OT: Probability Puzzle

primelord · 10-09-2003, 04:57 PM

The last puzzle I posted didn't go over so well. But this one is less confusing and has a clearer answer. Usual disclaimer that I did not create this puzzle I am just re-posting it for your enjoyment.

You have $3. Your opponent has an infinite bank roll. You flip an unfair coin for a dollar a pop. The coin is weighted 2 to 1 in your favor. What is the probability that you will ever go broke if you play forever?

Maple Leafs · 10-09-2003, 09:45 PM

Quote:

Originally posted by primelord
What is the probability that you will ever go broke if you play forever?

Zero, because you'll still have your unfair coin.

Woot! Do I win?

John Galt · 10-09-2003, 09:56 PM

100% - It is the gambler's ruin.

Easy Mac · 10-09-2003, 09:58 PM

AnalBumCover · 10-09-2003, 11:00 PM

mckerney · 10-09-2003, 11:14 PM

Easy Mac · 10-09-2003, 11:25 PM

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Easy Mac · 10-09-2003, 11:27 PM

hmmm weird format, anyway, thats pi to 10000 places. Here's a link to 50 million http://www.cecm.sfu.ca/projects/ISC/data/pi.html. Some people have too much fucking time.

TredWel · 10-09-2003, 11:30 PM

After a lot of writing and mathematical hand-waving, I have come to the answer of 1/9.

Please don't ask me to show my work. You wouldn't like it.

primelord · 10-09-2003, 11:35 PM

Quote:

Originally posted by AnalBumCover
1/8

Ding Ding Ding!

ABC is correct.

panerd · 10-10-2003, 09:23 AM

I would agree with John Galt. If you are playing it infinitely, at some point you will hit a string of millions of loses in a row. So you will lose.

John Galt · 10-10-2003, 09:27 AM

I'm not sure about how the 1/8 answer is derived, but let me at least defend the 100% answer.

In a 50/50 game, the casino will always win because they have a virtually infinite bankroll and you don't. Eventually you will hit your bottom point (at which point you can't continue), but they will never hit their break point. That is the principle of a gambler's ruin.

In your example, the time frame is truly infinite and the opponent's bankroll is also infinite. While your odds on a single event bet are much better than 50/50, the result will eventually be the same. That is to say, over an infinite time frame, you will will have a losing streak long and bad enough to hit your break point at 0. "In the long run, we all are dead."

How is 1/8 derived?

primelord · 10-10-2003, 09:44 AM

Ok here is where the 1/8 comes from.

Let's say instead of 3 dollars you only had 1 dollar. That means on the first flip you have a 1/3 chance of going broke. There would then be a 2/3 (it always comes back to 2/3 doesn't it?

) chance that you will win the first flip and double your bankroll 2 $2. If you win the first flip the chance of going broke from that point on would be wahtever it was when we had 1 dollar squared since now we would have to lose 1 dollar twice in order to go broke. So let's say x is the chance we will go broke.

x = 1/3 + 2/3(x)^2

Multiply both sides by 3 and rearrange

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

Factoring gives you

(2x - 1)(x - 1) = 0

Leaving you with x = 1/2 or x = 0

x = 0 can't be correct because we have alread determined you have a 1/3 chance of going broke on the first flip. So that means if there is a 1/2 chance of losing your 1 dollar bankroll then the chance of losing a 3 dollar bankroll is (1/2)^3 = 1/8 since you have to lose the 1 dollar bankroll 3 times.

John Galt · 10-10-2003, 09:49 AM

How does that solution account for the infinite time frame? It seems to me to create an average expectation of what is going to happen, but over an infinite time frame, doesn't EVERYTHING happen?

primelord · 10-10-2003, 09:59 AM

Quote:

Originally posted by John Galt
How does that solution account for the infinite time frame? It seems to me to create an average expectation of what is going to happen, but over an infinite time frame, doesn't EVERYTHING happen?

Well what I expected you to point out to me was that x = 1 actually does make the original formula true. It's impossible to have a 1 probability of going broke no matter how many flips you have because you will always have a non zero probablity of not going broke on any number of flips. The probability of going broke can't be 1 because then the probability of going broke and the probability of not going broke would add up to be greater than 1 and that is obviously not possible.

So you can make as many flips as you like, but the probability of losing that last dollar will never exceed 1/2.

John Galt · 10-10-2003, 10:12 AM

Quote:

Originally posted by primelord
Well what I expected you to point out to me was that x = 1 actually does make the original formula true. It's impossible to have a 1 probability of going broke no matter how many flips you have because you will always have a non zero probablity of not going broke on any number of flips. The probability of going broke can't be 1 because then the probability of going broke and the probability of not going broke would add up to be greater than 1 and that is obviously not possible.

So you can make as many flips as you like, but the probability of losing that last dollar will never exceed 1/2.

Why can't the probability of not going broke be zero? This is the principle of a gambler's ruin. If you don't leave the table when your money is high (which you can never know) and you keep betting over an infinite time frame, you will eventually hit bottom.

To put it this way, there is no end point for winning - you just keep playing, but there is an end point for losing (zero) - over an infinite time frame, you will hit the losing end point, but never the winning end point (because there isn't one).

primelord · 10-10-2003, 10:57 AM

Sorry I got called into a couple of meetings. I'll address this in a bit.

primelord · 10-10-2003, 11:28 AM

Ok here is the problem with the gamblers ruin is that is based on a game where you have a negeative expected value. When you play blackjack the house has the edge so despite the fact that you could possibly win 1000 straight hands and build up a huge bankroll the odds are always against you and given enough time you will go bust. (This is of course ignoring the fact that counting cards gives you a slight edge.)

In this game your EV is always positive. The probability of you not going broke can never be 0 because no matter what you do when you get down to that last dollar the probability of losing that dollar is still going to be 1/2. As your banroll decreases the probability of ruin does increase, but it will never reach 1.

primelord · 10-10-2003, 11:32 AM

Another note just because the time frame is infinite doesn't mean every combination is garaunteed to come up. If you are flipping a coin it is possible that it will land on heads every time forever. Is it likely that will happen? No, but it is possible. So you couldn't say that the probability that the coin will hit tails is 1 even if you had an infinite number of flips. With each filp the odds you will get a tail does increase towards one, but it will never get there.

albionmoonlight · 10-10-2003, 11:33 AM

One thing that may help, John, is that the chance of going up $1,000,000 and then losing the coin flip 1,000,001 times is taken into account in coming up with the 1/8 answer.

1 out of every eight people who do this are going to go broke. The majority of them will get a string of bad flips early and that will be that. However, if you had a billion monkeys doing this for a billion years, you would have the occasional monkey who would bust after being up several million dollars. The key is that you would still only have 1 out of every eight monkeys bust.

John Galt · 10-10-2003, 11:43 AM

Speaking of monkies. I was just about to bring up the Infinite Monkey Thereom.

In that case you are dealing with an extremely low probability event (that a monkey will randomly create Hamlet on a typewriter if given enough time). Yet, subscribers to the theory believe that it is a certainty over an infinite time frame that Hamlet will be written.

Imagine it this way - the growth in your earnings can be graphed as a wave slanted up with random oscillations. At some point an oscillation (or series of oscillations) will be large enough to wipe out your earnings.

I understand that your formula creates a graph with an asymptote, but I think that graph is flawed. It would be great for predicting the number of wipeouts in a ridiculously long time frame (1 billion years will have 1/8 bankrupt), but a truly infinite time frame will create 100% bankrupt.

You have basic algebra (and basic calculus if you graph it out) on your side, but I'm defending a competitive notion of infinity that I think you are skipping past.

Bee · 10-10-2003, 11:54 AM

Can I borrow that coin after you guys finish with it?

primelord · 10-10-2003, 12:22 PM

While it seems logical that if you put an infinite number of monkies on a typewriter or a set number of monkies, but gave them infinite time hamlet would likely be recreated. However it would still not be a mathmatical certainty. With each addition monkey the number grows closer to 1, but it will never get there.

Assuming the monkies are truly picking keys at random then it is mathmatically possible they will type nothing but a's. All of them. It would be highly unlikely, but a possibility. And with each monkey added the likelihood of them typing nothing but a's decreases towards 0, but again will never get there.

John Galt · 10-10-2003, 01:39 PM

Quote:

Originally posted by primelord
While it seems logical that if you put an infinite number of monkies on a typewriter or a set number of monkies, but gave them infinite time hamlet would likely be recreated. However it would still not be a mathmatical certainty. With each addition monkey the number grows closer to 1, but it will never get there.

Assuming the monkies are truly picking keys at random then it is mathmatically possible they will type nothing but a's. All of them. It would be highly unlikely, but a possibility. And with each monkey added the likelihood of them typing nothing but a's decreases towards 0, but again will never get there.

I basically agree, but your point illustrates our different results. My asymptote approaches 1 - your asymptote approaches 1/8.

dixieflatline · 10-11-2003, 06:37 PM

Sorry I came in late to the discussion but I wanted to give a different approach to show why Primelord is right. Let's go back to considering that you start with just one dollar. The probability you will lose is a third. If you win you have two dollars and it would take two straight losing flips to bankrupt you. But if you win you have three dollars and it would take three straight losing flips and so on. Writing this in a series:

1/3+1/3²+1/3³+...

Each term then is the chance we directly bankrupt(or lose all our money on the least amount of flips) based on how much money we have. If we win the flip then it's the next term that is our probability.

Now the question is does this series converge? The answer is yes. This series is called a geometric series(because the power is increased by one each term). Series of this form converage like this:

a+a²+a³+...=-a/(a-1)

As long as a is between 0 and 1.

The proof is kind of long bust doesn't use any calculus just function notation trickery. Here is a link with the proof:
http://www.moneychimp.com/articles/finworks/fmgeom.htm

Plugging in 1/3 for a and we get 1/2. Note the extraneous solutions Primelord got are missing here. Then since we have three dollars the answer is 1/2³ or 1/8.

Which begs the question at what point does this breakdown and your probability of going bankrupt is 1? Well that point occurs when the series converges to 1. So set -a/(a-1) equal to 1 and solve for a. The answer is 1/2.

So if you were playing with a fair coin then the house will take all of your money eventually. This is probably what John Galt was refering to as gamblers ruin and the house doesn't need and advantage for this to occur.

10-09-2003, 09:56 PM	#3
John Galt Pro Starter Join Date: Oct 2000 Location: The Internets	100% - It is the gambler's ruin. __________________ I do mind, the Dude minds. This will not stand, ya know, this aggression will not stand, man. - The Dude

10-09-2003, 09:58 PM	#4
Easy Mac Registered User Join Date: Nov 2001 Location: Here	7/11

10-09-2003, 11:00 PM	#5
AnalBumCover College Starter Join Date: Nov 2002 Location: La Mirada, CA	1/8 __________________ ABC's Game Giveaway list

10-09-2003, 11:30 PM	#9
TredWel High School Varsity Join Date: Oct 2000 Location: Old Forge, PA	After a lot of writing and mathematical hand-waving, I have come to the answer of 1/9. Please don't ask me to show my work. You wouldn't like it. __________________ There are three things I have learned never to discuss with people...religion, politics, and the Great Pumpkin. - Linus Van Pelt Last edited by TredWel : 10-09-2003 at 11:30 PM.

10-10-2003, 09:27 AM	#12
John Galt Pro Starter Join Date: Oct 2000 Location: The Internets	I'm not sure about how the 1/8 answer is derived, but let me at least defend the 100% answer. In a 50/50 game, the casino will always win because they have a virtually infinite bankroll and you don't. Eventually you will hit your bottom point (at which point you can't continue), but they will never hit their break point. That is the principle of a gambler's ruin. In your example, the time frame is truly infinite and the opponent's bankroll is also infinite. While your odds on a single event bet are much better than 50/50, the result will eventually be the same. That is to say, over an infinite time frame, you will will have a losing streak long and bad enough to hit your break point at 0. "In the long run, we all are dead." How is 1/8 derived? __________________ I do mind, the Dude minds. This will not stand, ya know, this aggression will not stand, man. - The Dude

10-09-2003, 11:14 PM	#6
mckerney Coordinator Join Date: Oct 2000	2/3

10-09-2003, 11:25 PM	#7
Easy Mac Registered User Join Date: Nov 2001 Location: Here	3.1415926535897932384626433832795028841971693993751058209749445923078164062862 089986280348253421170679821480865132823066470938446095505822317253594081284811 174502841027019385211055596446229489549303819644288109756659334461284756482337 867831652712019091456485669234603486104543266482133936072602491412737245870066 063155881748815209209628292540917153643678925903600113305305488204665213841469 519415116094330572703657595919530921861173819326117931051185480744623799627495 673518857527248912279381830119491298336733624406566430860213949463952247371907 021798609437027705392171762931767523846748184676694051320005681271452635608277 857713427577896091736371787214684409012249534301465495853710507922796892589235 420199561121290219608640344181598136297747713099605187072113499999983729780499 510597317328160963185950244594553469083026425223082533446850352619311881710100 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10-09-2003, 11:27 PM	#8
Easy Mac Registered User Join Date: Nov 2001 Location: Here	hmmm weird format, anyway, thats pi to 10000 places. Here's a link to 50 million http://www.cecm.sfu.ca/projects/ISC/data/pi.html. Some people have too much fucking time.

10-10-2003, 09:23 AM	#11
panerd Grizzled Veteran Join Date: Jul 2001 Location: St. Louis	I would agree with John Galt. If you are playing it infinitely, at some point you will hit a string of millions of loses in a row. So you will lose.

10-10-2003, 09:44 AM	#13
primelord Pro Rookie Join Date: Oct 2000	Ok here is where the 1/8 comes from. Let's say instead of 3 dollars you only had 1 dollar. That means on the first flip you have a 1/3 chance of going broke. There would then be a 2/3 (it always comes back to 2/3 doesn't it? ) chance that you will win the first flip and double your bankroll 2 $2. If you win the first flip the chance of going broke from that point on would be wahtever it was when we had 1 dollar squared since now we would have to lose 1 dollar twice in order to go broke. So let's say x is the chance we will go broke. x = 1/3 + 2/3(x)^2 Multiply both sides by 3 and rearrange 2(x)^2 - 3x + 1 = 0 Factoring gives you (2x - 1)(x - 1) = 0 Leaving you with x = 1/2 or x = 0 x = 0 can't be correct because we have alread determined you have a 1/3 chance of going broke on the first flip. So that means if there is a 1/2 chance of losing your 1 dollar bankroll then the chance of losing a 3 dollar bankroll is (1/2)^3 = 1/8 since you have to lose the 1 dollar bankroll 3 times. __________________ .

10-10-2003, 09:49 AM	#14
John Galt Pro Starter Join Date: Oct 2000 Location: The Internets	How does that solution account for the infinite time frame? It seems to me to create an average expectation of what is going to happen, but over an infinite time frame, doesn't EVERYTHING happen? __________________ I do mind, the Dude minds. This will not stand, ya know, this aggression will not stand, man. - The Dude

10-10-2003, 10:57 AM	#17
primelord Pro Rookie Join Date: Oct 2000	Sorry I got called into a couple of meetings. I'll address this in a bit. __________________ .

10-10-2003, 11:28 AM	#18
primelord Pro Rookie Join Date: Oct 2000	Ok here is the problem with the gamblers ruin is that is based on a game where you have a negeative expected value. When you play blackjack the house has the edge so despite the fact that you could possibly win 1000 straight hands and build up a huge bankroll the odds are always against you and given enough time you will go bust. (This is of course ignoring the fact that counting cards gives you a slight edge.) In this game your EV is always positive. The probability of you not going broke can never be 0 because no matter what you do when you get down to that last dollar the probability of losing that dollar is still going to be 1/2. As your banroll decreases the probability of ruin does increase, but it will never reach 1. __________________ .

10-10-2003, 11:32 AM	#19
primelord Pro Rookie Join Date: Oct 2000	Another note just because the time frame is infinite doesn't mean every combination is garaunteed to come up. If you are flipping a coin it is possible that it will land on heads every time forever. Is it likely that will happen? No, but it is possible. So you couldn't say that the probability that the coin will hit tails is 1 even if you had an infinite number of flips. With each filp the odds you will get a tail does increase towards one, but it will never get there. __________________ .

10-10-2003, 11:33 AM	#20
albionmoonlight Head Coach Join Date: Oct 2000 Location: North Carolina	One thing that may help, John, is that the chance of going up $1,000,000 and then losing the coin flip 1,000,001 times is taken into account in coming up with the 1/8 answer. 1 out of every eight people who do this are going to go broke. The majority of them will get a string of bad flips early and that will be that. However, if you had a billion monkeys doing this for a billion years, you would have the occasional monkey who would bust after being up several million dollars. The key is that you would still only have 1 out of every eight monkeys bust.

10-10-2003, 11:43 AM	#21
John Galt Pro Starter Join Date: Oct 2000 Location: The Internets	Speaking of monkies. I was just about to bring up the Infinite Monkey Thereom. In that case you are dealing with an extremely low probability event (that a monkey will randomly create Hamlet on a typewriter if given enough time). Yet, subscribers to the theory believe that it is a certainty over an infinite time frame that Hamlet will be written. Imagine it this way - the growth in your earnings can be graphed as a wave slanted up with random oscillations. At some point an oscillation (or series of oscillations) will be large enough to wipe out your earnings. I understand that your formula creates a graph with an asymptote, but I think that graph is flawed. It would be great for predicting the number of wipeouts in a ridiculously long time frame (1 billion years will have 1/8 bankrupt), but a truly infinite time frame will create 100% bankrupt. You have basic algebra (and basic calculus if you graph it out) on your side, but I'm defending a competitive notion of infinity that I think you are skipping past. __________________ I do mind, the Dude minds. This will not stand, ya know, this aggression will not stand, man. - The Dude

10-10-2003, 11:54 AM	#22
Bee Pro Starter Join Date: Oct 2000 Location: Fairfax, VA	Can I borrow that coin after you guys finish with it?

10-10-2003, 12:22 PM	#23
primelord Pro Rookie Join Date: Oct 2000	While it seems logical that if you put an infinite number of monkies on a typewriter or a set number of monkies, but gave them infinite time hamlet would likely be recreated. However it would still not be a mathmatical certainty. With each addition monkey the number grows closer to 1, but it will never get there. Assuming the monkies are truly picking keys at random then it is mathmatically possible they will type nothing but a's. All of them. It would be highly unlikely, but a possibility. And with each monkey added the likelihood of them typing nothing but a's decreases towards 0, but again will never get there. __________________ .

10-11-2003, 06:37 PM	#25
dixieflatline High School Varsity Join Date: Sep 2003	Sorry I came in late to the discussion but I wanted to give a different approach to show why Primelord is right. Let's go back to considering that you start with just one dollar. The probability you will lose is a third. If you win you have two dollars and it would take two straight losing flips to bankrupt you. But if you win you have three dollars and it would take three straight losing flips and so on. Writing this in a series: 1/3+1/3²+1/3³+... Each term then is the chance we directly bankrupt(or lose all our money on the least amount of flips) based on how much money we have. If we win the flip then it's the next term that is our probability. Now the question is does this series converge? The answer is yes. This series is called a geometric series(because the power is increased by one each term). Series of this form converage like this: a+a²+a³+...=-a/(a-1) As long as a is between 0 and 1. The proof is kind of long bust doesn't use any calculus just function notation trickery. Here is a link with the proof: http://www.moneychimp.com/articles/finworks/fmgeom.htm Plugging in 1/3 for a and we get 1/2. Note the extraneous solutions Primelord got are missing here. Then since we have three dollars the answer is 1/2³ or 1/8. Which begs the question at what point does this breakdown and your probability of going bankrupt is 1? Well that point occurs when the series converges to 1. So set -a/(a-1) equal to 1 and solve for a. The answer is 1/2. So if you were playing with a fair coin then the house will take all of your money eventually. This is probably what John Galt was refering to as gamblers ruin and the house doesn't need and advantage for this to occur.

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