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Roulette

 
 
Reply Wed 26 Jul, 2006 05:31 am
What is the chance to guess the correct number on which the ball stops on roulette? What is the math?
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Type: Discussion • Score: 1 • Views: 738 • Replies: 3
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fresco
 
  1  
Reply Wed 26 Jul, 2006 06:51 am
Roulette has 37 numbers (including zero).
Chances of guessing correct number =1/37.
Since the house pays only 35-1 on single numbers the odds are in their favour.
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Tryagain
 
  1  
Reply Wed 26 Jul, 2006 09:50 am
In most of the world, roulette wheels have 36 numbers (1 through 36)
and a 0 and 00 ("double zero"), so there are 38 outcomes. This is
certainly true in the United States, and I'm pretty sure it's true
of Canadian wheels as well. In the casino at Monte Carlo, however,
there are the same numbers, 1 through 36, and only a single 0.

Results for the US-type wheel.

Suppose you bet on black. 18 of the 36 numbers are red and the other
18 are black. The 0 and 00 are green. Thus, in the long term, all the
outcomes are equally likely, so of the 38 possible results, 18 will be
favorable. On average, you will win 18 times out of 38, or your
probability of winning is 18/38 = 9/19 - slightly less than half.

A typical two to one bet would be to bet on a group of 12 numbers
(there's a way to do this - bet on a column of numbers). Thus you will
win 12 times out of the 38 possible, for a probability of winning of
12/38.

You can work out any other probability the same way. Just figure out
how many of the results represent a win, and divide that by 38.

Now what's interesting is your "expected return". That turns out to be
the same, no matter what bet you make.

For example, if you're betting on black, and you bet one dollar every
time, on average, you'll win 18 times and lose 20. So for the 18
wins, you'll get back the 18 dollars you bet plus 18 dollars you won.
After 38 bets, on average, you will have bet 38 and gotten back 36 -
about a 5% loss. If you make the two to one bets, you'll win 12 times
out of 38. On each win, besides getting back your original bet, you
get 2 dollars extra, so for each of the 12 wins, on average, you get
3 dollars back for a grand total of 36 back for the 38 you bet. It's
the same payoff - 36 dollars back for every 38 you bet.

If you just bet on a single number and you win, you get 35 additional
dollars back, which is 36 total back if you include the original bet.
Of course this only happens once in 38 times, so again, you get about
36 back, on average, for every 38 you bet.

Hence every bet on the roulette wheel, on average, loses you about
5 percent. In fact, the loss, 2/38 = 5.2631579 percent. This is
called the "house advantage".

You can do the same calculations for the wheels in Monte Carlo - the
payoffs are the same, but there are only 37 possibilities. Which is why the house advantage at Monte Carlo is only 2.7027 percent.
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misogi
 
  1  
Reply Wed 23 Aug, 2006 09:31 am
37:1 if you're including 0 and 00

There are only 38 slots and numbers on the wheel...
I don't see where math comes in. It's obvious.
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