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A cricket partnership can score zero to six runs per ball with equal probability. What is the chance of 30 runs being scored in an over of six balls, assuming no outs or extras?

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  • $\begingroup$ For non-cricket lovers, there are usually $6$ balls in an over $\endgroup$
    – Henry
    May 25, 2017 at 13:57
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    $\begingroup$ maybe it's easier to work out how many distinct ways he can drop the 6 runs, if each score drops 0-6 runs (order could be important) $\endgroup$
    – Cato
    May 25, 2017 at 13:57
  • $\begingroup$ What does "throwing a ball" mean here? Might it be "bowling a no-ball"? $\endgroup$
    – Henry
    May 25, 2017 at 13:59
  • $\begingroup$ is this a 'real world' simulation? the prob of any score 0-6 is 1/7 - is that right? That isn't realistic though, because a 5 is rarer than most other scores. I don't know what 'throw' means either - only a fielder throws the ball - are you simulating detailed play? $\endgroup$
    – Cato
    May 25, 2017 at 13:59
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    $\begingroup$ Also note that many of the users on this site don't know the rules of cricket... $\endgroup$
    – Mickey
    May 25, 2017 at 14:01

1 Answer 1

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If you consider that it is equally likely for any of the (0,1,2,3,4,5,6) to occur on each ball. Than the ways in which 30 run can be scored could be calculated by the permutation of (6,6,6,6,6,0), (6,6,6,6,5,1), (6,6,6,6,4,2), (6,6,6,6,3,3) add all the ways than divide by total no of ways the over can happen which is 7^6. Total no. of ways to score 30 would be 6!/5!+6!/4!+6!/4!+6!/(4!*2!). which equals 81, so your desired probability is equals 81/7^6. I am new to stack exchange and don't know how to write mathematical symbols properly so feel free to edit my answer.

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