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# Rainman

Member Since 30 Jul 2012
Offline Last Active Yesterday, 12:54 AM

### In Topic: Comparing gambling systems

25 November 2013 - 11:24 AM

...........

Theorem: the probability of winning (or even getting at least a fixed non-zero return) approaches 0 as the number of rolls approaches infinity.

Proof: There are six equally likely outcomes of one roll, let's denote them a1 through a6 (a1 being 0.7 and ascending to a6 being 1.5). Let fi denote the relative frequency of the outcome ai during a sequence of rolls. For example, if we roll 12 times and get 1*a1, 3*a2, 0*a3, 1*a4, 2*a5, and 5*a6, then f1 = 1/12, f2 = 3/12, f3 = 0/12, f4 = 1/12, f5 = 2/12, and f6 = 5/12. The exact number of outcomes ai is the relative frequency fi times N. The final outcome of our game will be G = 0.7f1N*0.8f2N*0.9f3N*1.1f4N*1.2f5N*1.5f6N = (0.7f1*0.8f2*0.9f3*1.1f4*1.2f5*1.5f6)N.

The expected value of each fi is 1/6. By the law of large numbers, P(0.1666<fi<0.1667, for all i) approaches 1 as N approaches infinity. But if 0.1666<fi<0.1667, then G = (0.7f1*0.8f2*0.9f3*1.1f4*1.2f5*1.5f6)N < (0.70.1666*0.80.1666*0.90.1666*1.10.1667*1.20.1667*1.50.1667)N < 0.9998N. So the probability approaches 1 that your outcome is smaller than a value which approaches 0 as N approaches infinity. Conversely, for any given E>0, the probability approaches 0 that your outcome is at least E.

I like the reasoning, but the proof by example uses a slanted interval. The true Median of 1/6 is not in the middle of the interval 0.1666 to 0.16667. What happens if you use a truly median interval, e.g., from 9/60 to 11/60? It seems to be leading to the opposite conclusion.

I chose a slanted interval because it's shorter to write out. It has no bearing on the conclusion. If you insist on a median interval, we can use the interval 99999/600000<x<100001/600000 instead. The probability of the relative frequencies being in this interval approaches 1. If 99999/600000<x<100001/600000, then we can certainly conclude that 0.1666<x<0.1667, which I have already proven to imply that the payoff is less than 1 and tends to 0.

There is that little technicality in the above post. And then there are couple other points to make. Suppose, each of the 6 individual variants deviate within a chosen interval with a probability approaching 1. Don't we need to show that the ratio of all combinations of the deviating variants to all possible combinations (of 6N, where N tends to infinity) also approaches 1?

And then, of course, that “Law of Big Numbers”. Who passed, that law? When was it ratified? How is it enforced? The entire proof is riding on it.

But, I suppose, that is beyond the scope of this forum, unless there is some clever and funny way of demonstrating that Normal Distribution thingy without using any integrals. So never mind that.

If I accept the proof for ending up with nothing after riding my winnings infinite number of times, there are only few differences remaining in the interpretation of the OP and gambling philosophies.

1) Does BMAD's casino compel its patrons to play forever once they started? I hope -- not. At least the OP does not mention it. (Could be in the fine print, though:)

2) While I am being held to playing a pure form of riding the winnings (Post#38), somehow, Rainman is allowed a modification of betting just a half of his bankroll on each turn (Post#40). Which is a very good prudent winning form, but I like my modification (Post#33) with the minimum bet requirement of \$1 and a bankroll of \$20 better -- it seems faster and more exciting.

3) Suppose we have limited time, limited resources, and BMAD allows his patrons to leave the game whenever they want to cash in their winnings. Then what is better: showing up at the casino with a bunch of singles and betting exactly \$1 on each turn until closing time; or walking in with just \$1 and riding your entire bankroll on each turn until satisfied with the amount won (more than a million) or rolling the die for 1000 times, whichever comes first?

The riding seems more appealing to me. And 1000 consecutive rides look very good. After all, it is a long long way from 1000 to infinity.

With 1000-roll riding, you cannot lose more than \$1;

and your chance of winning more than a Million \$\$ looks like 4%, or better*.

Whereas betting \$1 at a time, you are basically working for \$33 all day long.

So choose.

*My simulation does not stop at the Million \$\$, but keeps rolling until 1000 rolls are done. In so doing it loses in some cases the millions procured in the interim, or wins some crazy amounts to the tune of 1012, which BMAD's casino is not able to honor.

Again, I invite anyone, who feels up to the challenge, to post theoretical justification (or disproof) to the statistics of a 1000-roll ride presented here.

http://en.wikipedia....f_large_numbers

It is a proven theorem.

### In Topic: Comparing gambling systems

25 November 2013 - 03:14 AM

In other words, even if you show that the probability of the ratio of variants within any given interval around the Median tends to 1, you'd still have to show that the resulting Payoff within that interval is zero.

Done that. Twice. If the ratios are all within the interval 0.1666<x<0.1667 then the resulting payoff is always less than 1, and it tends to zero as N tends to infinity. An upper bound for the payoff is: Payoff < (0.70.1666*0.80.1666*0.90.1666*1.10.1667*1.20.1667*1.50.1667)N < 0.9998N < 1. In plain words: over time, one hundred percent of the variations will be close enough to the median variation, that their payoff is zero.

To understand why this matters, do not concern yourself with the nominal values of the stakes and payoffs, but rather with their effect on your bankroll.

If your bankroll is \$1, and you bet \$1 and hit 0.7, your bankroll is reduced from 1 to 0.7, a change by the factor 0.7. Here the nominal change equals the bankroll change.

However, if your bankroll had been \$2 instead, and you bet \$1 and hit 0.7, your bankroll would be reduced from 2 to 1.7, a change by the factor 0.85. Because you only bet half your bankroll, the change is not as dramatic. If you bet half your bankroll instead of your entire bankroll, the payoffs would result in a change of 0.85, 0.9, 0.95, 1.05, 1.1, or 1.25. The geometrical mean of these changes is (0.85*0.9*0.95*1.05*1.1*1.25)1/6 ~ 1.008, which means you now have a positive expected effect on your bankroll. If you bet half your bankroll every time, you can expect it to grow exponentially (1.008N).

Now we might ask, when is the inflection point between a good game and a bad game? How much of your bankroll can you bet before you expect to lose money?

Well, if you bet the fraction x of your bankroll, the payoffs would be 0.7x, 0.8x, 0.9x, 1.1x, 1.2x, and 1.5x. Adding the 1-x you have remaining on the side, the change in bankroll would be 1-0.3x, 1-0.2x, 1-0.1x, 1+0.1x, 1+0.2x, or 1+0.5x. The geometrical mean would be GM = ((1-0.3x)(1-0.2x)(1-0.1x)(1+0.1x)(1+0.2x)(1+0.5x))1/6, which can be solved for x, at GM=1. The solution in the interval 0<x<1 is a bit over 0.989, which means the game is good for you as long as you bet between 0% and 98,9% of your bankroll.

But why am I not concerned with the nominal values of the winnings, but rather their effect on my bankroll? Consider this: if you had a \$1,000,000 total fortune (this includes your house and valuables), would you risk it all for a 50% chance to triple it? Would you risk it all for an 80% chance to double it? Would you risk \$999,000 of it for a 60% chance to double it? Very few people would be silly enough to take those risks, even though the EV is very good. The true difference between having one million and being dead broke is a lot more dramatic than the difference between having two millions and having one million, but the nominal difference is the same.

### In Topic: Comparing gambling systems

24 November 2013 - 04:22 PM

Prime, if you have a bankroll of \$20 and can run method 2 twenty times, then of course you're going to win. No one here is arguing against that. But method 2 in OP does not say anything about you being able to reset your stake at \$1 whenever you feel like it. You get one shot. One. Shot. For all purposes your bankroll is \$1. If you lose you lose. The most likely result is you will lose. Your simulations all show that. If you run 10 simulations of 1000 rolls each and 6 of them are losing, it means 6/10 people will lose. The other four people will win. As N grows larger the probability of winning approaches 0, as I have proven mathematically. Which means, if all 7,000,000,000 people in the world run method 2 for long enough, the probability approaches 1 that they will all lose.

### In Topic: Leslie Young in Grave Secrets of the Elderly

23 November 2013 - 06:07 PM

Oh, pish

### In Topic: Comparing gambling systems

22 November 2013 - 10:24 PM

So you're saying you wouldn't play a lottery with a 1/1,000,000 chance of winning \$5,000,000 with a \$1 ticket?

+1

Precisely. For \$1 you'd be buying \$5 worth of chances (provided the pot can't be split between several winners.) That's how I play lottery. In gambling EV is what counts.

.....

Theorem: the probability of winning (or even getting at least a fixed non-zero return) approaches 0 as the number of rolls approaches infinity.

Proof: There are six equally likely outcomes of one roll, let's denote them a1 through a6 (a1 being 0.7 and ascending to a6 being 1.5). Let fi denote the relative frequency of the outcome ai during a sequence of rolls. For example, if we roll 12 times and get 1*a1, 3*a2, 0*a3, 1*a4, 2*a5, and 5*a6, then f1 = 1/12, f2 = 3/12, f3 = 0/12, f4 = 1/12, f5 = 2/12, and f6 = 5/12. The exact number of outcomes ai is the relative frequency fi times N. The final outcome of our game will be G = 0.7f1N*0.8f2N*0.9f3N*1.1f4N*1.2f5N*1.5f6N = (0.7f1*0.8f2*0.9f3*1.1f4*1.2f5*1.5f6)N.

The expected value of each fi is 1/6. By the law of large numbers, P(0.1666<fi<0.1667, for all i) approaches 1 as N approaches infinity. But if 0.1666<fi<0.1667, then G = (0.7f1*0.8f2*0.9f3*1.1f4*1.2f5*1.5f6)N < (0.70.1666*0.80.1666*0.90.1666*1.10.1667*1.20.1667*1.50.1667)N < 0.9998N. So the probability approaches 1 that your outcome is smaller than a value which approaches 0 as N approaches infinity. Conversely, for any given E>0, the probability approaches 0 that your outcome is at least E.

I think, you've misunderstood what I was asking to prove.

With N rolls there are 6N total variations. There are TW variations ending in a payoff P > 1 and TL variations ending with the final payoff P < 1. TW+TL=6N.

Find the limit of TW/6N as N tends to infinity.

I agree that "Expected Outcome" tends to zero as the number of rolls N tends to infinity. I just don't see it all that relevant to winning in this game. If this game was offered in casinos, I would have most certainly won millions in a matter of days starting with a bankroll of \$100 or so. And I wouldn't let the entire \$100 ride in just one sitting. Money management still counts. However, exponential accumulation of winnings should be the predominant scheme to win big.

Another sure thing is -- the Casino offering that game would go bust very quickly.

I proved that the probability of a randomly selected variation ending in a payoff P>1 approaches 0 as N approaches infinity. TW/6N is the probability of a random variation ending in a payoff P>1. Hence I have proven that TW/6N approaches 0 as N approaches infinity. Simultaneously, I also proved that this is true not just for payoffs P>1, but for payoffs P>E, as long as E>0. For example, if we let Tp be the number of variations that give at least a penny back, then Tp/6N also approaches 0 as N approaches infinity.

I will try to explain my proof step by step:

Given a randomly created variation of N rolls, we can count the relative frequency of each result. Let f1 be the relative frequency of 0.7-rolls, which is defined as the number of 0.7-rolls divided by the total number of rolls N. It follows that the number of 0.7-rolls is f1*N. Similarly, let f2 be the relative frequency of 0.8-rolls, f3 the relative frequency of 0.9-rolls... and so on until f6, being the relative frequency of 1.5-rolls.

Now, consider the interval 0.1666<x<0.1667. Clearly 1/6 is an internal point in this interval. Because the expected values of f1 through f6 all equal 1/6, the law of large numbers guarantees the following as N approaches infinity: the probability approaches 1 that f1 through f6 will all be within this interval. A direct implication* of f1 through f6 all being within this interval is that the payoff is less than 1. Replacing the clause "f1 through f6 will all be within this interval" with its direct implication "the payoff is less than 1" in the bold-face text above, we get that the probability approaches 1 that the payoff is less than 1**. If the probability of A approaches 1, then the probability of (not A) approaches 0. So the probability approaches 0 that the payoff is not less than 1. Consequently the probability that you get a payoff P>1 approaches 0.

Since all variations are equally likely, the probability that you get a payoff P>1 is equal to the number of such variations divided by the total number of variations, i.e. TW/6N. So replacing "the probability that you get a payoff P>1" with its equal "TW/6N", we get the sought result: TW/6N approaches 0.

*I proved this implication in my original proof (the inequality part), but I will try to explain that proof as well. So what I'm trying to prove is the implication "if f1 through f6 are all within the interval 0.1666<x<0.1667, then the payoff is less than 1".

We can write the payoff as P = 0.7A*0.8B*0.9C*1.1D*1.2E*1.5F, where A is the number of 0.7-rolls, B is the number of 0.8-rolls, etc. But as already stated, the number of 0.7-rolls is f1*N, the number of 0.8-rolls is f2*N, etc. So replacing A through F with f1*N through f6*N, and extracting the common exponent N, we get the payoff as P = (0.7f1*0.8f2*0.9f3*1.1f4*1.2f5*1.5f6)N. This function will increase as f1,f2, and f3 decrease (because they are exponents for base numbers less than 1), and it will increase as f4,f5, and f6 increase (because they are exponents for base numbers greater than 1). So to maximize the payoff, we should minimize f1 through f3, and maximize f4 through f6. Setting f1 through f3 as 0.1666 and f4 through f6 as 0.1667, we get an upper bound for the payoff: P < (0.70.1666*0.80.1666*0.90.1666*1.10.1667*1.20.1667*1.50.1667)N, which can be verified by calculator to imply P < 0.9998N, which in turn implies P < 1.

**If B is a direct implication of A, then P(B) is greater than or equal to P(A). Since P(A) approaches 1, and P(A) <= P(B) <= 1, it follows that P(B) approaches 1 as well. So we can replace the original clause with its direct implication.