19 replies to this topic

[bonanova]

If I repeatedly throw a fair 6-sided die, what is the probability that the running total will at one point equal 137?

[bushindo]

If I repeatedly throw a fair 6-sided die, what is the probability that the running total will at one point equal 137?

 

With some recursive code, I get

 

Spoiler for

0.2857143

[CaptainEd]

Spoiler for Bushindo's surprising value of

2 / 7

Edited by CaptainEd, 14 January 2013 - 09:32 PM.

[Prime]

It seems to stand the reason...

Spoiler for just a guess

Since with each roll we add with equal probability between 1 and 6 to the running sum, the average step is 3.5. Then the probability would be 1/3.5 just like Bushindo calculated.

I'll try for exact formula later.

[BobbyGo]

Spoiler for I must be way off

I got an anwer of 0.2536043somethingsomethingmorenumbers

[bonanova]

Bushindo has seven decimal places; the Captain has it exactly, if surprisingly; and Prime has the insight.
A triple tag team solution! Good job all. So who gets the "solved" tag .... ?

The nod goes to Prime's explanation.

[Prime]

Thanks Bonanova. My explanation does look sound. But now, after I have taken some time to reflect, I must disprove and disown my solution.

Sure, the fraction that we found here is good for everyday gambling needs, but it is not the exact probability.

I suspect, Bushindo’s method would have produced the exact answer, had it not succumbed to rounding error.

 

Spoiler for Proof of falsehood, but not the answer.

Any number may or may not be an intermediate running total in this experiment. But there is one exception - zero! Zero running total will occur with 100% probability right before the very first roll.
Consequently, the probabilities for the first six numbers (1, 2, 3, 4, 5, and 6) to become intermediate totals are:
P₁= 1/6;
P₂ = (1+1/6)/6 = 7/6²;
P₃ = (1+1/6+7/6²)/6 = 7²/6³;
And in the like manner,
P₄  = 7³/6⁴;
P₅ = 7⁴/6⁵;
P₆ = 7⁵/6⁶;
Note, all the above probabilities are different from 2/7. First four are smaller, and the last two are higher.

From that point on we must change the method for probability calculation. Since a running total n can be obtained by a single roll from 6 preceding totals,
P_n = (P_n-1 + P_n-2 + P_n-3 + P_n-4 + P_n-5 + P_n-6)/6. Note, for P_n = 2/7, the sum of its six predecessors must be equal 12/7.

Let’s figure out the exact probability for P₇ using addition of the fractions, like they taught me at school. The common denominator for the first six probabilities is 6⁶, whereas the numerator of the sum becomes:
6⁵ + 7*6⁴ + 7²*6³ + 7³*6² + 7⁴*6 + 7⁵. Note, the numerator is modulo 1 with respect to 6.
Thus P₇ = (6⁵ + 7*6⁴ + 7²*6³ + 7³*6² + 7⁴*6 + 7⁵)/ 6⁷.

For P₈ we add up P₂ through P₇ and divide it by 6. To bring P₂ through P₆ fractions to the common denominator of 6⁷, we must multiply their numerators by various powers of 6. Whereas the numerator of P₇ remains modulo 1 with respect to 6 and so will the numerator of P₈.

Thus by induction we know that the exact fraction for P₁₃₇ has 6¹³⁷ in its denominator and a very big who knows what in the numerator, which is modulo 1 with respect to 6, however. The numerator of that fraction is not divisible by 2 and its denominator is not divisible by 7. Therefore, it cannot be exactly equal to 2/7. That big fraction cannot be simplified at all - the numerator and denominator have no common factors.

Whoever finds the numerator - will have solved the problem.

In Excel spreadsheet, the probabilities for intermediate totals oscillated around 2/7, until the application ran out of precision having reached 15 digits past decimal point at the intermediate total of 108.

[CaptainEd]

Prime, your "disproof" notwithstanding, we should congratulate you on your insight that has identified the limit towards which any experiment to 137 should tend (IMH and Naive Opinion). I'm interested to understand about Bushindo's recursive code. My contribution was only to respond to Bushindo's numeric result, as it matches the cyclic stream of digits I learned in childhood.

[BobbyGo]

Thanks Bonanova. My explanation does look sound. But now, after I have taken some time to reflect, I must disprove and disown my solution.

Sure, the fraction that we found here is good for everyday gambling needs, but it is not the exact probability.

I suspect, Bushindo’s method would have produced the exact answer, had it not succumbed to rounding error.

 

Spoiler for Proof of falsehood, but not the answer.

Any number may or may not be an intermediate running total in this experiment. But there is one exception - zero! Zero running total will occur with 100% probability right before the very first roll.
Consequently, the probabilities for the first six numbers (1, 2, 3, 4, 5, and 6) to become intermediate totals are:
P₁= 1/6;
P₂ = (1+1/6)/6 = 7/6²;
P₃ = (1+1/6+7/6²)/6 = 7²/6³;
And in the like manner,
P₄  = 7³/6⁴;
P₅ = 7⁴/6⁵;
P₆ = 7⁵/6⁶;
Note, all the above probabilities are different from 2/7. First four are smaller, and the last two are higher.

From that point on we must change the method for probability calculation. Since a running total n can be obtained by a single roll from 6 preceding totals,
P_n = (P_n-1 + P_n-2 + P_n-3 + P_n-4 + P_n-5 + P_n-6)/6. Note, for P_n = 2/7, the sum of its six predecessors must be equal 12/7.

Let’s figure out the exact probability for P₇ using addition of the fractions, like they taught me at school. The common denominator for the first six probabilities is 6⁶, whereas the numerator of the sum becomes:
6⁵ + 7*6⁴ + 7²*6³ + 7³*6² + 7⁴*6 + 7⁵. Note, the numerator is modulo 1 with respect to 6.
Thus P₇ = (6⁵ + 7*6⁴ + 7²*6³ + 7³*6² + 7⁴*6 + 7⁵)/ 6⁷.

For P₈ we add up P₂ through P₇ and divide it by 6. To bring P₂ through P₆ fractions to the common denominator of 6⁷, we must multiply their numerators by various powers of 6. Whereas the numerator of P₇ remains modulo 1 with respect to 6 and so will the numerator of P₈.

Thus by induction we know that the exact fraction for P₁₃₇ has 6¹³⁷ in its denominator and a very big who knows what in the numerator, which is modulo 1 with respect to 6, however. The numerator of that fraction is not divisible by 2 and its denominator is not divisible by 7. Therefore, it cannot be exactly equal to 2/7. That big fraction cannot be simplified at all - the numerator and denominator have no common factors.

Whoever finds the numerator - will have solved the problem.

In Excel spreadsheet, the probabilities for intermediate totals oscillated around 2/7, until the application ran out of precision having reached 15 digits past decimal point at the intermediate total of 108.

 

I'm glad I wasn't too far off in my thinking.  I had a similar approach, but only considered 131 - 136 when trying to calculate the odds for 137.  I had wrongly assumed each number 131 - 136 would have an equal likelyhood of being hit irrespective of of the likelyhood of any of the previous numbers.

 

I do have a question about one part of your disproof:

 

 

Spoiler for Question

 

Let’s figure out the exact probability for P₇ using addition of the fractions, like they taught me at school. The common denominator for the first six probabilities is 6⁶, whereas the numerator of the sum becomes:
6⁵ + 7*6⁴ + 7²*6³ + 7³*6² + 7⁴*6 + 7⁵. Note, the numerator is modulo 1 with respect to 6.
Thus P₇ = (6⁵ + 7*6⁴ + 7²*6³ + 7³*6² + 7⁴*6 + 7⁵)/ 6⁷.

 

 

If the common denominator for the first six probabilities is 6⁶, why does the equation for P₇ have a denominator of 6⁷?

 

Shouldn't the equation be:

P₇ = (6⁵ + 7*6⁴ + 7²*6³ + 7³*6² + 7⁴*6 + 7⁵)/ 6⁶

(Which equals my attempted answer of about .2536043952903521)

 

If this is right, I would assume that the common denominator would cap at 6⁶ as, eventually, all probabilities would be averages of probabilities with a 6⁶ denominator; including P₁₃₇.

[bushindo]

Prime, your "disproof" notwithstanding, we should congratulate you on your insight that has identified the limit towards which any experiment to 137 should tend (IMH and Naive Opinion). I'm interested to understand about Bushindo's recursive code. My contribution was only to respond to Bushindo's numeric result, as it matches the cyclic stream of digits I learned in childhood.

 

Here is the recursive code I used. The logic behind it is already described by Prime's excellent and insightful analysis.

Spoiler for

Let P(N) be the probability that the total sum would eventually fall on number N. This code sets P(0) = 1, P( N ) = 0 for negative N, and then iterates recursively using a known formula as described in the following code

 
function P( N )
{
    if( N == 0 )
    {     return( 1 )    }    
    else if( N < 0 )
    {    return( 0 )    }
    else
    {
        A = 1/6*( P( N - 1 ) + P( N - 2 ) + P( N - 3 ) + P( N - 4 ) + P( N - 5 ) + P( N - 6 ) )
        return( A )
    }
}