# 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.

• "You roll a load of dice at the same time..." So, if I roll a $3,7,6,5,4$ does that count as five successes or one success? Your calculations seem to neglect this fact. – JMoravitz Feb 25 '17 at 23:41
• You seem to be wishing to calculate the probability of at least one success (as indicated by your adding these up). It is much easier to calculate the probability of no success and subtract this away from one. No success occurs when all five dice are $8$ or higher and occurs with probability $0.3^5=0.00243$, so at least one success is $0.99757$. The probability for exactly one success corresponds to all dice being $7$ or higher and having at least one seven which has probability $0.4^5-0.3^5=0.00781$. – JMoravitz Feb 25 '17 at 23:46
• JMoravitz, that would be 5 successes - I am afraid that I am not sure how to make my calculations account for this (you will have to pardon my limited skill in this regard) as for your second comment, I am not sure I understand - does that mean for 2 successes the probability is (0.5^5)-(0.4^5)-(0.3^5)? I tried calculating that for all possibilities, but the value was still not 100% Thanks for the rapid reply though! – monteckine Feb 26 '17 at 0:11

$10^5$ 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 $3.36340$