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Pascal Meets Pickling

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Mr. Big Food and I put up 12 pints and three quarts of four different kinds of pickled things this afternoon. It took about four hours– we are experienced and efficient picklers. Music was playing in the background and we seldom run out of extraneous things to chat about, e.g., poor Randy. We guess he’s not even going to be playing at State Fairs any more. It’s one thing to be drunk, but quite another to be publicly drunk and publicly naked all at the same time. I have no sympathy.

That said, we’d sort of reached the end of the line. We fell one ring/lid rig short. 

Reusing Ball-Mason jars is fine because they are sterilized with boiling water, and although glass degrades, it does not degrade nearly as fast as the little rubber seals of the lids. So we use new rings and lids to increase the probability of a perfect seal. This afternoon, we had to reuse one ring/lid. I found one that looked to be in excellent shape and put it with the three others and four jars (doing a batch of four) in the sink to be sterilized. Mr. Big Food asked if I knew which was the crappy old ring/lid. I did not.

As we packed each quart jar with spices and cucumbers and filled it with brine and put the ring/lid on, we asked, what is the probability that we will pull the crappy old ring/lid out of the boiling water in the sink, as we pick one ring/lid at a time? And, if there are only enough sliced cucumbers to make three– rather than four– quart jars of dill pickles, what is the probability that the crappy old ring/lid will be left in the sink and not sealing a jar of dill pickles?

There are at least two ways to approach this problem. One– Mr. Big Food’s– is step-by-step.
The odds of picking the one crappy old ring out of four is clearly one-in-four. If it’s not picked on the first pick, then the odds are one-in three. Then one-in-two if we haven’t already picked the old ring/lid on the first or second picks. The key condition is if we haven’t already picked it. (And remembering we are talking about sampling without replacement.)

The probability of not picking the old ring/lid on the first pick is 3/4. The probability of picking one of three of 3/4 is 1/3 x 3/4 = 3/12. If we still don’t pick the old ring (2/3) the probability of picking it when there are only two left is 3/4 x 2/3 x 1/2 = 6/24 =1/4.

Or you could think of it this way. What’s the probability that the crappy old ring/lid will be the last ring/lid in the sink? (Without replacement.) I’d say one-in-four.  

One-in-four all the way down. Bayes be damned. (I think. I’m not schooled in Bayesian probability.)