Calling math guru's...

[SP33Demon]

Ok, I was playing some American roulette (has the 0 and 00's) the other day and thought of a possible good way to win, but I know that there's probably a catch and am having trouble figuring out some odds.

There are 38 numbers, and 1-36 is split into 3rd's at 2:1 odds. If you cover 2/3rd's of the numbers, that's 24/38 = 63.157%. At 2:1 odds, you will net a bet (say $5). Somehow I don't think that you would have a 63% chance of netting a bet, and that since the 0 and 00's are their own "sections" altogether that somehow screws up your odds of netting a bet. So it's probably not 63% at 2:1 odds, but what would your true chance of winning be?

 

[Pepsi90919]

42

 

[SP33Demon]

Originally posted by: Pepsi90919
42

Cool, could you post how you got that number? Also, I'd be curious to know what the highest % bet in roulette would be, but I can look that up.

 

[Eeezee]

Originally posted by: SP33Demon

Originally posted by: Pepsi90919
42

Cool, could you post how you got that number? Also, I'd be curious to know what the highest % bet in roulette would be, but I can look that up.

Google search for Douglas Adams

 

[Chaotic42]

Originally posted by: SP33Demon

Originally posted by: Pepsi90919
42

Cool, could you post how you got that number? Also, I'd be curious to know what the highest % bet in roulette would be, but I can look that up.

He's just posting a number from a book because he doesn't know how to calculate the odds.

 

[jonmullen]

well you see 42 is the answer to the universe so ofcourse it is the answer to your problem

 

[silverpig]

Originally posted by: SP33Demon
Ok, I was playing some American roulette (has the 0 and 00's) the other day and thought of a possible good way to win, but I know that there's probably a catch and am having trouble figuring out some odds.

There are 38 numbers, and 1-36 is split into 3rd's at 2:1 odds. If you cover 2/3rd's of the numbers, that's 24/38 = 63.157%. At 2:1 odds, you will net a bet (say $5). Somehow I don't think that you would have a 63% chance of netting a bet, and that since the 0 and 00's are their own "sections" altogether that somehow screws up your odds of netting a bet. So it's probably not 63% at 2:1 odds, but what would your true chance of winning be?

Naw you won't win. That's like covering all 3 thirds for 1:1 payout. Bet $20, lands on 13, you win your $20 back. Bet $20, land 00, you're out.

Going to thirds you cover 1-24 with $20. Bet, lands in the 2/3 you have, you win $30. You can expect to do this twice for every time it comes up 25-36, so again, you're no worse off. But then there's those pesky 0 and 00 where you'll lose.

 

[InverseOfNeo]

Black!

 

[chuckywang]

Originally posted by: SP33Demon
Ok, I was playing some American roulette (has the 0 and 00's) the other day and thought of a possible good way to win, but I know that there's probably a catch and am having trouble figuring out some odds.

There are 38 numbers, and 1-36 is split into 3rd's at 2:1 odds. If you cover 2/3rd's of the numbers, that's 24/38 = 63.157%. At 2:1 odds, you will net a bet (say $5). Somehow I don't think that you would have a 63% chance of netting a bet, and that since the 0 and 00's are their own "sections" altogether that somehow screws up your odds of netting a bet. So it's probably not 63% at 2:1 odds, but what would your true chance of winning be?

Your odds of winning will be 24/38. There are 24 numbers that will net you money, and the rest of the numbers will make you lose your money. Don't overthink it. If you're asking for the expected amount of money you will win if you make that bet, that's another question entirely.

 

[SP33Demon]

Originally posted by: chuckywang

Originally posted by: SP33Demon
Ok, I was playing some American roulette (has the 0 and 00's) the other day and thought of a possible good way to win, but I know that there's probably a catch and am having trouble figuring out some odds.

There are 38 numbers, and 1-36 is split into 3rd's at 2:1 odds. If you cover 2/3rd's of the numbers, that's 24/38 = 63.157%. At 2:1 odds, you will net a bet (say $5). Somehow I don't think that you would have a 63% chance of netting a bet, and that since the 0 and 00's are their own "sections" altogether that somehow screws up your odds of netting a bet. So it's probably not 63% at 2:1 odds, but what would your true chance of winning be?

Your odds of winning will be 24/38. There are 24 numbers that will net you money, and the rest of the numbers will make you lose your money. Don't overthink it. If you're asking for the expected amount of money you will win if you make that bet, that's another question entirely.

Ok, I've thought it out and here's what I've come up with:

24/38 gives you a 63% chance to net 100% profit in any given spin. (For 100 rolls, $315 profit)

2/38 gives you a 5.26% chance to lose 200% in any given spin. (For 100 rolls, $52.60 loss)

So if the bet is $5, and you play 100 times you should net $262.40 if the percentages played out perfectly over time. Am I missing any other variables to this equation?

 

[imported_Tick]

Originally posted by: SP33Demon

Originally posted by: chuckywang

Originally posted by: SP33Demon
Ok, I was playing some American roulette (has the 0 and 00's) the other day and thought of a possible good way to win, but I know that there's probably a catch and am having trouble figuring out some odds.

There are 38 numbers, and 1-36 is split into 3rd's at 2:1 odds. If you cover 2/3rd's of the numbers, that's 24/38 = 63.157%. At 2:1 odds, you will net a bet (say $5). Somehow I don't think that you would have a 63% chance of netting a bet, and that since the 0 and 00's are their own "sections" altogether that somehow screws up your odds of netting a bet. So it's probably not 63% at 2:1 odds, but what would your true chance of winning be?

Your odds of winning will be 24/38. There are 24 numbers that will net you money, and the rest of the numbers will make you lose your money. Don't overthink it. If you're asking for the expected amount of money you will win if you make that bet, that's another question entirely.

Ok, I've thought it out and here's what I've come up with:

24/38 gives you a 63% chance to net 100% profit in any given spin. (For 100 rolls, $315 profit)

2/38 gives you a 5.26% chance to lose 200% in any given spin. (For 100 rolls, $52.60 loss)

So if the bet is $5, and you play 100 times you should net $262.40 if the percentages played out perfectly over time. Am I missing any other variables to this equation?

Yes. Yet bet $10 one each third. Yet get back $20 on a win. 20=20.

 

[chuckywang]

Let's say you bet x on 1-12 and x on 13-24. With a 24/38 probability, you will end up with 3x, but with a 14/38 probability, you will end up with 0. Therefore, the expected amount of money you have after this bet is: 3x*24/38+0*14/38 = 1.895*x. Since you started out with 2x amount of money, you expect to lose 5.26% of your money with each bet.