Probabilty question on roulette for class-Need help ASAP

Here is the problem I have.
In the game of roulette, the ball is rolled 25 times, find the probabilities of the following events.
a. The ball falls into the green two or more times
b. The ball does not fall onto any green slots
c. The ball falls into black slots 15 or more times
d. The ball falls into red slots 10 or fewer times.

I assume you know there are 18 red, 18 black, and 2 green slots. Thanks a ton!

Hey NHQ,

I don't really feel right doing your homework for you, but I don't mind giving you some help.

You're probably studying the Binomial Distribution, right?

In an experiment consisting of n independent trials in which an event has a probability p in a single trial, the probability P(x) of obtaining x successes is:
P(x) = C(n,x)*(p^x)*((1-p)^(n-x))
where C(n,x) = n! / (x!(n-x)!)

Hopefully this isn't greek to you. The calculations are straight forward but can be tedious (especially for parts (c) and (d)).

To find p for each event, take the number of appropriate color slots and divide by the total number of slots. For example, if you were looking at the chance of landing in a Christmas color (red or green), you'd calculate that p = (18 red) + (2 green) / (38 total) = 20/38.

To make the calculations less complicated, you can often figure out chance of the opposite of what you want, and subtract that result from 100%. For example, imagine you're going to get either $0, $10, or $20 for Christmas. If I told you that there was a 30% of getting $0 and a 20% of getting $10, then you'd be able to figure out that you have a 100%-30%-20% = 50% chance of getting $20.


a) To find P(x>=2), you'd normally have to add P(2) + P(3) + P(4) + .... + P(24) + P(25). That's a lot of adding. You can use the subtration trick to reduce the calculation you need to do: P(x>=2) = 1 - P(x<2) = 1 - P(0) - P(1).

b) Haven't we already done this?

c) Yuck. I don't think we can cut any corners here. Just calculate P(x>=15) = P(15) + P(16) + ... + P(14) + P(25). I don't suppose your teacher gave you a table of all of the P(x) values?

d) Ditto. P(x<=10) = P(0) + P(1) + ... + P(9) + P(10)


Hope this helped some,
PP

[BruceZ]

I thought of a shortcut to c and d where you only need to compute 2 terms. It is an approximation, but it should be very close. You should have 13 or more approximately 50% of the time, and 12 or less approximately 50% of the time. So if you want the probability of 10 or less, compute the probability of exactly 11 and 12, and subtract this from 50%. Similarly for 15 or more, compute the probability of exactly 13 and 14 and subtract this from 50%. Perhaps this is why the problem picked the numbers it did. It would be exact if the probability of red or black was exactly 50%. There is also an approximate formula for the binomial "tail" which you may have been given.

If you need an exact answer you could a) compute 11 terms 0-10 by hand and add them together for d, then compute 4 more terms 11-14 and subtract terms 0-14 from 1 for part c b) write a program or spreadsheet to do this c) use a table of the binomial "tail" if you have one d) use a table of the binomial disribution if you have one that applies to this case, and add the right numbers togther from the table.

[BruceZ]

I had a couple minutes while some coffee was brewing, so I ran this in excel. A couple minutes is all it takes since excel has the function COMBIN for doing combinations. I won't give the result, but I will say that you should not use my suggested approximation as it doesn't give good results. The problem is that when you change the probability from 1/2 to 18/38, the split between <=12 and >=13 is not close to 50-50 anymore, it is about 60-40.

[Homer]

I also posted an answer under the HoldEm board... but it is not nearly as good as that given by the above poster. Good luck.

[lorinda]

As an aside, it's a good job you clarified which type of roulette you were using, We only have one green in England, and I am led to believe that more and more tables in America are going the same way [img]/forums/images/icons/laugh.gif[/img]

[baggins]

they're crazy if they do. 2 greens instead of one is a considerable house edge.

[lorinda]

Competitive markets,
one green is still plenty to bust the customer [img]/forums/images/icons/laugh.gif[/img]

[BruceZ]

I understand that where both are offered, many players still play 2 greens because that is what they are familiar with! Of course if you play roulette you are not the brightest bulb on the porch anyway (unless you play biases or use physics).

[PseudoPserious]

There was a book about a team of physicists who made a computer that predicted the octant that a roulette ball would land in, based on wheel spin, tilt, how hard the guy shot the ball in, etc. It was a pretty good read, although I can't for the life of me remember the title ("The Ever-Present Daemons?" something like that).

PP