Rolling 10-sided dice with shifting probabilities 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.
Thanks
monte
----------edit----------------
Ok, I think I may have it. Is it possible for someone to check that this is correct?
Thanks!
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.
 A: $10^5$ equally probable cases is small enough to enumerate and count, so I think the actual numbers are (on the assumption that the dice have faces $1,2,3,4,5,6,7,8,9,10$, and the first success is a die $7$ or less, the second is the first success and another die $6$ or less, the third is the first two and yet another die $5$ or less, etc)
n  At least n  Exactly n  
   successes   successes

0   100000        243
1    99757       3211
2    96546      15521
3    81025      34301
4    46724      34436
5    12288      12288

These are not quite your numbers (you have $243, 2554, 12150, 30619, 38104, 16330$) so let's find the number of cases with exactly one success as a check:


*

*either you have one or more $7$s and the rest $8,9,10$; this can be done $4^5-3^5=781$ ways

*or you have exactly one $6$ or smaller and the rest $8,9,10$; this can be done ${5 \choose 1} \times 6\times 3^4 = 2430$ ways


and $781+2430 = 3211$ rather than $2554$
That would make the expected number of successes $3.36340$
