Services & Resources / Wolfram Forums
-----
 /
MathGroup Archive
1998
*January
*February
*March
*April
*May
*June
*July
*August
*September
*October
*November
*December
*Archive Index
*Ask about this page
*Print this page
*Give us feedback
*Sign up for the Wolfram Insider

MathGroup Archive 1998

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: A comment on the Gambler's problem



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:┌ble[Random[Integer,{1,6}],{3}]

Then,

Length[Select[Table[triple,{10000}],Plus@@#┘&]] //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_]:║,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@@#┘&]]

There are 25 sum nine and 27 sum 10.

Hope this helps.

Tomas Garza
Mexico City



  • Prev by Date: Re: postscript file
  • Next by Date: Re: Graphics Output
  • Prev by thread: Re: calculation (Perhaps Solve could do more with rational exponents???)
  • Next by thread: Q. about Solve as applied to vector equations