orresponding probability of winning is (win) = p (no 0 or 00) X p (number selected) : (1 2/38) X (1/36) = 1/38 = 0.0263, and the probability of losing is p (lose) = 1 p (win). This is a two-event source See B. Osgood, Mathematics of the Information Age (2004), p. 64, at www-ee.stanford.edu/~osgood/ ophomore%20College/Math%20of%20Info.htm. See interactive example, while safely taking bets with "free money" at (for instance) www.mondo- casinos.com/gratuit/roulette/index.php (note: this reference is for study purposes and does not constitute in any way a recommendation of gambling). If the roulette outcome is strictly random (can one always be sure of this?), the odds on the bank to pocket all the bets with the 0 or 00 outcomes are 1/37 = 2.7% (French) and 2/38 = 5.4% (American). For the individual player, the odds on winning any single-number bet are (1 - 1/37)/36 = 1/36 : 2.7% (French) and (1 - 2/38)/36 = 1/38 = 2.6%. This means that (with respect to this player) the bank wins with probabilities 97.3% (French) and 97.4% (American). The odds on winning are obviously greater with the other bets (a)-(c), namely (1 - 1/37)/2 = 48.6% (French) and (1 - 2/38)/2 = 47.3% (American) for (a)-(c) and (1 - 1/37)/3 = 32.4% (French) and (1 - 2/38)/3 = 31.5% (American) for (d), but we note that they are lower than 50% (a)-(c) or 33% (d), which, on average, is always in favor of the bank. We note that in the real game, the odds on winning are less than 1/36, which makes the player's situation "unfair." It is possible to play with a high probability of making zero gains, but this comes with a low probability of making maximal loss. Indeed, if we place one token on each of the 36 numbers, two events can happen: (a) One of the 1-36 numbers comes out: we win 36 tokens, which is the exact amount we bet, and the net gain is zero; this has a high probability of 36/38 = 94.75 % < 100%. (b) The 0 or 00 comes out, the bank wins the 36 tokens (our loss is maximum); this has a low probability of 2/38 = 5.25% > 0%. Similar conclusions apply wh