Maths probability question

[DaveF]

For the 2nd one, 149/5000.

 

[Visage]

Once you have picked one number the other matching draws are no longer possible so they are irelevant. Therefore if i pick the number 50 (that was a 1/100 chance) the ONLY matching pair is another 50 and the chance of picking that is 1/100. As the condition is that you MUST have both of these they are multiplied to make 1/10000. GCSE maths !

I've gtg now - i'll work on the second later

You should probably have read the thread to avoid looking dim.

 

[Monkey Puzzle]

Hey guys

Simple maths question but it's playing with me


Picking 2 random numbers, between 0 and 99 (range of 100).

What's the probability in them both being the same? What's the maths?

If we pick 3 random numbers, what's the probability of 2 being the same? Again what's the equation?

Trying to figure out if picking random numbers for some program is a good idea

The first part is 1/100.

The last part: I'm presuming you want precisely 2 matching number. In which case you use the binomial equation I think.

 

[neon.uk]

You should probably have read the thread to avoid looking dim.

I know. As soon as i walked out the door i knew the mistake i'd made. The probability of picking the first number is 1/1 as any number can be picked, so sorry, yes, it is 1/100 time to edit my post XD. GCSE maths a year early - god i should have spotted that .

 

[EffBee]

Hey guys

Simple maths question but it's playing with me


Picking 2 random numbers, between 0 and 99 (range of 100).

What's the probability in them both being the same? What's the maths?

If we pick 3 random numbers, what's the probability of 2 being the same? Again what's the equation?

Trying to figure out if picking random numbers for some program is a good idea

Probability is a wonderful thing from what I remember at school. You have to couch your question exactly. Do you mean at least two numbers the same or precisely two numbers the same. Any number or specific number?

I think for this you have to look at the three cases: 1st and 2nd same, third different; 1st and third same second different; 2nd and third same, first different.

I haven't done this stuff for thirty years and I'm glad. My head hurts just thinking about this simple example.

I am very happy to be corrected by anyone who thinks this is wrong

 

[DaveF]

I think for this you have to look at the three cases: 1st and 2nd same, third different; 1st and third same second different; 2nd and third same, first different.

If you want the probability of a duplicate, just find the probability P that all 3 are different and your answer is 1-P.

If you want the probability that exactly 2 are the same, you're basically right, but you should be able to convince yourself that all 3 cases have the same probability, so you can just calc one of them and multiply by 3.

 

[Nash]

Incorrect. 1:100

No, not that either in this instance. He already said a predetermined number is found so there's no version of the first iteration in the OP this time. The number is 47. There's a 1 in 100 chance of it being 47 if you pick a random number from 100 choices. If it's 47 on the second random chance it'd be 1 in a 100 times the previous iteration so 1 in 100,000 actually.

 

[ErNciLator]

aha cool in the end i got to

prob of 2 matching = 1-(99/100)^(n(n-1))/2

where n is the number of numbers picked (bit of google, bit of asking)