Don't Use Formula For This

Thaty wasn't supposed to be posted, I was still working on it and se an error already.


6^5 = total number of combinations given one toss. For each of the 5 slots the 3 can fall into, the other slots together hold 5^4 combos. So we have 5(5^4)/6^5 giving the fraction of ways a 3 will fall on one toss.


The total combos given 5 tosses are 6^5^5.


That's how far I am now but I keep getting disconnected and its time for poker

can't post til tomorrow isp keeps disc. and it led me to believe I had posted when I hadn't ...got Message Posted screen but it was like a ghost post, so my post referring to DS's was actually intended to refer to my own (invisible post)

Here's how I was hoping you would do it.


Throw five dice. The chances the first one is a three and the rest aren't, is 1/6 times 5/6 to the fourth power. 08038. But that three could also be on one of the other four dice. Total chances for exactly one three is .402


Do above five times. Chances that the first throw of five dice contains exactly one three and the other four throws, don't is .402 times .598 to the fourth power. .514 Since the throw with exactly one three could also be the 2nd 3rd 4th or 5th as well, the total chances are five times as much or .257.

That is nice and compact and I like it.


It took me maybe 5 minutes to really follow your explanation through and see where you got the numbers (like where you got .598 had me baffled briefly, until it suddenly struck me that .598 and .402 added up to 1.0).


I am basically new to counting problems, but I will practice a few examples with variations on this theme over the next few days.


Thank you.