# probability in roulette!

So I have read how to play roulette...still a little confused, and now I'm faced with a probability question about it which makes the problem a little harder. Please help me reason this where possible.

Let's say you're in a debt of $500. You have$250 and you're going to use that to try to pay back your debt. What are the advantages of:

1. Betting all on black

So if I bet all on black, I have a 0.5 chance of winning \$250 (and then having$500). So simply advantages are quick return if I win...and just being bankrupt otherwise. Probably a risk not taking.

1. Betting $50 on black 5 times (and if you haven't won, then you're just bankrupt). Would this be an expectation question? I can make X~Bern(p). Would I have to use any conditional cases here? 1. Betting in increments of$x until you make it all back or you go bankrupt.
• The best thing you can is not to play. – Emanuele Paolini Jun 12 '14 at 6:06
• On the basis of statistics you will always lose. So choose 1. and save some time! – draks ... Jun 12 '14 at 6:17
• If I were a player, I'd say the same. But I'm trying to figure out how to best quantify this reasoning behind these choices are. Thanks – sofreakinlost Jun 12 '14 at 6:22
• Note that the chance that black occurs is less than one half, since you need to incorporate the green tile. – Marc Jun 12 '14 at 6:23
• surely that law of large numbers would make both options equal chances? – martin Jun 12 '14 at 7:45

If you really want to pay off your debt and insist on gambling to do so, choice 1) is your best bet. There's something called the house edge, which basically says that if you play the smartest way possible (which doesn't really come into play with roulette...), then the house, on average, is almost guaranteed to make a certain percentage of your money every time you spin the wheel.

Because of this, option 1) is the most likely to let you walk away with the most money possible. Choices 2) and 3) lead very naturally to a well-studied idea called the gambler's ruin, which is the mathematics behind why making sequences of bets, when the house has an edge, will eventually make you go broke.

To think about option 1) where you actually have a 50% chance of winning (0 and 00 don't exist in this world) you indeed have a 50% chance of walking away with $500. If you try to bet twice, you only have a 25% chance of winning both rounds, 25% chance of going bust, and a 50% chance of walking away even. Even if the bets are smaller, you can see that this doesn't compare favorably to a 50% chance of winning just once. The advantages of choices 2) and 3) would be free drinks and all the fun you can handle! Until you go broke, that is... Your roulette odds are fair (contrary to Vegas where they include extra 0 and 00 in favor of the house). So if you bet once you have 50% chance to win. If you bet by increment of$50 you still have 50% chance to win. If we look at the expected value for your bets it will look like this:

-------|-----|------------
|-50  | 50
P(X=x) |0.5  |0.5
-------|-----|------------


so -50*0.5+50*0.5=0 that means that however you play you will end up with no change.

However, if we look at Vegas odds, it goes like this: there 38 spots (18 black, 18 red and 2 green). Betting all on red or black gives you 18/38=0.473% chance of winning.

if we go on betting by chunks of 50:

-------|-------|-------|-----
|-50    | 50    |
P(X=x) |0.526  |0.473  |
-------|-------|-------|-----


your expected value is -2.65 which means, you will lose on average \$2.56 on each bet. This is a very simplicity approach without looking into Markov's Chain for the Gambler's Ruin.