If there are four numbers in a box, and you pick two of them one after another, what is the probability that the second number that you pick is less than the first number that you pick?
OT- Probability problem
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Very very rusty on probability - have to say more info is needed - you need to know that the numbers are unique & what initial number is pulled
If -
(a.) the numbers in the box are:... [1]...[2]...[3]...[4]
and -
(b.) you pull out ... [1]
then the the probability that the second number that you pick is less than the first number is ZERO
however if -
(c.) you pull out ...[4]
then the the probability that the second number that you pick is less than the first number is 100%
If -
(a.) the numbers in the box are:... [1]...[2]...[3]...[4]
and -
(b.) you pull out ... [1]
then the the probability that the second number that you pick is less than the first number is ZERO
however if -
(c.) you pull out ...[4]
then the the probability that the second number that you pick is less than the first number is 100%
OK ... after further thought -
if you reach in the box & grab a number without checking it & put it in a bowl marked "A"
and then immediately reach in the box again and grab another number & put it in a bowl marked "B"
Then the probability that the second number (in bowl"B") is less than the first number (in bowl "A") is 50%
if you reach in the box & grab a number without checking it & put it in a bowl marked "A"
and then immediately reach in the box again and grab another number & put it in a bowl marked "B"
Then the probability that the second number (in bowl"B") is less than the first number (in bowl "A") is 50%
Why do you have to reach in "immediately" after? I'm pretty sure the probabilities don't change if you (EDIT) "lock" everything up, come back 50 years later, and draw the second number.
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To take this example to conclusion.
a. There's a 25% chance that you pull a 1, in which case the probability that the second number is less than the first number is 0%
b. There's a 25% chance that you pull a 2, in which case the probability that the second number is less than the first number is 33%
c. There's a 25% chance that you pull a 3, in which case the probability that the second number is less than the first number is 67%
d. There's a 25% chance that you pull a 4, in which case the probability that the second number is less than the first number is 100%.
(25%*0)+(25%*33%)+(25%*67%)+(25%*100) = 50%
if the numbers are all 1 or higher...there will always be more possible numbers that are higher (infinite) than whatever number you pull first.
Do the numbers have to be positive numbers? If not, then there is an infinity of numbers below as well as above, aren't there? So I would think -- I'm *not* a probability whiz -- that if negative numbers are possible, the odds are 50% -- I think.
Yes you are correct - "immediately" was only a sort of mental placeholder for me - to express the idea that the you select both "A" and "B" before you make any assessments of the odds or know the value of either
- you don't pick "A", look at it , and then are asked to calculate the odds of "B" being less than what you know to be the value of "A"
.... or at least that is how I came to look at it - it is not my puzzle
- that is basically how I deconstructed it
- only I took my crayons & drew out each of the 12 possible combinations & added on my fingers to count the 6 less than outcomes
I'd go with the monkey if the safety on his MAC 10 is off. The dog, if the monkey is dead before the fight starts.*
*Exception: live unmuzzled Pit Bull vs. live, unarmed monkey. Pit Bull wins everytime.
Are they 4 consecutive, if not, what is the highest, lowest. I believe these all come into play if you get to see the first selected number before guessing the percent probability. If all unique and you select the second without seeing the first, then it should be 50%
If one assumes there to be an infinite set of numbers, then no matter what number one picks first, there ought to still be an infinite set of numbers greater than the number picked and an infinite set of numbers less than the number picked. That sounds like 50/50, but I don't recall if infinity works that way or not (I think it depends on whether we're talking integers or real numbers). I'm also not sure what the significance of the 4 numbers is.
I don't think it matters whether there is an infinite set of number or limited set, or if the numbers are integers or not.
There are 4 numbers in the box. You can label the numbers A, B, C, D, with A having the lowest value, followed by B and C, and D with the highest value. It doesn't matter what the values are; A still has a lower value than B.
If your first pick is A, there are 0 options for your second pick to be a lower value and 3 options for your second pick to be a higher value. If your first pick is B, there is 1 option for your second pick to be a lower value, etc. You can work this all out, like Doctor Worm did above, and you come to 50% (assuming no duplications).
It gets a little more complicated if you allow duplication. If your first pick is B, you don't know if the remaining numbers are AAA, AAB, AAC, etc. But if you work it out, you get 37.5% probability that the second ball has a lower value than the first ball.
Just a quick heads up--probabilities aren't expressed with a percentage sign. An event that happens 50% of the time is expressed as having a probability of .5. I can feel a bro-down coming on in this thread...
Unless it were Peter Gabriel's dog, in which case it would shock the monkey. I guess that would also be true for Terry Shea's dog.
I was thinking more of the general case where the first two numbers are randomly generated, but didn't state that well. My point about how many numbers in the box still stands, I think. Let's just say there were two numbers, as long as they're different, so one is greater and one is lesser. Depending which one you pick, it's 100% that the 2nd one will be greater or lesser, such that before you pick the first one, the probability of the 2nd one being lesser is 50% (or p = 0.5 to make RaRa happy).
Yes, the 4 numbers are unique. Say they are 1, 2, 3 and 4 and they are not put back.
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