I am having an issue working out the probabilities for rolling a series of 10-sided dice. The issue comes in that I need the required rolls to change as the number of dice increases.
The rules of the game are simple, you roll a load of dice at the same time, then allocate them to try and score the maximum number of successes.
The scores required are as follows, 1 dice must be 7 or less to be a success, then one more must be under 6, then one more under 5 - and so on.
If the dice were rolled one at the time, this wouldn't be a problem for me, the issue comes in that all dice are rolled at the same time.
I am sorry that I am bad at explaining what I mean, so I will give an example with 5 dice.
All dice are rolled at the same time
1 dice has to be under 7 the next has to be under 6 then 5 then 4 then 3
SO the chance of 0 successes is 0.3^5, as for any dice to roll less than a 7 would automatically be a success. - or about 0.243%
The chance of getting 1 success is 0.7*(0.4^4), as one dice has to be less than 7 to get the success, but then none of the others can be under 6 (or else there would be 2 successes). - or about 1.792%
The chance of getting 2 successes is 0.7*0.6*(0.5^3) - or about 5.25%
The chance of getting 3 successes is 0.7*0.6*0.5*(0.6^2) - or about 7.56%
The chance of 4 is 0.7*0.6*0.5*0.4*0.7 - or about 5.88%
and to get all 5 successes is 0.7*0.6*0.5*0.4*0.3 - or about 2.52%
The problem is that all these added together is only 23.245% I have no idea how to work out the actual chances of rolling in this way.
I am sure it is probably something quite simple, which I am missing.
Ok, I think I may have it. Is it possible for someone to check that this is correct?
for N dice, the chance of getting no successes is the following formula
(1-(A1/10))^N - where A1 is the highest value needed to score a success (in the case above, A1=7)
1 success is (1-((1-(A1/10))^N)*((1-(A2/10))^(N-1)) - where A2 is the second highest value needed to score a success (in the case above, A2=6)
2 successes is (1-((1-(A1/10))^N)(1-(1-(A2/10))^(N-1))((1-(A3/10))^(N-2)) - where A3 is the third highest value needed to score a success (in the case above, A3=5).
This gives the values No successes - 0.243% 1 success - 2.554% 2 successes - 12.15% 3 successes - 30.619% 4 successes - 38.104% All successes - 16.33%
Which all add up to 100%
If correct, I'd like to thank JMoravitz for the help. I probably wouldn't have thought of trying to work it out in this way without his input.