# One Vs. Two

Why the probability of picking any ball numbered (1 to 4) in ONE draw
out of a box containing 37 balls which is (4/37)= (0.108108), DOES NOT equal to the probability of picking any  ball numbered (1-2) at LEAST once[with replacement] out of the same box of 37 balls in TWO draws which is [1-(35/37)x(35/37)]=( 0.106975)???
I mean In the first case we are talking about picking a ball from 4 balls. In the second case, even though we HALVED the number of balls to choose out of from (4 to 2) we DOUBLED the number of draws, should it that make both probabilities EQUAL??? Any MATHEMATICAL explaination!!!!!!!
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Commented:
Because in case two you put the first drawn ball back in so that there is a (small) probability that it will be drawn again.
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Author Commented:
So What if you Draw again?? Remeber we said at LEAST once that Means if we draw the right ball on the first draw we STOP. If not we go to the other trial!!!
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Commented:
When you have 2 separate drwas, you should not count the times when you pick a (1-2) ball in BOTH draws. Hence your total probability is given by

P(A) + P(B) - P(A and B) = 2/37 + 2/37 -(2/37)^2 = .105186
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Commented:
i think it is because u just can't do it like that...if you are doing the four out of 37 then it would look like this:

(4/37)=.108108108108...

but to get it equal you would have to take not only half as many balls your looking for but also half as many balls in the bag or whatever it is...so it would be like this:

(2/(37/2))=.108108108108...

-----------------------------------------------------------
here is an example to prove this works:

(11/22)=.5

now to make it equal the next equation it would be:

((11/2)/(22/2))= .5

so it works for any numbers and this could also be proved algebraically

-----------------------------------------------------------

(a/b)=c

((a/2)/(b/2))=c

therefore:  (a/b) has to equal ((a/2)/(b/2))

using the simple formula ((a/b)/(c/d))=(ac/bd)

you can conclude that ((a/2)/(b/2)) = (2a/2b) which equals (a/b)... PROVEN

-----------------------------------------------------------

so that is how it is supposed to be done to make them equal each other...i think the formula is like "Product of the means over the product of the extremes"... well that is how i was always told to remeber it... everything comes back to it in this problem

hope this helps...post any more questions here

Mark

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Associate Director - Product EngineeringCommented:
In your first draw, you took out one ball numbered 1 to 4 out of 37. In the second case, since you are taking two draws, you can choose the same ball twice, since it is with replacement.

Mayank.
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Author Commented:
Mr. hdhondt, Even if your calculation is write the problem remains0.108108 does not equal to .105186???
Now for Mr.simpsons17371, Of course 12/24=6/12=3/6. BUT the question is why is it when we HALVE the number of balls and DOUBLE the number of draws, we do not get EQUAL probabilites of the original experiment?? I am looking for the mathematical relation between Number of balls AND number of DRAWS!!!!!!!
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Commented:
The average number of balls drawn is the same
33/37 chances of drawing 0 balls + 4/37 chances of drawing 1 ball = .108108
vs
1225/1369 chances of drawing 0 balls + 140/1369 chances of drawing 1 ball + 4/1369 chances of drawing 2 balls = .108108
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Commented:
To put it another way, the probabilities of your 2 cases are NOT the same, although you may think they should be.

The difference is that, when you try to pick one of 4 balls, you're only picking once. When you are twice picking 1 of 2 balls you can double up on the possible draws. These doubled up ones have to be discarded, since this CANNOT happen if you only pick once.

Note that 2 * (1 - 35/37) = 0.108108, i.e. the same as 1 - 33/37

The mathematical relationship is as I have stated: the probability of picking (1-2) in the first draw, PLUS the probability of picking (1-2) in the second draw, MINUS the probability of doubling up on the picks.
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progCommented:
how about if you picked any ball form a set numbered 1-38 the chances of getting a ball numbered 1-38 is then 1

If you pick twice, with replacement. then the chance of getting any ball 1-19 cannot be 1/2 + 1/2 = 1

it is 1/2 + 1/2 - 1/2 * 1/2 = 3/4
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Commented:
Because of the way the 2 are calculated:

To calculate the chances of picking balls 1-4 out of 37 balls
(4/37)=.1081081
or 1-(33/37)=.1081081

To calculate the chances of picking balls 1-2 out of 37 AT LEAST ONCE with replacement:
(1-(35/37)^2)=.105186

If the formula were (1-(35/37))*2 then the probability would be equal to 1-(33/37) and equal .1081081, but that isn't the mathmatical way of calculating the probability of multiple events.
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Author Commented:

I think the best way to look at the issue is to see if there is any mathematical explaination, or otherwise, for the fact that probabilities DO NOT function in a linear and proportional manner. Is there any explaination for that FACT!!!
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progCommented:
when drawing one ball at a time, the progression of probability cannot be linear, because the probability then has to reach and the exceed one.

e.g balls numbered 1-37 - draw one ball, chance of it being 1 = 1/37, draw two balls at once, prob = 2/37

up until if you draw all 37 balls at once, the probability of getting ball 1 = 37/37 = 1

now if you draw one ball at a time 37 times, the probability cannot be 1 - because how can you guarantee that 1 would have been drawn.

in fact it is 1 - (36/37)^37 = .637

I think what you are looking for is an intuitive reason, not a mathematical reason - I think what you need to accept there is that drawing two balls at once is a different scenario to making two separate drawings, because you are replacing the ball drawn in the separate drawings case

when drawing two balls at once you are eliminating the first ball drawn from the second draw - increasing your chance of success, hence the probability is higher.
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Commented:

balls numbered 3 and 4 stand out and
are more likely to be drawn than balls
numbered 1 and 2. hence the higher
probability for the single draw of balls
(1-4) case.

disproving this statement is left as an
excercise for the student.

leo
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Commented:
hello  again cig,

I did take another look at the question
and was going to do a simulation to get
results.... but in thinking about the design
of a simulation it became obvious that you should
not expect the same probability for both cases.

in case one you draw one number...in case two you draw
2 numbers ... in case 2 you could get the same number
twice.

work out the probability of getting the same number
twice for case two and look at that probability with
respect to your case 1, case 2 probabilities.

leo
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Commented:
You have confuse most of us successfully.

you are correct in the first calculation:
the posibility of picking a ball numbered 1 to 4 from 37 balls
= 4/37
= 0.108108108

Very sorry to say, your second calcualtion( [1-(35/37)x(35/37)]=0.106975 ) is WRONG !!!

Posibility of picking ball numbered 1 to 2 in both picks with replacement
= (1 - (35/37)) * (1 - (35/37))
= (2/37) * (2/37)
= 0.00292184

Posibility of picking ball numbered 1 to 2 in at least 1 pick with replacement
= (2/37) + (2/37)
= 2 * (2/37)
= 4/37
= 0.108108108

Conclusion:
the Posibility of picking ball numbered 1 to 2 in at least 1 pick in 2 pick with replacement IS SAME with posibility of picking ball numbered 1 to 4 in 1 pick.
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progCommented:
b612

>>the Posibility of picking ball numbered 1 to 2 in at least 1 pick in 2 pick with replacement IS SAME with posibility of picking ball numbered 1 to 4 in 1 pick

no its not

first scenario = 1 - (35/37)^2 =.10519

2nd scenario = 4/37 = .108108
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Commented:
done?
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Commented:
>first scenario = 1 - (35/37)^2 =.10519

To deighton, first scenario is not true, let me try to prove it:

posibility of pick up ball numbered 1 to 2 in 1 pick
= 2/37

since it is with replacement, the second pick should have same posiblity of picking up the ball numbered 1 to 2
= 2/37

both pick have posibilty of 2/37, so we can conclude that:

Posibility of picking ball numbered 1 to 2 in both picks with replacement
= (2/37) * (2/37)
= 0.00292184

Posibility of picking ball numbered 1 to 2 in at least 1 pick out of 2 picks with replacement
= 2 * (2/37)
= 4/37
= 0.108108108
<> 1 - (35/37)^2

Pls prove that I am wrong and not just telling me I am wrong. Thank You.
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Commented:
Sorry for giving wrong answer twice and pls ignore them. I am now trying to prove that I am wrong.

The following equalation is wrong since I miss out a fact that once you get the ball in first pick, you wont go for the second. perhaps this is the reason of the different of posibility in 2 different scenerios.

//THIS IS WRONG, PLS IGNORE
Posibility of picking ball numbered 1 to 2 in at least 1 pick out of 2 picks with replacement
= 2 * (2/37)
= 4/37
= 0.108108108
<> 1 - (35/37)^2

The correct calculation should be like this:
posibility of get balls number 1 to 2 inat least 1 pick out of 2 picks
= posibility of get in 1st pick + (not get in 1st pick * get in 2nd pick)
= 2/37 + (35/37 * 2/37)
= 0.105186267

Since the posibility of go for 2nd pick in 2 picks scenario is less than 1.0, and the posibility of go for 1st pick in 1 picks scenarion is 1.0. So the possibility of 2 picks scenarion is little bit lower.

Hope this time i am correct.Thanks.
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Commented:
I feel that my previous explaination is not too clear so i would like to elaborate it here.

1. If you calculated it on paper and not strait away calculate with calculator, you will find the following result:
1 - (35/37)^2 = 144/1369
2/37 + (35/37 * 2/37) = 144/1369
both of tem is equal to 0.105186267

2. What I means by less than 1.0 and equal to 1.0 is that:

In the 1 pick scenarion, it is 100% you will utilise all your chances to pick up a ball (1 pick in the 1 pick scenario)

In the 2 pick scenarion, it is NOT 100% you will utilise all your chances to pick up a ball (2 picks in the 2 pick scenario). The posibility is less than 100% since you might get the ball in first draw and stop without 2nd attempt.

So the posibility of 2 picksscenario is lower.

Hope this make it clear. Thanks.
= 1.0 -

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Commented:
Just a simple example to prove the 2 probabilities do not have to be equal.

Instead of 37 balls, let's use just 4. In that case, the probability of picking balls (1-4) in 1 draw is obviously 1. Equally obviously, there us no guarantee that in 2 draws we will pick balls (1-2) at least once (might might pick balls 3 or 4 both times). Hence that probability HAS to be less than 1.

Make sense?
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Commented:
yes
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Commented:
I have discover something, I found that even we draw for 2nd time after get the ball in 1st draw, the posibility is still remain unchange.

Posibiblity of AT LEAST ONE
= get in 1st draw only + get in 2nd draw only + get in both draws
= (2/37 * 35/37) + (35/37 * 2/37) + (2/37 * 2/37)
= (70 + 70 + 4)/1369
= 144/1369
= 0.105186267

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progCommented:
b612, i agree with your calculations no, sorry I'd misread your first calculation regarding picking a ball 1-2 in both draws.
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Commented:
Doesn't this all prove that your proposition regarding the equivalence of the events (1 of 4 vs 2*(1 of 2)) reasonable as it may seem, is in fact not the case - and thus a candidate for a scam?

...Bill
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Commented:
I am very regret of reading this question, now I am unable to control myself from stop thinking about this question.

I hope this calculation bring inspriration to others expert to solve the problem.

1 pick scenario
= 1 - 33/37
= 4/37

2 picks scenario
= 1 - (35/37)^2
= 1 - 1225/1369
= 144/1369

Since the prosibility of not get the balls is different
33/37 <> 1225/1369

So the posibility on get the balls is also different.

But why? Why the prosibility of not get the balls is different?
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Commented:
There's no reason why the two scenarios *should* produce the same result.
In the first case you have 37 possible outcomes, of which only 4 are acceptable. In the second, with *two* drawings from the same set, you have 37^2 (1369) possible outcomes, of which 144 are acceptable.

The possible outcomes of the second scenario are, if we designate Y as drawing 1 or 2, and N as drawing a ball numbered between 3 and 37

N&N: 1225 = 35X35
N&Y:      70 = 35X2
Y&N:      70 = 2X35
Y&Y:        4 = 2X2

Another way to look at it is to say that, in scenario #1, the probability of *not* succeeding is 33/37, and in scenario #2 it is (35^2)/(37^2). The probability of not succeeding in two drawings is the product of the probabilities of not succeeding in one drawing, not the sum.
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