View Full Version : No Formulas or Advanced Math

08-07-2002, 08:09 AM
You have three dollars. Your opponent has infinite wealth. You flip an unfair coin for a dollar a pop. Thc coin is weighted two to one in your favor. What is the probability that you will ever go broke if you play forever?

08-07-2002, 08:57 AM
It's a certainty you will go broke in an infinite contest.

08-07-2002, 09:07 AM
I would say 100%.

08-07-2002, 09:38 AM
Well if you're going to play forever you can't ever go broke /images/smile.gif I suspect what you mean (though one is never quite sure with you) is what is the chance that you will go broke sometime before you play forever.

Suppose instead of 3 dollars we only had 1 dollar. Then there would be a 1/3 chance that we go broke on the first flip. There would be a 2/3 chance that we win the first flip and double our bankroll to 2 dollars. If that happens, the chance of going broke from that point on would be whatever it was when we had 1 dollar squared since now we have to lose 1 dollar twice in order to go broke. So if p is the chance we will lose 1 dollar, we have

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

Or multiplying both sides by 3 and rearranging:

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

factoring gives

(2p - 1)(p - 1) = 0

So p = 1/2 or p = 0.

We reject p=0 since we already said there was a 1/3 chance of losing it on the first flip.

So there is a 1/2 chance of losing a 1 dollar bankroll. The chance of losing a 3 dollar bankroll then is (1/2)^3 = 1/8 since we have to lose the 1 dollar bankroll 3 times.

08-07-2002, 10:05 AM

08-07-2002, 10:05 AM
*(2p - 1)(p - 1) = 0

So p = 1/2 or p = 0. *

Wrong ! ... otherwise it looks GOOD !

You even got the semantics right.

But what about * No Formulas or Advanced Math * ;-)

08-07-2002, 12:42 PM
100% - you'll never win forever

08-07-2002, 01:28 PM
Why would you necessarily go broke in the long run? I would agree with all of you if this was a -EV game, but it isn't. If the answer is 100%, every blackjack card counter, +EV VP player, and sports better should give up and go home. There is a certain risk of ruin involved in this (and a high one with such a small bankroll), but I just can't see how you will inevitably go broke 100% of the time. All comments (read as: flames) are welcome.

-- Homer

08-07-2002, 01:56 PM
Yeah, I saw that after I posted and had to think about how to dismiss p=1.

There cannot be a probability of 1 for going broke after N flips for any N no matter how large. This is because there is always a non-zero probability that we do not go broke after any N flips. For example, it is always possible to win all N times with probability (2/3)^N. In fact, as long as our number of losses does not exceed our number of wins by 3, we will not go broke, and this will have a non-zero probability for any number of flips. So we can't have p=1 or else the probability of going broke plus not going broke would be greater than 1 which is nonsense. It would also imply that our opponent with a negative expectation would have a positive expectation of 3 dollars after N flips which we know can't be right.

So you can make as many flips as you please, and the probability of losing our last dollar will never exceed 1/2, and the probability of losing 3 dollars is 1/8.

08-07-2002, 04:08 PM
But if you have to play _forever_ sooner or later you'll have a swing that is big enough to eat your entire bankroll... I think...

Sincerely, Andreas

08-07-2002, 04:15 PM
Forever is a long time, eventually, the 1 quadzillion to 1 swing will happen, and you will get tapped out, but I think that for any finite period of time, there is probably a liklehood that if you can survive the early flips, you will be able to survive a long time. There must be a method for estimating your survival for a given period of time, and it reminds me of a calculus problem, but I have to go to work in about 5 minutes, so I cant add more now.

Good luck,

Play well,

Bob T.

08-07-2002, 04:36 PM
It will never be 100% as long as you have an edge on every play. It approaches 100% when your bankroll is 1 bet and your edge is close to even. It goes down exponentially as your bankroll increases (doubling bankroll squares risk of ruin). The games you mention are more difficult to analyze due to variable bets, variable payouts, and bets when your ev is negative. Still, if you have an overall edge, it is always possible to make your risk of ruin as small as you want by having a large enough bankroll relative to your bet size.

08-07-2002, 04:39 PM
Many people are of the misconception that the likelihood of tapping out become greater the longer you play. Actually, if you don't go bust fairly early, it is even less likely that you ever will. That is, the probability of going bust has a maxima.

08-07-2002, 04:40 PM
I computed the probability of ruin in 200 coin flips or less. I got


That would seem to indicate that 1/8 is the best answer.

08-07-2002, 04:53 PM
The reason this number is so close to 1/8 is that if you make it to 200 flips, you will almost certainly never go broke after that point.

08-07-2002, 10:28 PM
Need to brush up on my calculus, but the answer is whatever 1/3^3 + 1/3^5 + 1/3^7 + ... comes out to be (it's definitely not 1).

08-07-2002, 10:31 PM
OK, the answer is 1/24 (or 0.0416667).

08-07-2002, 10:39 PM
Also, if my math is correct, if you start with X dollars, and bet $1 per coin flip with this coin weighted 2/3 in your favor, the probability that you'll ever go broke is:

0.375 * (1/3)^(x-1)

Start w/ $1 = 3/8 chance you'll ever go broke

Start w/ $2 = 1/8 chance you'll ever go broke

Start w/ $3 = 1/24 chance you'll ever go broke

Start w/ $4 = 1/72 chance you'll ever go broke


08-07-2002, 11:06 PM
Here's how I came up with that answer

Start: $3 (All of the time)

1 Flip: $2 (1/3) $4 (2/3)

2 Flips: $1 (1/9) $3 (4/9) $5 (4/9)

3 Flips: $0 (1/27) $2 (6/27) $4 (12/27) $6 (8/27)

4 Flips: $0 (1/27 - from before) $1 (6/81) $3 (24/81) $5 (32/81) $7 (16/81)

5 Flips $0 (1/27 + 1/243) $2 (36/243)...

So the $0 component will eventually be (1/27 + 1/243 + 1/2187 + 1/19683 + ...)

08-07-2002, 11:14 PM
Same process as above, just realized a math error...

Probability will be (1/27 + 2/81 + 4/243 + 8/729 + ... + (2^x)/[3^(x+3)])

This reduces to 1/9.

08-08-2002, 02:46 AM
Responding to your own post 4 times. Good work!

On the other hand, you didn't do such a good job with the, "No formulas or advanced math" part. /images/smile.gif

08-08-2002, 03:56 AM
Because you have a one third chance of losing on any given bet, the odds that you will lose three times in a row are one third to the 3rd power, or one third times one third times one third with equals one eighteenth. This means that you will have a five and a half percent chance of going broke.

08-08-2002, 04:59 AM
Opps, My math was wrong. 3*3*3 = 27. 1/27 = .037 which rounded up is 4%

08-08-2002, 06:09 AM
I think the probability for you to go broke is 6.7%

08-08-2002, 08:28 AM
Call the probability of going broke with one dollar x. That means the probability with two dollars is x squared and with three dollars its x cubed. Why? Because the probability that a guy with two bucks falls to one buck is exactly the same as falling from one to zero when the opponent's bankroll is limitless. So to go from two to zero is x times x.

When you have one buck you can go broke two different ways. Either immediately, which has a probability of 1/3, or win the first bet and then go broke, which has a probability of 2/3 times x squared. Together the chances add up to x. x = 1/3 + (2/3)(x squared). So x is 1/2. X cubed is 1/8.

08-08-2002, 11:29 AM

08-08-2002, 01:24 PM
x = 1/3 + (2/3)(x squared) sure looks like a formula. So this is not an acceptable answer to your own problem!

08-08-2002, 06:39 PM
its not a formula or advanced math, its a simple algebraic equation /images/smile.gif

08-08-2002, 07:00 PM
Go back to Day 1 of Introduction to Calculus and review limits. Well, to be fair, maybe this has an added dimension that makes it a bit more complicated than Day 1 stuff. Still, it should be intuitively obvious what the answer is once you combine terms infinite and probability.

08-08-2002, 11:22 PM

08-09-2002, 12:25 AM
Mr. S,

I agree with BruceZ that you have to explain why a probabiliy of one is not correct. (It fits the formula given x = 1/3 + 2/3 x*x.)

email: scundal at YA HOO !!

(BruceZ, I still don't quite believe your explanation of why x is not 1. ( 7 August 2002 post) Could you email me your reasoning?)

08-09-2002, 04:17 PM
i think i agree with you here. the key is the term infinite.

david says that "When you have one buck you can go broke two different ways. Either immediately, which has a probability of 1/3, or win the first bet and then go broke"

this is true, but incomplete. what if you win the first and second bets, then go broke? what if you win the first 500 bets, then begin a disastrous 503-bet losing streak? there are simply more ways to lose in an infinite series than we are accounting for.

infinite string of bets implies the possibility of ALL bankroll-crushing downswings.

and in answer to Homer J Simpson (i think), the reason that poker players and blackjack players and other people who wager consistently with +EV don't quit and go home is that:

A) we don't play in an infinite game. we will only play a finite number of hands in our lives. the probability of going broke in a finite number of hands is MUCH less than 100% given the proper bankroll.

B) even if we do go broke on an unfortunate downswing of 'bad luck', we can always establish a new bankroll.

08-09-2002, 07:36 PM
You have completely missed the point. Winning the first bet and then going broke accounts for all the cases you mention. Think about it.

08-10-2002, 03:39 PM
i guess i see your point. then maybe you could explain to me how the probability of winning the first bet and then going broke is 2/3 times x squared? i understand where we're going, i just don't follow the math we are doing to get there. its not that i don't do math well. i just don't know how we got this particular equation.

08-11-2002, 02:04 PM
If you win the first bet, you will have doubled your bank to 2 dollars. The probability of losing that bank is x^2 since the probability of losing 1 dollar is x, and we have to lose 1 dollar twice to lose 2 dollars. Since we have to win the first flip in order to even make it to the second flip, and the probability of that is 2/3, we multiply the x^2 by 2/3. Also see my answer which says basically the same thing.