*To*: mathgroup@smc.vnet.net*Subject*: [mg12045] Re: [mg12006] A comment on the Gambler's problem*From*: Tomas Garza <tgarza@mail.internet.com.mx>*Date*: Fri, 24 Apr 1998 01:52:16 -0400

Daniel Sanders wrote: > "In the early 1600s, Galileo was asked to explain the fact that, > although the number of triples of integers from 1 to 6 with sum 9 is > the same as the number of such triples with sum 10, when three dice are > rolled, a 9 seemed to come up less often a 10 supposedly in the > experience of gamblers." > A few of you observed and proved Mathematica-ly that the number of ten > sums exceeds the number of nine sums in three throws of the dice. True > enough, but the folks in Galileo's time didn't understand the concept > of a probability space, and in particular an ordered n-tuple. The > number of such permutations in the throw of three dice does sum to ten > more frequently than it sums to nine, but the number of combinations is > the same, and that is what the gambler's counted on. There are six > combinations of 3-tuple that sum to nine and ten. > Nine:{{1,2,6},{1,3,5},{1,4,4},{2,2,5},{2,3,4},{3,3,3}} > Ten:{{1,3,6},{1,4,5},{2,2,6},{2,3,5},{2,4,4},{3,3,4}} > And this another programming problem, that of a replacement rule that > takes a finite probability space and kicks out a designated subset. In > particular the 3-tuple set of the three dice and the number of > combinations that sum to nine or ten. Daniel, First, one way to run a simulation of the outcome of rolling three dice. Define triple:Úble[Random[Integer,{1,6}],{3}] Then, Length[Select[Table[triple,{10000}],Plus@@#Ù&]] //Timing will give you the frequency of sums equal to nine in 10000 throws of the three dice, together with the time used for the calculation (which in my 233 MHz PC took 3.8 seconds). Similarly for sum ten. In an actual run the ratio obtained was 1116:1265. Second, you can construct the sample space corresponding to this experiment, i.e., the Cartesian product of three sets {1,2,3,4,5,6} as follows. In[1]:[a_,b_,c_]:º,b,c} In[2]:ampleSpace Partition[Flatten[Outer[f,{1,2,3,4,5,6},{1,2,3,4,5,6},{1,2,3,4,5,6}]], 3]; triples with sum nine, by Length[Select[sampleSpace,Plus@@#Ù&]] There are 25 sum nine and 27 sum 10. Hope this helps. Tomas Garza Mexico City