3 ms·
If Jeff Atwood argument would be right, couldn't it than be applied to roulette as well? If you play two games for example, there are 4 different orders of bla
by BvS 18y ago
If Jeff Atwood argument would be right, couldn't it than be applied to roulette as well?
If you play two games for example, there are 4 different orders of black/red possible (neglecting 0/00):
b,b
b,r
r,b
r,r
Each has the same possibility of 1/4. If you place 1$ in the first round on black, your chance to win is 1/2. If it is black (or red), only 3 of the original 4 option are left and two will lead to red...
- randallsquared 18y agoNo, you've reintroduced ordering.
- mpk 18y agorandallsquared is correct. You are misunderstanding sequences of related events. Running the roulette wheel twice is a sequence of unrelated events. The (ideal) roulette wheel is totally random. The results of run #1 are unrelated to the results of run #2. However, in this case of the man and woman both having two children, one of which is a boy, but the man having as an oldest child a boy we are talking about two different scenarios. The events are suddenly related. If you want it spelled out, the matrices are, The childbirth options are this, BB BG GB GG However, for the man it's been stated that his oldest child is a boy, so it becomes, BB BG XX XX For the woman (who has as a condition 'at least one boy, but not necessarily the eldest') it is, BB BG GB XX (Where 'B' is Boy, 'G' is Girl and 'X' is a non-option).
- jfarmer 18y agoIf the FIRST round is black, only two of the four possibilities are left, one of which leads to red. Similarly, if Atwood's problem stated that the FIRST child is a girl then the probability that the second is a boy is 1/2. The problem, however, states that one of the children is a girl -- it could be the first or the second. In this case the probability that the other child is a boy is 2/3.