version of birthday paradoxon

[bhelmes]

A peaceful day for all,

Let S1 the probability that two boules are the first time in one bucket
by throwing one boule after the other by random in 100 buckets.

This is the classical birthday paradoxon and there is an answer from the page https://en.wikipedia.org/wiki/Birthday_problem

Let S2 the probabilty if you throw two boules at the same time in two different buckets.

Let S3 the probability if you throw two boules at the same time in two buckets which are placed as neighbours.

Is the probabilty S2=S3 or is it different.

Would be nice to get your opinion to it.

Nice greetings from the primes
Bernhard

[retina]

Originally Posted by bhelmes

Let S1 the probability that two boules are the first time in one bucket
by throwing one boule after the other by random in 100 buckets.

This is the classical birthday paradoxon and there is an answer from the page https://en.wikipedia.org/wiki/Birthday_problem

But I have very bad aim so it is likely that I will completely miss all the buckets with high probability (whether or not it is my birthday). So S1 approaches zero, quite different from what the WP page suggests.

[science_man_88]

Originally Posted by bhelmes

A peaceful day for all,

Let S1 the probability that two boules are the first time in one bucket
by throwing one boule after the other by random in 100 buckets.

This is the classical birthday paradoxon and there is an answer from the page https://en.wikipedia.org/wiki/Birthday_problem

Let S2 the probabilty if you throw two boules at the same time in two different buckets.

Let S3 the probability if you throw two boules at the same time in two buckets which are placed as neighbours.

Is the probabilty S2=S3 or is it different.

Would be nice to get your opinion to it.

Nice greetings from the primes
Bernhard

if there are 100 buckets and two things to place ( or throw) into them then in theory the odds assuming you make 100% of throws is are 99/100 the third one though seems to be dependant on setup as an example if they form a filled in square then one bucket not on the edge has 8 neighbours the ones on the edge have less. if they form a 10 by 10 square then there's 64 with 8 neighbours and 32 with 5 and 4 with just 3. so the odds on the whole depend on the number of ways of placing it randomly versus specifically.

[CRGreathouse]

Originally Posted by bhelmes

Let S2 the probabilty if you throw two boules at the same time in two different buckets.

Let S3 the probability if you throw two boules at the same time in two buckets which are placed as neighbours.

Is the probabilty S2=S3 or is it different.

I think the probabilities you're asking for are: given a number n, what is the chance that your n-th throw with this method causes two boules to land in the same bucket for the first time?

In that case S2 does not equal S3 since S2(2) = binomial(98,2)/binomial(100,2) = 197/4950 < 295/99 = (97*3 + 4)/99 = S3(2).

[retina]

Originally Posted by CRGreathouse

... 295/99 = (97*3 + 4)/99 = S3(2).

Probabilities can be above 1?

[CRGreathouse]

Originally Posted by retina

Probabilities can be above 1?

Bleh, should have been 99^2. The line should have been

S2(2) = binomial(98,2)/binomial(100,2) = 197/4950 > 295/9801 = (97*3 + 4)/99^2 = S3(2).