4 ms·
Hm I'm not sure if it's correct but here's my logic (here on mastodon https://social.ciaranmc.com/@ciaran/111850662374227629 https://social.ciaranmc.com/@ciaran
by CiaranMcNulty 3y ago
Hm I'm not sure if it's correct but here's my logic (here on mastodon https://social.ciaranmc.com/@ciaran/111850662374227629 https://social.ciaranmc.com/@ciaran/111850662374227629)
There are N-1 red balls out of 99 remaining, so the probability the next ball is red is (N-1/99).
The possible values of N are evenly distributed [1...100] (we discount the N=0 case because we drew a red ball)
The total probability that the next ball is red is the sum of the individual probabilities divided by cases:
P = SUM[1...100](n-1/99) / 100
= SUM[1...100](n-1) / 99 * 100
= SUM[0...99](n) / 99 * 100
= 4950 / 9900
= 5.0
So red or green are equally likely
- 3PS 3y agoYour mistake is here: > The possible values of N are evenly distributed [1...100] (we discount the N=0 case because we drew a red ball) These probabilities over N are not actually even anymore; the fact that you drew a red ball means that higher values of N are more likely.
- jcl 3y agoThere are actually 10,100 equally likely situations: 101 choices of N x 100 choices of ball. When you completely discount the N=0 case, it’s because all choices of ball are green. You similarly need to almost completely discount the N=1 case, because almost all choices of ball in that situation are green, etc.