of Hearts, turning over a new card in a column containing four face-down cards but building in-suit
  • Ace of Clubs onto the Two of Clubs, keeping the above options open.

Although a win-rate of 33%  is to be taken with a pinch of NaCl we could still conclude that:

  • If you do obtain an empty column before dealing from the stock then you are probably in good shape
  • If you don’t get the empty column then you are not necessarily in bad shape. I know from previous experience I can clear all eight suits about half the time sans ROT13(haqb). If you were close to an empty column but didn’t quite make it, you are probably doing okay.

Until next time, happy Spider Solitairing 😊

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