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WSOPwinner10
05-12-2005, 12:50 AM
1st Question:
This question is stupid. Let's say you are a 1/6 shot to pull a red marble out of a bag of 5 blue marbles and 1 red marble. What are the odds you pull at least one red marble out if you get two tries. So, you draw the first try. IF you get a red marble, you don't need to do the second trial - you already got your red marble. If you don't draw a red marble, you replace your blue marble into the bag and redraw. Are the chances you draw a red marble if you draw two times just 2/6 = 1/3?

2nd Question:
This is a question I would love to have answered. I would love the formula if possible also. Let's say you have a quarter. The most probable outcome, if you flip it ten times, is 5 heads and 5 tails. However, what are the odds that this actually happens.

A similar situation: You have a quarter. You flip it 100 times. What are the odds you land 60 heads and 40 tails.

I hope I made that question clear, as that problem has been bugging me for a while.

Thanks in advanced

BruceZ
05-12-2005, 01:23 AM
[ QUOTE ]
1st Question:
This question is stupid. Let's say you are a 1/6 shot to pull a red marble out of a bag of 5 blue marbles and 1 red marble. What are the odds you pull at least one red marble out if you get two tries. So, you draw the first try. IF you get a red marble, you don't need to do the second trial - you already got your red marble. If you don't draw a red marble, you replace your blue marble into the bag and redraw. Are the chances you draw a red marble if you draw two times just 2/6 = 1/3?

[/ QUOTE ]

No. 1/6 + (5/6)*(1/6) =~ 30.6%

Or 1 - (5/6)*(5/6) =~ 30.6%.

[ QUOTE ]
2nd Question:
This is a question I would love to have answered. I would love the formula if possible also. Let's say you have a quarter. The most probable outcome, if you flip it ten times, is 5 heads and 5 tails. However, what are the odds that this actually happens.

[/ QUOTE ]

C(10,5)*(1/2)^5*(1/2)^5 =~ 24.6%.

C(10,5) is the number of ways to pick the 5 heads, and is equal to 10*9*8*7*6/(5*4*3*2*1).

[ QUOTE ]
A similar situation: You have a quarter. You flip it 100 times. What are the odds you land 60 heads and 40 tails

[/ QUOTE ]

C(100,60)*(1/2)^60*(1/2)^40 =~ 1.1%

WSOPwinner10
05-12-2005, 01:41 AM
[ QUOTE ]
2nd Question:
This is a question I would love to have answered. I would love the formula if possible also. Let's say you have a quarter. The most probable outcome, if you flip it ten times, is 5 heads and 5 tails. However, what are the odds that this actually happens.

[/ QUOTE ]

C(10,5)*(1/2)^5*(1/2)^5 =~ 24.6%.

C(10,5) is the number of ways to pick the 5 heads, and is equal to 10*9*8*7*6/(5*4*3*2*1).

[ QUOTE ]
A similar situation: You have a quarter. You flip it 100 times. What are the odds you land 60 heads and 40 tails

[/ QUOTE ]

C(100,60)*(1/2)^60*(1/2)^40 =~ 1.1%

[/ QUOTE ]

the responses to both questions were exactly what I was looking for. Thanks a lot. I am not sure if I understand this whole C(10,5) concept. It appears to have something to do with factorials. Is there a button for "C" on a calculator. I am a little unclear of what "C" actually is.

Regardless, good response. Thanks a lot

BruceZ
05-12-2005, 01:45 AM
[ QUOTE ]
the responses to both questions were exactly what I was looking for. Thanks a lot. I am not sure if I understand this whole C(10,5) concept. It appears to have something to do with factorials. Is there a button for "C" on a calculator. I am a little unclear of what "C" actually is.

[/ QUOTE ]

Number of combinations of 5 things chosen out of 10 things or "10 choose 5". It is equal to 10!/5!/(10-5)! You may have a C(x,y) button on your calculator, or !, or you can type "10 choose 5" into google (without the quotes), or use the COMBIN function in Excel.

WSOPwinner10
05-12-2005, 01:48 AM
I understand now. Thanks for the tip.

irchans
05-13-2005, 04:21 AM
It's funny how Pi can show up in a formula that has nothing to do with geometry. If you flip a coin 2*n times, the probability of getting n heads and n tails is approximately

1/Sqrt[Pi*n].

(A better approximation that is not as nice is Exp[-1/(8*n)] * 1/Sqrt[Pi*n].)