How do you calculate odds?

[hardcore_gamer]

So I am suffering from a bit of a math problem.

Calculating odds is easy enough if you are only calculating the odds of a single thing happening, like say the odds of drawing an ace out of a deck. But what if I want to calculate the odds of drawing 2 aces out of a deck in a row before it actually happens?

Basically I am asking how to calculate the odds of many things happening in a row before any of them actually happen.

?

[Linguica]

Drawing one ace: 4/52

Drawing a second ace: 3/51

Drawing 2 aces in a row: (4/52) * (3/51) =~ 0.45%

[hardcore_gamer]

Linguica said:

Drawing one ace: 4/52

Drawing a second ace: 3/51

Drawing 2 aces in a row: (4/52) * (3/51) =~ 0.45%

What does the * symbol mean?

[Kontra Kommando]

hardcore_gamer said:

What does the * symbol mean?

multiply

[hardcore_gamer]

Kontra Kommando said:

multiply

Ah I see. But why is there a ~ symbol after the =?

I hate sounding like a school kid but I was never good at advanced math.

[40oz]

it basically means "about 0.45%"

the real number has like 20 some decimal places which would make a really long number to write and the difference in negligible.

[hardcore_gamer]

40oz said:

it basically means "about 0.45%"

the real number has like 20 some decimal places which would make a really long number to write and the difference in negligible.

Ah I see!

But if you draw the first card and then another person draws the second one and you don't know what card he took? Then it becomes impossible to calculate odds doesn't it?

I case you people haven't guess already I am trying to calculate the odds of drawing certain cards in a game of poker :D

[Maes]

Enter the concept of "conditional probability" and "branching".

The probability of you pulling the third ace right after another person B has pulled the second one are 2/50.

Person B has a 3/51 chance of pulling an ace, and you have a 2/50 probability of pulling another ace after B, conditional on B's pullin

P({Pull an Ace}|{B pulls 2nd ace before you})= (2/50)*(3/51).

Essentially no different than pulling two aces in a row, no matter who does the pulling. Easiest ace to pull, ofc, is the first one. Fourth is the hardest.

If you need to speculate on the other person's performance, then smart's money is on that he DIDN'T pull an ace (probabilities of 48/52, 48/51, 48/50 and 48/49, respectively.

So if you're playing for money, make sure that the payouts balance the odds of losing (e.g. betting against pulling the 1st ace needs paying odds better than 52/4, to be worthwhile).

[Doominator2]

Eughh, reminds me of the boring days in the 10th grade learning about probability.

[geekmarine]

Being obsessed with programming, those were some of my favorite days of math class. I loved anything that I could apply to programming and/or video games. I mean, maybe I was unique in terms of my take on math, but I almost instantly saw how plotting points on a graph correlated to pixels on a computer screen. Hell, that's the primary reason I loved physics, too - my physics teacher was horrible, but the moment I figured out the formula for gravity so that I could simulate it in a computer program, I went friggin' nuts.

[Bucket]

Everything has a 50/50 chance of happening. Either it does or it doesn't. Two total outcomes, one of which will happen.

Bam, give me my statistics degree.