Math question - odds

[byosys]

A friend and I were debating (read: arguing) over the following:

You are given a 1/9 chance of "winning" (what is irrevelant). What are the chance that you "win" atleast 3 of the first 4 matches?

My orginal logic was its (1/9)^3 in that you can win any three matches, then the fourth is irrevelant (because you have already satasfied the conditions) and therefore the odds on the last match is 1/1. Since you can move that 1/1 around (ie 1/9*1/1*1/9*1/9 etc), then the chances are still (1/9)^3. However, a friend pointed out that the chances of winning the first 3 matches is also (1/9)^3, so I obviously can't be right.

His best guess is .005 or 1/199 if you don't round. I can't give you the logic behind that one as I didn't think of it. The math he gave me was:
((1 / 9)^4) + (4 * ((1 / 9)^3) * (8 / 9))

Thanks all.

BTW for the record this isn't HW. I'm in linear algebra and never really understood odds with any accuracy.

 

[Syringer]

At least 3 of 4 the more means winning 3 or 4 of the matches.

What you need to use is the binomial theorem:

(4 3)*(1/9)^3*(8/9) + (4 4)*(1/9)^4)

(4 3) indicates 4 choose 3, which is 4.

 

[notfred]

Chance that you win 0 matches:
8/9 ^ 4 (you have an 8/9 chance of losing, and have to do it 4 times in a row)

Chance that you win 1 match:
1/9 * 8/9 * 8/9 * 8/9 <--- that's the chance that you win only the first match.
8/9 * 1/9 * 8/9 * 8/9 <--- that's the chance that you win only the second match.
etc. for the third and fourth.

The chances of any one of those 4 things happening are computed by jsut adding them all up, so you get:
4(1/9 * 8/9^3)
As the chance of winning exactly once.

By the same method, you get :
4(1/9^2 * 8/9^2)
As the chance of winning exactly twice.

And:
4(1/9^3 * 8/9)
As the chance of winning exactly 3 times.

Chances that you win all 4 are:
1/9^4

So you, want at least 3 times? You add up the calculations for 3 and 4:
4(1/9^3 * 8/9) + 1/9^4 = 0.00502972108

 

[chuckywang]

Originally posted by: Syringer
At least 3 of 4 the more means winning 3 or 4 of the matches.

What you need to use is the binomial theorem:

(4 3)*(1/9)^3*(8/9) + (4 4)*(1/9)^4)

(4 3) indicates 4 choose 3, which is 4.

^The man speaks the truth.

 

[esun]

Here's an easy way to think about it. Consider we are rolling a 9-sided die four times, and want a specific number to come up at least 3 times. How many ways can that happen? There are 4 choose 3 such ways. How many ways total can the 4 rolls of a 9-sided die come up? That's 9^4. Therefore, by my calculation, we have 4 choose 3 ways to win out of 9^4 possible rolls, so the answer is (4 3) / 9^4, where I am using Syringer's notation for "choose".

 

[OdiN]

50% Either it will happen or it won't.

 

[altonb1]

Originally posted by: byosys
A friend and I were debating (read: arguing) over the following:

You are given a 1/9 chance of "winning" (what is irrevelant). What are the chance that you "win" atleast 3 of the first 4 matches?

My orginal logic was its (1/9)^3 in that you can win any three matches, then the fourth is irrevelant (because you have already satasfied the conditions) and therefore the odds on the last match is 1/1. Since you can move that 1/1 around (ie 1/9*1/1*1/9*1/9 etc), then the chances are still (1/9)^3. However, a friend pointed out that the chances of winning the first 3 matches is also (1/9)^3, so I obviously can't be right.

His best guess is .005 or 1/199 if you don't round. I can't give you the logic behind that one as I didn't think of it. The math he gave me was:
((1 / 9)^4) + (4 * ((1 / 9)^3) * (8 / 9))

Thanks all.

BTW for the record this isn't HW. I'm in linear algebra and never really understood odds with any accuracy.

So in normal conversation, when you have nothing better to do, THESE are the the types of things you debate/argue over?

I think the obvious answer is you need to get out more, find a girl, and get laid. The current odds are that none of my suggetions are currently being realized.

 

[Schfifty Five]

Originally posted by: OdiN
50% Either it will happen or it won't.

The village idiot everyone.

 

[byosys]

So in normal conversation, when you have nothing better to do, THESE are the the types of things you debate/argue over?

I think the obvious answer is you need to get out more, find a girl, and get laid. The current odds are that none of my suggetions are currently being realized.

No. He was going on about how he "figured out a pattern" for the color of the flop in some online poker game where you could place side bets on the color of the flop. He won 3/4 times he bet and we wanted to figure out the odds of that happening. He orginally said something like 1/142 or thereabouts, which is obviously really low and we both realized that. And for the record, were both semi geeks and have gfs. So stop trolling.

Anyway...Thanks for the help guys and girls. I complealtly forgot about that n choose m thing. As for the rest...well I'm still trying to decifer it and fully understand it. Odds are just one of thoes things that I'll never get. Give me integrals or differentials over odds any day.