# Probability that Scouting ashes continue to circulate?

I am intersecting here the oft-repeated example of Ceasar's dying breath (link) with a Scouting tradition (link) that dates back to at least the first World Jamboree, held 1920 in London, of spreading ashes from prior campfires on each new campfire, and collecting the ashes the next morning. It is absolutely unquestioned Scouting wisdom that every campfire carries the spirit, comraderie, and memories from every other campfire which shares the ashes tradition. But what about the physical matter?

With the following assumptions:

1. My old ashes and the ashes of other attendees were sprinkled on the fire just as it began to pick up,
2. The fire burned for several hours, being stoked and stirred occassionally,
3. I returned to the cold pile of ashes in the morning, and used my pocket knife to scoop some into my vial,
4. I have done this repeatedly, having at least once crossed paths with Scouts claiming unbroken chains back to the first World Jamboree.

What is the probabilty that my vial of ashes contains at least one molecule from the first World Jamboree campfire? Or even my prior campfire?

Can the Ceasar logic be applied? Is this a 'Kolmogorov zero-one law' example?

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What is a Kolmogorov example? – Did Oct 17 '11 at 18:13
@Didier Piau: en.wikipedia.org/wiki/Kolmogorov_zero-one_law (edited question accordingly). – cobaltduck Oct 17 '11 at 18:16

If you have an ounce of ashes, you have about $10^{24}$ atoms that you contribute to the fire. If at the end you withdraw $1\%$ of the ashes (say there are $6$ pounds of ashes and you take an ounce), there is a factor $100$ dilution each time. So after $12$ or so fires you run the risk of not having any of the original atoms left.
So if I am understanding you, a fire may have on the order $10^{26}$ atoms of ash at the end, and I remove about $10^{24}$ atoms. If I model this as a binomial with $n = 10^{24}$ and $p = .01$, then my expected number of atoms from the original fire is $10^{22}$. After two fires, this goes down to $10^{20}$ and so on, until by fire number 12, my expected value goes below $10^{0}$. Makes sense, if my interpretation is correct? – cobaltduck Oct 17 '11 at 20:24