4 ms·
when you roll 2 dice, you get a lot more evens than odds. i imagine the same principle holds. if an odd (Hydrogen) forms with another odd, you get even. Hydr
by autokad 11y ago
when you roll 2 dice, you get a lot more evens than odds.
i imagine the same principle holds. if an odd (Hydrogen) forms with another odd, you get even. Hydrogen+helium=odd, but helium + helium = even. as the evens outnumber the odds, even more evens are forming with evens.
- Someone 11y agoNo, you don't 1+3+5+5+3+1 = 2+4+6+4+2 (1/36 probability of a sum of 2, 3/36 of a sum of 4, etc)
- ethbro 11y agoWas going to say, read that, did it in my head. D1 = {E,O} D2 = {E,O} *Where E and O are balanced in terms of possible outcomes that satisfy Outcomes: {EE, EO, OE, OO} Odd outcomes: {EO, OE} Half.
- autokad 11y agooh yeah your right
- autokad 11y agonot sure why i was down voted for admitting I was wrong, assholes.
- learnstats2 11y agoThis is a really interesting thought. Assuming you can only add two atoms at random, there are four options: odd + odd = even (odd reduced by 2, evens increased by 1) odd + even = odd (evens reduced by 1) even + odd = odd (evens reduced by 1) even + even = even (evens reduced by 1!) When the evens outnumber the odds, odd+odd will be rare, so the percentage of evens will tend to decrease until it reaches 50%. No matter many of each type you start with, the equilibrium position is 50-50% odd and even. However: there may be an exception to this. If all the odds are in one place, then odd-odd may be more likely than random. If you have all evens in one place, you can't get back to having odds again - you can't (ever) increase the number of odds that already exist to balance things out, you can only decrease the evens. 0-100% odd and even could also be an equilibrium position.