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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

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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.

  • 0
    "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. – 2017-02-25
  • 0
    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 $0.4^5-0.3^5=0.00781$. – 2017-02-25
  • 0
    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! – 2017-02-26

1 Answers 1

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$10^5$ 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 $3.36340$