Chebby
Guest
LobstaMan:Chebby,
I understanding picking the same four numbers, but the same four numbers exactly in the same order? I think the probability of that scenario is much less than 50%. But, I suck at math so maybe you're right. lol.
LobstaMan
You are right, I was wrong. It's actually 118
See the birthday problem paradox . I was surprised when I found out it was this low too. Just plug 10000 (0000 to 9999) instead of 365.
That is because you are asked about the prob of 2 RANDOM NUMBERS occuring in a set and NOT 2 SPECIFIC NUMBERS.
It basically comes down to p(k)=1-P(10000,k)/(10000^k) where the P stands for permutations (10000! / (10000-k)!). You have to solve to k, but because I already did, you can just plug 118 in to see if it works.
P.S: sorry for taking this thread off topic, just thought it would be something fun to know.