Question 1139615
.


            In my post, I'd like to propose another solution.



<pre>
We have 100 balls numbered from 1 to 100, inclusively; so we have 100 distinguishable balls.


Of them, we can form  {{{C[100]^3}}} = {{{(100*99*98)/(1*2*3)}}} = 161700 different triples.


Jack's 24 tickets cover 24 such triples - so the probability to win for Jack is 


    {{{24/161700}}} = 0.000148 = 0.0148%  (approximately).     <U>ANSWER</U>
</pre>