In this lesson we will examine the issues of rank imbalances. The current diagram shows the state of play after we dealt a second row of cards from the stock.
One thing you may have noticed is we seem to have a lot of Jacks but not many Tens. To be more precise we have only one Ten but six Jacks. That’s a delta of 5. If we try to construct the entire “histogram” for all card ranks we will also find several other discrepancies between other pairs of adjacent ranks such as seven Fours but not as many Threes or Fives.
Of course all players would know that such imbalances are less than convenient, but how best to deal with such imbalances?
One thing to note is that we can’t affect the probability of turning over specific cards. For instance, there are two Jacks remaining and 104-49=55 cards unseen. The probability that the next exposed card is a Jack is always 2/55 no matter how well or badly we play (if we deal from the stock we can always pretend 10 cards appear sequentially instead of simultaneously). But we can mitigate the effects to some extent. For example if two Jacks are buried under a King then the effects of too many Jacks will be attenuated, but if the only Ten was buried under two Kings then that is obviously much worse.
It is beyond the scope of this post to discuss in detail how to deal with rank imbalances, but a general principle is that the more flexibility you have, the better your chances will be – and the best way to retain flexibility is to procrastinate whenever possible.
Consider the following questions:
- Can we get back our empty column? (a good question to ask whenever at least one column has no face-down cards!)
- Can we increase the number of in-suit builds?
- How many guaranteed turnovers do we have? (Note that an empty column usually equates to one more turnover, but not always).
- What would be your next play?
As an aside, here’s a question for the math geeks among you. Do you think we would be justified in complaining about our bad luck seeing that out of 49 exposed cards there are six Jacks but only one Ten?
One might try to compute the probability that out of 49 cards we will get at least six Jacks and at most one Ten. Computer simulation says the chances are 0.61 per cent.
Not so fast. Note that I specified “Jacks versus Tens” after seeing the current game state, which is clearly unfair. Either we have to guess a pair of adjacent ranks (e.g. Fours vs Threes) or alternatively include all ranks. In the latter case we might ask “what is the probability that out of 49 cards we will get at least six repeats of X and at most one Y for some pair of adjacent ranks X and Y?” Remember that X can either be Y+1 or Y-1. In this case the probability is about 12.5%.
If you’re really nit-picky you might also ask “why 6X versus 1Y? Why not 7X vs 2Y etc”, but you get the gist.
There is also the issue of selective memory. We might have played eight games and we only remember the one game with way too many Jacks and only a solitary Ten. And by some strange coincidence, 12.5% happens to equal the fraction 1/8.
This is probably too much detailed mathematic specificity for the average Joe Bloggs, but the point I wish to make is don’t complain that the game is rigged unless you really know your statistics better than your alphabet.
Now, going back to the lesson …