You need: 20 dried beans (food), about 40 coins or buttons (animals), a plate (the habitat), paper and pencil.
Rules: each animal needs 2 beans to survive a year. Every animal that survives has 1 baby (so survivors double). Animals that can't get 2 beans don't survive.
- Put 20 beans on the plate. Start with 2 animals.
- Feed the animals 2 beans each. Count the survivors. Double them for the next year. Put all 20 beans back on the plate.
- Record in a table: Year, animals at start, animals that survived, animals for next year.
- Play 6 years. Make a line graph of the survivors.
- Harsh winter: for years 7 and 8, only 12 beans are available. What happens?
Expected results: survivors 2, 4, 8, 10, 10, 10 (from year 4 on, 16 or 20 animals arrive but only 10 can get 2 beans). With 12 beans, only 6 can survive.
Discuss: What is the carrying capacity with 20 beans? With 12 beans? Which limiting factor did the harsh winter change, and is it density-dependent or density-independent?
Safety: keep beans and coins away from very young children (choking hazard).
What to look for: the child finds a carrying capacity of 10 animals (20 ÷ 2), then 6 (12 ÷ 2), draws a graph that rises and levels off, and explains that the winter is a density-independent factor that lowered the carrying capacity.