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Bless you, Blaise! (Posted on 2022-12-31) Difficulty: 3 of 5
Chevalier de Méré (who was more gambler than mathematician) originally thought that rolling a 6 in 4 throws of a die was equiprobable to rolling a pair of 6's in 24 throws of a pair of dice.

In real life he would win the first bet more than half the time, but lose the second bet more than half the time.
He never understood, why.

de Méré asked his mathematician friend, Blaise Pascal, to explain this problem, which he did.

​Over 330 years later I offer this issue to you.
Please provide your answer:

a.intuitively, i.e. prior to solving.
b. Final answer based upon your reasoning.

Enjoy wearing BP’s shoes!

No Solution Yet Submitted by Ady TZIDON    
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Some Thoughts answers | Comment 1 of 2
a. I'd think the probability of rolling a 6 in 4 throws of one die would be a little over 1/2, but not fully 2/3--something like 55 to 60 percent. The deMere reasoning is that since two 6's is 1/6 as probable as one 6, it's given six times as many rolls. The problem is that the reasoning applies to the expected value of the number of occurrences, and that is inflated by the extra value of higher numbers of occurrences, with less benefit to the probability of getting at least one occurrence so the second wager would have somewhat lower chances of success.

b. The probability of rolling at least one 6 in 4 throws of a single die is

   1 - (5/6)^4 = 671/1296 ~= 0.517746913580247,
   
somewhat lower than I thought, though indeed a little over 1/2.

In the second proposition, the probability of success in each trial drops to 1/36.

   1 - (35/36)^24 = 1106543156715777/2251799813685248
                ~=  0.491403876130903
                
A table shows the probabilities of given numbers of hits:

  n      prob       contribution
                    to expected value
                    
  0    0.50859612   0.00000000 
  1    0.34875163   0.34875163 
  2    0.11458982   0.22917964 
  3    0.02400930   0.07202789 
  4    0.00360139   0.01440558 
  5    0.00041159   0.00205794 
  6    0.00003724   0.00022343 
  7    0.00000274   0.00001915 
  8    0.00000017   0.00000133 
  9    0.00000001   0.00000008 
 10    0.00000000   0.00000000 
 11    0.00000000   0.00000000 
       ----------   ----------
       1            0.66666667    
       
so indeed it's the expected value of the number of hits that is still 2/3, and it is that expected value that is the same in the two, rather than the probability of at least one occurrence.       

  Posted by Charlie on 2022-12-31 08:56:11
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