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Message: <blockquote data-quote="Coredump" data-source="post: 2634187" data-attributes="member: 6939">This just didn't sound right to me.... and it has caused me to bang my head against an excel spreadsheet, and some notepaper, to prove it to myself.You cannot add medians together to get an 'overall' median. It does not work that way. The median of rolling a 1 on a D10 is 6.591 rolls, and rolling a 1 on a D12 is 7.968. But to roll a 1 on a D12, then roll a 1 on a D10 will have a median *higher* than 15.Unfortunately, my math skills are too rusty to provide a totally rigorous proof, I can show I deduced this.The trick it to determine how many of the combined rolls will be above the proposed median. According to the assumption above, a bit less than 50% of the rolls should be above 15.Above and below the median is a 50-50 chance. So to roll above the median on a D12 (I call that 12H) is going to happen half the time, and below the median is half the time (12L). Since we are doing these rolls in order, we get 4, equal, possible types of outcomes.12H-10H (25%), 12H-10L (25%), 12L-10H (25%), 12L-10L (25%).Of those, the first and last are easy.Looking at the 12H-10H combinations, if both rolls are above the median, than the total will be above the median, then 100% of those combinations will be above the median. That means that 100% of that 25% segment will be above the median (Total above=25%)Looking at the 12L-10L combinations, if both rolls are below the median, than the total will be below the median, then 0% of those combinations will be above the median. That means that 0% of that 25% segment will be above the median (Total above=25%+0%=25%)That leaves us with going high-low, and low-high.Looking at the 12H-10L combinations, we can start by ignoring the 10L roll for now. The 12H rolls will give results of 8 and up. Any roll of 14 or higher will put the total at or above 15, and thus above the median. (This is about to get a bit confusing..) About 66% of the *12H* rolls will be at 14 or above, thus MORE THAN 66% of the 12H-10L combniations will be above the median. That means that more than 66% of that 25% segment will be above the median; 66% of 25% gives us 16.5% (Total above=25%+16.5%=41.5%)[NOTE: this is quite an underestimation. For illustration: This only counts if the 12H roll is 14 or higher, yet if it is 13, the 10L roll *has* to be a one, or the total will be above the median. There are lots of 'above the median' combinations that we are not counting.]**Looking at the 12L-10H combinations, we can start by ignoring the 12L roll for now. The 10H rolls will give results of 7 and up. Any roll of 14 or higher will put the total at or above 15, and thus above the median. (This is about to get a bit confusing..) About 50% of the *10H* rolls will be at 14 or above, thus MORE THAN 50% of the 12H-10L combniations will be above the median. That means that more than 50% of that 25% segment will be above the median; 50% of 25% gives us 12.5% (Total above=41.5%+12.5%=54%)[NOTE: Same as above]So we can that even while underestimating the results, we are getting more than 54% of the combination totals will be at 15 or above. Thus the median cannot be 14.5 as projected. We cannot just add medians together to get an overall median. And this is just two dice, the more dice we add together, the higher the resultant median will creep.**Some may ask why I didn't figure the rest out and come up with a true median. Two reasons, This started as an excercise to see if adding medians is okay, and it is not. Second, do figure the others out will take a lot more tedious maths....Hope this made sense, and I realize it is probably more info than anyone really cared about, but hey, it was fun.  <img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f600.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":D" title="Big grin    :D"  data-smilie="8"data-shortname=":D" /></blockquote>

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