16 replies to this topic

[CaptainEd]

Spoiler for

so you have a 1/4 chance of going broke the first roll obviously.

you have a 1/4 chance of just getting your dollar back, and then another 1/4 chance of going broke.(1/16)

you have a 1/4 chance of getting an additional dollar, and then a 1/4 chance for each of losing them. (1/64)

and so on.

I think the same words would apply equally well if the die had sides reading 0,1,1,1, rather than 0,1,2,3, yet I feel the probability of going bust should be different. Perhaps that's the trick that I'm falling for.

[CaptainEd]

Spoiler for Here's an argument

Let P(n) be the probability of going bust, starting with N dice, where a die is said to go bust if all of its descendants go bust. The descendants of a die are those dice that are put into the gambler's stake after a successful wager, and (recursively) the descendants of those dice.

 

I believe P(2) = P(1)^2, because in order to go bust with both dice, each die needs to go bust, and they are independent.
Similarly, P(3) = P(1)^3.

 

So, what is P(1)?

P(1) = 1/4 (1 + P(1) + P(2) + P(3))
= 1/4 (1 + P(1) + P(1)^2 + P(1)^3)

 

so 3P(1) = 1 + P(1)^2 + P(1)^3

P(1)^3 + P(1)^2 -3P(1) + 1 = 0

The roots of this equation are 1, sqrt(2)-1, and sqrt(2)+1.

In the interval of interest, we have two roots: sqrt(2)-1 and 1.

I'm inclined to rule out 1 (without good justification--this is infinity we're talking about...), yielding a probability of (eventually) going bust of 0.414.

[bushindo]

You glue two small square pyramids together at their bases to form an eight-sided object that becomes a fair die. You mark the opposite pairs of faces with 0 1 2 and 3.

You begin the game with $1. You roll the die; the number that shows is your new stake. That is, with equal probability you lose you dollar, you keep your dollar, you double your dollar or you triple your dollar.

A minute later, you bet each of your dollars, if any remain after the first roll, with another roll of the die; one roll for each dollar that you have, and collect your winnings, if any. After another minute passes, each of your dollars, if any, suffers the fate of another roll. To be clear, in all cases each dollar is wagered individually. As the minutes turn into years, you eventually become rich, or you go bust.

What is the probability that you go bust?

I like the conditions that Prime gave in his last game. You can simulate the game if you like. But a submitted solution must comprise an answer and a method, both of which are correct. Enjoy!

 

 

Here's an approach

Spoiler for

Define P(N) as the probability of going bust if you currently have N dollars. Obviously, P(0) = 1, and the quantity of interest is P(1). Note that P(N) for large N should be approaching 0.

The procedure to find P( 1 ) is as follows,

Construct a series of equations for P(N) as in the following

P(N) = a₀ * P(0 ) + a₁ * P(1 ) + a₂ * P(2 ) + … + a_n P(N)

where a_n is easily defined based on the properties of the rolling dice. For instance, the first 3 equations will be

 

P(1) = (1/4) * P(0 ) + (1/4) * P(1) + (1/4) * P(2 ) + (1/4) P(3)
P(2) = 0.0625 * P(0 ) + 0.1250* P(1 ) + 0.1875* P(2 ) +  0.2500* P(3 ) +  0.1875* P(4 ) +  0.1250* P(5 ) +  0.0625 * P(6 )
P(3) = 0.015625* P(0 ) + 0.046875* P(2 ) + 0.093750* P(2 ) + 0.156250* P(3 ) + 0.187500* P(4 ) + 0.187500* P(5 ) + 0.156250* P(6 ) + 0.093750* P(7 ) + 0.046875* P(8 ) + 0.015625* P(9 )

The key now is that we will assume P(N) = 0 for N >=  30. This is a reasonable assumption because a rough order-of-magnitude guess for P(30) = (1/4)^(30) =  8e-19, so that's close enough to zero.

So, we we construct 30 equations P(N) for  0 <= N <= 29 as described above. This system of 30 equations can be rewritten as a matrix equation
A * p = p

where A is a matrix constructed about of the coefficients a_n, and p is the 30-dimensional vector (P(0), P(1), P(2), … , P(29 ) ).

p is obviously an eigenvector of A corresponding to the eigenvalue 1. It is then straightforward to find p using eigen decomposition. The value for P(1) is .4142136. The bust probabilities for all starting dollar values between 1 and 29 are

 

 

 [1] 4.142136e-01 1.715729e-01 7.106781e-02 2.943725e-02 1.219331e-02
 [6] 5.050634e-03 2.092041e-03 8.665518e-04 3.589375e-04 1.486768e-04
[11] 6.158394e-05 2.550890e-05 1.056613e-05 4.376635e-06 1.812861e-06
[16] 7.509115e-07 3.110374e-07 1.288356e-07 5.336514e-08 2.210427e-08
[21] 9.155624e-09 3.792164e-09 1.570594e-09 6.504336e-10 2.693280e-10
[26] 1.114981e-10 4.614419e-11 1.908862e-11 7.891683e-12
 

 

[bonanova]

CaptainEd, Good solve.
You're on a roll.

[bonanova]

You glue two small square pyramids together at their bases to form an eight-sided object that becomes a fair die. You mark the opposite pairs of faces with 0 1 2 and 3.

You begin the game with $1. You roll the die; the number that shows is your new stake. That is, with equal probability you lose you dollar, you keep your dollar, you double your dollar or you triple your dollar.

A minute later, you bet each of your dollars, if any remain after the first roll, with another roll of the die; one roll for each dollar that you have, and collect your winnings, if any. After another minute passes, each of your dollars, if any, suffers the fate of another roll. To be clear, in all cases each dollar is wagered individually. As the minutes turn into years, you eventually become rich, or you go bust.

What is the probability that you go bust?

I like the conditions that Prime gave in his last game. You can simulate the game if you like. But a submitted solution must comprise an answer and a method, both of which are correct. Enjoy!

  Here's an approach

Spoiler for

Define P(N) as the probability of going bust if you currently have N dollars. Obviously, P(0) = 1, and the quantity of interest is P(1). Note that P(N) for large N should be approaching 0.

The procedure to find P( 1 ) is as follows,

Construct a series of equations for P(N) as in the following

P(N) = a₀ * P(0 ) + a₁ * P(1 ) + a₂ * P(2 ) + + a_n P(N)

where a_n is easily defined based on the properties of the rolling dice. For instance, the first 3 equations will be

P(1) = (1/4) * P(0 ) + (1/4) * P(1) + (1/4) * P(2 ) + (1/4) P(3)P(2) = 0.0625 * P(0 ) + 0.1250* P(1 ) + 0.1875* P(2 ) +  0.2500* P(3 ) +  0.1875* P(4 ) +  0.1250* P(5 ) +  0.0625 * P(6 )P(3) = 0.015625* P(0 ) + 0.046875* P(2 ) + 0.093750* P(2 ) + 0.156250* P(3 ) + 0.187500* P(4 ) + 0.187500* P(5 ) + 0.156250* P(6 ) + 0.093750* P(7 ) + 0.046875* P(8 ) + 0.015625* P(9 )

The key now is that we will assume P(N) = 0 for N > 30. This is a reasonable assumption because a rough order-of-magnitude guess for P(31) = (1/4)^(31) =  2.168404e-19, so that's close enough to zero.

So, we we construct 30 equations P(N) for 1 <=0 <= 30 as described above. This system of 30 equations can be rewritten as a matrix equation
A * p = p

where A is a matrix constructed about of the coefficients a_n, and p is the 30-dimensional vector (P(1), P(2), P(3), , P(30 ) ).

p is obviously an eigenvector of A corresponding to the eigenvalue 1. It is then straightforward to find p using eigen decomposition. The value for P(1) is .4142136. The bust probabilities for all starting dollar values between 1 and 29 are  

 [1] 4.142136e-01 1.715729e-01 7.106781e-02 2.943725e-02 1.219331e-02 [6] 5.050634e-03 2.092041e-03 8.665518e-04 3.589375e-04 1.486768e-04[11] 6.158394e-05 2.550890e-05 1.056613e-05 4.376635e-06 1.812861e-06[16] 7.509115e-07 3.110374e-07 1.288356e-07 5.336514e-08 2.210427e-08[21] 9.155624e-09 3.792164e-09 1.570594e-09 6.504336e-10 2.693280e-10[26] 1.114981e-10 4.614419e-11 1.908862e-11 7.891683e-12 

 

I think you have it.
The Ap=p is the key and A can be the first part of your matrix.
Nice solve.
Cap'n beat you by only a few minutes. ;(

[bushindo]

Spoiler for Here's an argument

Let P(n) be the probability of going bust, starting with N dice, where a die is said to go bust if all of its descendants go bust. The descendants of a die are those dice that are put into the gambler's stake after a successful wager, and (recursively) the descendants of those dice.

 

I believe P(2) = P(1)^2, because in order to go bust with both dice, each die needs to go bust, and they are independent.
Similarly, P(3) = P(1)^3.

 

So, what is P(1)?

P(1) = 1/4 (1 + P(1) + P(2) + P(3))
= 1/4 (1 + P(1) + P(1)^2 + P(1)^3)

 

so 3P(1) = 1 + P(1)^2 + P(1)^3

P(1)^3 + P(1)^2 -3P(1) + 1 = 0

The roots of this equation are 1, sqrt(2)-1, and sqrt(2)+1.

In the interval of interest, we have two roots: sqrt(2)-1 and 1.

I'm inclined to rule out 1 (without good justification--this is infinity we're talking about...), yielding a probability of (eventually) going bust of 0.414.

 

Good solve, CaptainEd. Nice and compact solution =)

[CaptainEd]

Spoiler for Here's an argument

Let P(n) be the probability of going bust, starting with N dice, where a die is said to go bust if all of its descendants go bust. The descendants of a die are those dice that are put into the gambler's stake after a successful wager, and (recursively) the descendants of those dice.

 

I believe P(2) = P(1)^2, because in order to go bust with both dice, each die needs to go bust, and they are independent.
Similarly, P(3) = P(1)^3.

 

So, what is P(1)?

P(1) = 1/4 (1 + P(1) + P(2) + P(3))
= 1/4 (1 + P(1) + P(1)^2 + P(1)^3)

 

so 3P(1) = 1 + P(1)^2 + P(1)^3

P(1)^3 + P(1)^2 -3P(1) + 1 = 0

The roots of this equation are 1, sqrt(2)-1, and sqrt(2)+1.

In the interval of interest, we have two roots: sqrt(2)-1 and 1.

I'm inclined to rule out 1 (without good justification--this is infinity we're talking about...), yielding a probability of (eventually) going bust of 0.414.

 

Good solve, CaptainEd. Nice and compact solution =)

Thank you both so much!