Yikes, we basically have to start from square one, but you've come to the right place, so panic not.
Quote:
Originally Posted by whois1230
67 combinations/21 2 even 1 uneven combinations = 3,19047619047619
For starters, you meant to say 21/67, because you can have a 31.34% chance but you can't have a 319% chance!
.3134 is still wrong, though, mainly because you tallied the combinations instead of permutations. (You also double-counted 11 of them, because the total should have been 56. For instance you counted 123 and 132 separately, but they're the same combination.)
Even though the order doesn't matter for this problem, you can't use combos to calculate a probability when repetitions are possible. If you do, the permutations involving repeats will be over-represented. For example, there are 6 arrangements of 123, but there is only one arrangement of 555. If you use combos, 123 gets the same treatment as 555, which is bad. The only way around that is to use permutations.
But there are 6^3 = 216 permutations and listing them all out would be a royal pain in the ass. Making matters worse, you probably wouldn't get any credit for it on an exam!
Instead, we can make life much easier by using some mathematical shortcuts.
First, take into account that we don't care about the actual numbers that come up, only their parity (even/oddness). Since any given die has a 50% chance of landing odd or even, this problem reduces to flipping 3 coins (or equivalently, flipping one coin 3 times) and asking the probability of 2 heads and 1 tail.
Each flip has 2 possibilities, and there are 3 flips, so the total # of permutations is 2⋅2⋅2 or 2³
Next: how many ways are there to arrange HHT (or EEO)? Well, the T can be in one of 3 different places, so 3.
So the answer is 3/8
For 1b we take the same 3 HHT perms and then add the HHH, bringing the tally up to 4 successful perms and making the answer 1/2.
There is another way of thinking about it that will be helpful for some of the other problems: we can treat each die as a separate event and then calculate the compound probability. With that in mind, take a stab at #1c (which doesn't translate to fair coinflips).
Quote:
Originally Posted by whosnext
I was thinking of answering these using Combinatorics but decided against it since I didn't want to be yelled at by David Sklansky for not doing it the easy way by using fractions.
I was gonna joke that technically it's even easier (in the same way) to use brute force, but then OP beat me to it and attempted brute force!
Last edited by heehaww; 01-24-2020 at 05:42 PM.