i really should know how to solve this my self but for some reason i dont ....so : if you toss a coin there is a probability of 1/2 to get either side of the coin if you define one side of the coin as the winning side and you get two coin tosses and getting one coin to land on the winning side is enough to win the game you will have a probability of 4/1 to win because the likelihood of getting both tosses wrong is 1/4. so far this is simple but now imagine a three sided coin , still only one side is the winning side and getting the coin to land on the winning side once is enough to win the game but this time you get three tries. intuitively the likelihood of winning the game should be the same as with the two sided coin and two tries since, if you play the game infinitely often you will still get on average one coin to land on the winning side per game. i know the probability of winning the two games is not the same but i cant really explain why except by just calculating it... can you help me to understand this on a more intuitive level ?

excuse my bad english , its not my native language

## confused by simple probability ....

**Moderators:** gmalivuk, Moderators General, Prelates

### Re: confused by simple probability ....

Just a really quick response but you might gain by looking into the binomial distribution.

double epsilon = -.0000001;

### Re: confused by simple probability ....

i think binominal distribution doesn't apply here because the game is won not only if you get exactely one coin right but also if you get more than one coin right

- phlip
- Restorer of Worlds
**Posts:**7573**Joined:**Sat Sep 23, 2006 3:56 am UTC**Location:**Australia-
**Contact:**

### Re: confused by simple probability ....

So, you have a 3-sided die, one side marked "Win", the other two marked "Lose". Then you roll three of these, and if at least one comes up "Win" then you win, but if all three come up "Lose" then you lose. Am I reading this correctly?

Well, some hints:

If you roll one die, what's the chance of it coming up "Lose"?

If you roll three dice, what's the chance they all come up "Lose"?

If you roll three dice, what's the chance they don't all come up "Lose"?

Well, some hints:

If you roll one die, what's the chance of it coming up "Lose"?

If you roll three dice, what's the chance they all come up "Lose"?

If you roll three dice, what's the chance they don't all come up "Lose"?

Code: Select all

`enum ಠ_ಠ {°□°╰=1, °Д°╰, ಠ益ಠ╰};`

void ┻━┻︵╰(ಠ_ಠ ⚠) {exit((int)⚠);}

### Re: confused by simple probability ....

phlip wrote:If you roll one die, what's the chance of it coming up "Lose"?

If you roll three dice, what's the chance they all come up "Lose"?

If you roll three dice, what's the chance they don't all come up "Lose"?

if i role one die the probability of it coming up "lose" is 2/3

if i role three dice the probability that thy all coe mup "loose" is (2/3)³ or 0.296296296296....

if i role three dice the chance they don't all come up "lose" is 1-0.296296296296....

but still its not intuitive to me for some reason ... i fell so stupid right now ...

### Re: confused by simple probability ....

brötchen wrote:i think binominal distribution doesn't apply here because the game is won not only if you get exactely one coin right but also if you get more than one coin right

Instead of defining the successful outcome as winning, define it as losing. Then, with n = 3, p = 2/3 we have a binomial distribution that describes the number of loses you get. When X = 3 you have 3C3 * (2/3)^3 * (1/3)^0 = (2/3)^3 = the chance of getting three loses which equals the chance of not winning the game.

In fact, you don't even need to do that: with n= 3, p = 1/3 we have a binomial distribution that describes the number of wins you get. When X = 0 you have (the same calculation as above) which is the chance of not winning any of the three times, which is the chance of not winning the game.

Even if the winning condition was something weird, like 'with p = 1/3, n = 3, win two or three times' then you can solve it with a binomial distribution - just sum the chances of winning twice and winning three times, or sum the chances of winning zero and one times then subtract 1 from that. (There's no closed form for this kind of thing, as far as I know.)

### Re: confused by simple probability ....

Patashu wrote:When X = 3 you have 3C3 * (2/3)^3 * (1/3)^0 = (2/3)^3 = the chance of getting three loses

why are you defining x ? you dont use x anywhere in the formula. and what is C ? or is that a typo and C is actually suposed to be X ? even than i dont see why you would define and use a variable when you didnt show the general formula before hand what am i missing ?

### Re: confused by simple probability ....

In fewer words, the probability of winning is 1 minus the probability of losing.

It is easier in this case to compute the probability of losing n flips. If all it takes is one successful win over the course of n flips, then to compute that probability, all you need to do is compute the probability that each of the n flips is a loss. Then, subtract that number from 1.

It is easier in this case to compute the probability of losing n flips. If all it takes is one successful win over the course of n flips, then to compute that probability, all you need to do is compute the probability that each of the n flips is a loss. Then, subtract that number from 1.

### Re: confused by simple probability ....

brötchen wrote:Patashu wrote:When X = 3 you have 3C3 * (2/3)^3 * (1/3)^0 = (2/3)^3 = the chance of getting three loses

why are you defining x ? you dont use x anywhere in the formula. and what is C ? or is that a typo and C is actually suposed to be X ? even than i dont see why you would define and use a variable when you didnt show the general formula before hand what am i missing ?

X is the random variable he was talking about. He was saying to find the probability of X being 3 we get (insert what he typed). C isn't by itself. One notation for choose is nCk which means n choose k which is n!/(k!(n-k)!). So the 3C3 means 3 choose 3 which is 3!/(3!(3-3)!) = 1

Have you looked into the binomial distribution?

double epsilon = -.0000001;

### Re: confused by simple probability ....

brötchen wrote:Patashu wrote:When X = 3 you have 3C3 * (2/3)^3 * (1/3)^0 = (2/3)^3 = the chance of getting three loses

why are you defining x ? you dont use x anywhere in the formula. and what is C ? or is that a typo and C is actually suposed to be X ? even than i dont see why you would define and use a variable when you didnt show the general formula before hand what am i missing ?

X is a random variable that varies as a binomial distribution with n = 3 and p = 2/3. (I should have said P(X = 3) = that stuff on the right, actually.) Read about the binomial distribution on wikipedia: http://en.wikipedia.org/wiki/binomial_disiribution

### Who is online

Users browsing this forum: No registered users and 15 guests