Simple Math
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August 11th, 2014 at 7:19:24 PM
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If you have 50 ping pong balls in a bag, half
are white and half are black, what percentage
of the time will you blindly pull out 2 of the
same color.
Why can't I figure this out.
are white and half are black, what percentage
of the time will you blindly pull out 2 of the
same color.
Why can't I figure this out.
"It's not called gambling if the math is on your side."
August 11th, 2014 at 7:34:11 PM
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Quote: EvenBobIf you have 50 ping pong balls in a bag, half
are white and half are black, what percentage
of the time will you blindly pull out 2 of the
same color.
Why can't I figure this out.
Edit: thought it was 50 of each color.
24/49
August 11th, 2014 at 7:37:29 PM
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Quote: AxiomOfChoiceEdit: thought it was 50 of each color.
24/49
Not 25/50?
"It's not called gambling if the math is on your side."
August 11th, 2014 at 7:40:28 PM
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Zero thankfully because both white and black are not colors.
I am a robot.
August 11th, 2014 at 7:42:08 PM
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Quote: EvenBobNot 25/50?
your first ball is going to be either black or white. If you pull a white ball then there are 24 white balls left out of 49.
Expect the worst and you will never be disappointed.
I AM NOT PART OF GWAE RADIO SHOW
August 11th, 2014 at 7:42:40 PM
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Quote: onenickelmiracleZero thankfully because both white and black are not colors.
bazinga
Expect the worst and you will never be disappointed.
I AM NOT PART OF GWAE RADIO SHOW
August 11th, 2014 at 7:43:19 PM
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Quote: onenickelmiracleZero thankfully because both white and black are not colors.
When talking about pigments they are, look it up.
"It's not called gambling if the math is on your side."
August 11th, 2014 at 7:43:57 PM
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Recognizing the substantial number of people who visit this site looking for loopholes in the rules and such, I thought I should point out that there is a trick way to increase this percentage. Bob said, "... what percentage of the time will you blindly pull out 2 of the same color." All you have to do to increase the percentage is to blindly pull out a total number of balls that is greater than 2. Of course, many would consider that cheating or at least outside the allowed parameters, but it would indeed increase the percentage.
Assuming that only two balls are drawn and that they are drawn randomly, I agree with AxiomOfChoice. No matter what color ball is drawn first, there remain 24 ways to draw a ball of the same color, out of the 49 ways to draw a second ball.
Assuming that only two balls are drawn and that they are drawn randomly, I agree with AxiomOfChoice. No matter what color ball is drawn first, there remain 24 ways to draw a ball of the same color, out of the 49 ways to draw a second ball.
August 11th, 2014 at 7:44:47 PM
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Quote: EvenBobNot 25/50?
I assume that you are drawing without replacement. Is that incorrect?
ie, you pull 2 balls out at the same time, or pull one out, then another, without putting the first one back.
If you pull one out, put it back, then pull another (so you have a chance of pulling the same ball twice) then it's 25/50.
August 11th, 2014 at 7:46:35 PM
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probability of selecting first white ball = 25/50
probability of selecting second white ball = 24/49
probability of selecting two consecutive white balls = 25/50 * 24/49 = .244898
probability of selecting first black ball = 25/50
probability of selecting first black ball = 24/49
probability of selecting two consecutive black balls = 25/50 * 24/49 = .244898
probability of selecting two consecutive white or black balls = .244898 + .244898 = .489796
probability of selecting second white ball = 24/49
probability of selecting two consecutive white balls = 25/50 * 24/49 = .244898
probability of selecting first black ball = 25/50
probability of selecting first black ball = 24/49
probability of selecting two consecutive black balls = 25/50 * 24/49 = .244898
probability of selecting two consecutive white or black balls = .244898 + .244898 = .489796
August 11th, 2014 at 7:54:12 PM
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Try it another way. 50 b/w balls, withdraw 2 at once,
what are the chances of one them bring white.
what are the chances of one them bring white.
"It's not called gambling if the math is on your side."
August 11th, 2014 at 8:00:15 PM
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Quote: EvenBob... what are the chances of one them bring white.
Before people start arguing semantics or conflicting answers, you should clarify whether you mean "exactly one" or "at least one".
August 11th, 2014 at 8:06:52 PM
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Quote: DocBefore people start arguing semantics or conflicting answers, you should clarify whether you mean "exactly one" or "at least one".
At least one.
"It's not called gambling if the math is on your side."
August 11th, 2014 at 8:44:35 PM
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Quote: EvenBobAt least one.
P(at least one white) = 1- P(both black) = 1 - P(first is black)*P(second black given first is black) = 1 - (25/50)*(24/49)
August 11th, 2014 at 8:48:23 PM
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i.e., 37/49 ≈ 75.510%
August 11th, 2014 at 10:55:45 PM
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Quote: Doci.e., 37/49 ≈ 75.510%
I suspected it was 75% but didn't know the
math to arrive at it. Thanks.
"It's not called gambling if the math is on your side."
August 12th, 2014 at 10:38:30 AM
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Well, actually, it's a little more than 75%. The difference between the right answer and 75% is the effect of removal. The more balls there are, the smaller the effect of removal, so the closer the answer will be to 75%. 50 balls is enough that the effect of removal is small. With 2 balls, the answer is 100%.
