Simple Probability Question (that I have no idea how to answer)
 

In playing a set of four 10 person sit-and-gos, what is the probability of placing first in three, or at least three (10% chance of placing first)? How many sets of four would you have to play to have a 50% chance of doing this twice? Cross-posting in probabilities. Thanks, ThorGoT (who hasn't taken any math classes for 15 years).

Re: Simple Probability Question (that I have no idea how to answer)
 

I posted this in the 1-table tournaments also:

I think I have the answer. You have a 0.37% chance of winning a tournament at least 3 out of 4 times. This is assuming that you are exactly an average player, which means that you'll finish first exactly 10% of the time.

There are 10,000 different combinations of placing in 4 different tournaments: 10 X 10 X 10 X 10 = 10000. Your scenario has 37 different possibilities:

1 1 1 1

1 1 1 2
1 1 1 3
1 1 1 4
1 1 1 5
1 1 1 6
1 1 1 7
1 1 1 8
1 1 1 9
1 1 1 10

1 1 2 1
1 1 3 1
1 1 4 1
1 1 5 1
1 1 6 1
1 1 7 1
1 1 8 1
1 1 9 1
1 1 10 1

1 2 1 1
1 3 1 1
1 4 1 1
1 5 1 1
1 6 1 1
1 7 1 1
1 8 1 1
1 9 1 1
1 10 1 1

2 1 1 1
3 1 1 1
4 1 1 1
5 1 1 1
6 1 1 1
7 1 1 1
8 1 1 1
9 1 1 1
10 1 1 1

37/10000 = 0.37%.

So if you played 10000 tournaments, you would on average find 37 instances of when you won at least 3 out of 4 tournaments in a row (actually 10003 tournaments. Your first set of four would be tournaments # 1, 2, 3, and 4. Your second set of four would be tournaments # 2, 3, 4, and 5. Your last set of tournaments would be tournaments #10000, 10001, 10002, and 10003)

I just ran a simulation of 10003 tournaments. I won 1032 times, and 37 times I won at least 3 out of 4 tournaments. So there you go.

On average, this result should happen once every 270 sets of four tournaments (100%/0.37%). So if you played 270 sets of four, it should have happened once. There's a 50% chance that it would have happened twice.

Of course, the assumption that you are an average player is hopefully wrong. So it would happen a little more often for you.

Re: Simple Probability Question (that I have no idea how to answer)
 

[ QUOTE ]
On average, this result should happen once every 270 sets of four tournaments (100%/0.37%). So if you played 270 sets of four, it should have happened once. There's a 50% chance that it would have happened twice.


[/ QUOTE ]

This is not correct, though I agree with the 0.37%.

(0.1)^4 + 4*0.9*(0.1)^3 = 0.37%

If the probability of it occuring at least twice in N trials is 0.5, then the probability of it occuring 0 or 1 time in N trials is 0.5.

(1-0.0037)^N + N*(0.0037)*(1-.0.0037)^(N-1) = 0.5

Solving this for N gives N = 453, so we must play 453 sets of four tournaments. I used the Excel goal seek function to find this solution, or you can use trial and error. If you set N = 270, there will only be a 26.4% chance of it happening twice.

Re: Simple Probability Question (that I have no idea how to answer)
 

Thanks to both of you. I remember shortly after taking my last math class, being surprised at how people couldn't figure stuff out that was still fresh and new for me . . . now I am that person.