Russian roulette Suppose that you are forced to play Russian roulette, but that you have the option to pay to remove one bullet from the loaded gun before pulling the trigger. Would you pay more to reduce the number of bullets in the cylinder from four to three or from one to zero? According to Kahneman and Tversky, if you are like most people, you would pay more to reduce the number from one to zero than from four to three. Why? Reducing the number of bullets from four to three would reduce the probability of dying from 4/6 to 3/6. In this range, the probability-weighting function is relatively flat, meaning fairly unresponsive to changes in the underlying probability. Reducing the number of bullets from one to zero would reduce the probability of dying from 1/6 to 0. Here, there is a jump from π(1/6) > 1/6 to π(0) = 0. Thus, the value to you of reducing the number of bullets from one to zero exceeds that of reducing it from four to three.