ROR Derivation Question

[DavidC]

I'm looking at blackjack stuff right now, but I'm looking to really understand what I'm doing before I just start following the examples of people that know what they're doing... "Raise preflop," has never been good enough advice for me unless I knew why.

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Some variables:

Evh is "EV per hand" and is expressed in units.

Evh = B/Wr - Ha where B is the bonus, Ha is the house edge, and Wr is the wagering requirements of the bonus.

Evb is B - Wr*Ha: "The EV of the bonus." and is expressed in absolute terms. It's not relevant to this discussion.

Rw is "riskiness over reward" and is Var/Evh.

Rr is risk of ruin.

Br is bankroll in units.

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Rr = exp(-2*Br*Rw), therefore
Br = -Rw*ln(Rr)/2

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I'm looking to figure out why the Rr formula is the way it is. [img]/images/graemlins/smile.gif[/img]

Looking at an old Bruce post, I run into the following formula for a simple coinflip game:

more variables:

"p" is your chance of winning the flip and "q" is chance of losing.

Therefore p=1-q and q=1-p.

Now they calculate risk of ruin r as:

r = q + (1-q)*r^2.

I'm really confused by what r would be on the right side of this formula. Is it possible to give p,q, and r a value at this point on the right side of the formula in order to come up with a value for r on the left side?

I'm going to assume we can rewrite that formula (at this point) as r=q+p*r^2, letting {p,q} = 0.5

r = 0.5+r^2/2

How do we find out what r^2 is?

Edit: Here's the link to the Bruce post... this is basically where I'm stuck.

--Dave.

Edit: I've read through like three posts on this subject, as well as some stuff on the Central Limit Theorem with Statistical Fine Print and I'm pretty thoroughly confused. I like math, but this is a bit over my head right now.

[BruceZ]

[ QUOTE ]
Now they calculate risk of ruin r as:

r = q + (1-q)*r^2.

I'm really confused by what r would be on the right side of this formula. Is it possible to give p,q, and r a value at this point on the right side of the formula in order to come up with a value for r on the left side?

[/ QUOTE ]

r is the risk of ruin for a $1 bankroll if we play forever. It is the same on both sides of the equation.

You have found the correct derviation.

[DavidC]

[ QUOTE ]
[ QUOTE ]
Now they calculate risk of ruin r as:

r = q + (1-q)*r^2.

I'm really confused by what r would be on the right side of this formula. Is it possible to give p,q, and r a value at this point on the right side of the formula in order to come up with a value for r on the left side?

[/ QUOTE ]

r is the risk of ruin for playing forever. It is the same on both sides of the equation.

You have found the correct derviation.

[/ QUOTE ]

ok...

I'm still stuck at the "first neat trick", where r=q+(1-q)*r^2...

This shows that we can bust either by winning once then busting, or by busting outright on the first flip, if we have a bankroll of 1 unit.

However, there's some chance that we're going to win 3 flips and then bust also, right?

So would this look more like:

r = q + (p)*r^2 + p^2*r^3... etc.?

Eventually we could get a number here that looked kinda like a constant, because the changes would be so small...

[BruceZ]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
Now they calculate risk of ruin r as:

r = q + (1-q)*r^2.

I'm really confused by what r would be on the right side of this formula. Is it possible to give p,q, and r a value at this point on the right side of the formula in order to come up with a value for r on the left side?

[/ QUOTE ]

r is the risk of ruin for playing forever. It is the same on both sides of the equation.

You have found the correct derviation.

[/ QUOTE ]

ok...

I'm still stuck at the "first neat trick", where r=q+(1-q)*r^2...

This shows that we can bust either by winning once then busting, or by busting outright on the first flip, if we have a bankroll of 1 unit.

However, there's some chance that we're going to win 3 flips and then bust also, right?

[/ QUOTE ]

That is included in the second term where we win the first flip and then go bust after that. There are only 2 cases. The second term includes all cases where we win the first bet and then go broke at any time after that, no matter how many wins and losses that takes. The probability of this is r^2 because however we go bust, we must lose a $1 bankroll twice, and r is the probability of losing $1 once.

[DavidC]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
Now they calculate risk of ruin r as:

r = q + (1-q)*r^2.

I'm really confused by what r would be on the right side of this formula. Is it possible to give p,q, and r a value at this point on the right side of the formula in order to come up with a value for r on the left side?

[/ QUOTE ]

r is the risk of ruin for playing forever. It is the same on both sides of the equation.

You have found the correct derviation.

[/ QUOTE ]

ok...

I'm still stuck at the "first neat trick", where r=q+(1-q)*r^2...

This shows that we can bust either by winning once then busting, or by busting outright on the first flip, if we have a bankroll of 1 unit.

However, there's some chance that we're going to win 3 flips and then bust also, right?

[/ QUOTE ]

That is included in the second term where we win the first flip and then go bust after that. There are only 2 cases. The second term includes all cases where we win the first bet and then go broke at any time after that, no matter how many wins and losses that takes. The probability of this is r^2 because however we go bust, we must lose a $1 bankroll twice, and r is the probability of losing $1 once.

[/ QUOTE ]

Wow.

Okay, I'm going to continue with the derivation and see if I run into any other problems. Thanks for your help, Bruce.

[BruceZ]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
Now they calculate risk of ruin r as:

r = q + (1-q)*r^2.

I'm really confused by what r would be on the right side of this formula. Is it possible to give p,q, and r a value at this point on the right side of the formula in order to come up with a value for r on the left side?

[/ QUOTE ]

r is the risk of ruin for playing forever. It is the same on both sides of the equation.

You have found the correct derviation.

[/ QUOTE ]

ok...

I'm still stuck at the "first neat trick", where r=q+(1-q)*r^2...

This shows that we can bust either by winning once then busting, or by busting outright on the first flip, if we have a bankroll of 1 unit.

However, there's some chance that we're going to win 3 flips and then bust also, right?

[/ QUOTE ]

That is included in the second term where we win the first flip and then go bust after that. There are only 2 cases. The second term includes all cases where we win the first bet and then go broke at any time after that, no matter how many wins and losses that takes. The probability of this is r^2 because however we go bust, we must lose a $1 bankroll twice, and r is the probability of losing $1 once.

[/ QUOTE ]

Wow.

Okay, I'm going to continue with the derivation and see if I run into any other problems. Thanks for your help, Bruce.

[/ QUOTE ]

In textbook derivations of gambler's ruin for the coin flip game, this is usually done with a difference equation instead of a quadratic. I have given that derivation here, and it leads to the same result. This is just for the coin flip game against an opponent with a finite bankroll, and the derivation in Sileo's paper starts with the end result of this derivation. Note that against an opponent with an infinite bankroll, T->infinity, and the risk of ruin reduces to (q/p)^B, which agrees with the quadratic derivation. The quadratic equation was intended to be a simplification for people who might not be familiar with difference equations. The difference equation derivation is actually stronger, because the quadratic equation does not exclude r = 1 as a possible solution.