How many throws of a fair are required to have an even chance of at least one six?
Let’s break down the problem. 1) The probability of at least one six in n throws is (1 – the probability of getting no sixes in n throws). It is 1 – (5/6)n. 2) Even chance means 0.5. By combining the two pieces of information:
1 – (5/6)n = 0.5
Taking logs on both sides,
n ln (5/6) = ln(1/2)
n = ln(1/2)/ln (5/6) = 3.8
So, four throws.
