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# Comparing gambling systems

49 replies to this topic

### #41 Prime

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Posted 25 November 2013 - 03:25 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.

• 0

Past prime, actually.

### #42 Prime

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Posted 25 November 2013 - 08:15 AM

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.

• 0

Past prime, actually.

### #43 Rainman

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Posted 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.

• 0

### #44 bonanova

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Posted 28 November 2013 - 08:41 AM

This puzzle seemed to merit an analysis that went further than I had taken it.

Especially since most of my analysis was dead wrong.

• 0

Vidi vici veni.

### #45 bushindo

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Posted 29 November 2013 - 08:50 AM

This puzzle seemed to merit an analysis that went further than I had taken it.

Especially since most of my analysis was dead wrong.

Excellent insight, bonanova. The idea of using log is brilliant. However, my coding (and theoretical results) disagree with the simulation shown you and prime.

Spoiler for

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### #46 plasmid

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Posted 29 November 2013 - 05:14 PM

I get similar numbers as bonanova and prime.
I did 1000 trials of playing 1000 games per trial.
Average final bankroll ~ \$2E10
Median final bankroll = \$0.49
Number of winning trials = 465

If you're on a Windows machine with a Java JDK installed (free to download from Oracle if you don't have it already): copy the code below and paste it into a txt file called LetItRide.java, then on a command prompt in that directory type javac LetItRide.java to compile it, then type java LetItRide to run it.

Spoiler for java code

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### #47 bushindo

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Posted 30 November 2013 - 02:16 AM

I get similar numbers as bonanova and prime.
I did 1000 trials of playing 1000 games per trial.
Average final bankroll ~ \$2E10
Median final bankroll = \$0.49
Number of winning trials = 465

If you're on a Windows machine with a Java JDK installed (free to download from Oracle if you don't have it already): copy the code below and paste it into a txt file called LetItRide.java, then on a command prompt in that directory type javac LetItRide.java to compile it, then type java LetItRide to run it.

Spoiler for java code

Thanks for the code. I think I see what is causing the discrepancy between the simulation results. I made a mistake in transcribing the game probabilities...

Spoiler for

• 0

### #48 bonanova

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Posted 01 December 2013 - 01:49 PM

So it seems the following statements are true?

1. We can't have infinite pulls, but we can see trends as the number of pulls gets big.

2. Introducing multiple players makes the game much like a favorable lottery with a group of participants: almost all the individuals will lose. But if enough players participate, (significantly more than N if the winning odds are 1/N,) there should be at least one winner; and the winnings turn out to be great enough that the group will win. That is, the winnings of the few will cover the \$1 stakes of the many, with money left over.

3. Introducing multiple players (each with his own fresh \$1 stake and then looking at the aggregate payoffs to determine whether the game is won or lost) is indistinguishable from an individual person playing multiple times, periodically refreshing his \$1 stake and aggregating his winnings. This is the essence of a Method 1 game whose only difference is the fact that it is played using a different set of payoffs.

4. If the payoffs of multiple players are multiplied rather than aggregated, the probability with which the group loses is much closer to certainty than the loss probability of the individual players is. The group becomes in essence a single Method 2 player using a much larger number of pulls. Thus, the win probability of an individual player trends to zero with increased pulls.

5. Since (2) mimics Method 1, and (4) mimics Method 2, it seems justifiable to say that the win probability for an individual Method-1 player trends to unity, while the win probability for an individual Method-2 player trends to zero, even though that probability multiplied by his likely payoff (if he wins) is greater than his original stake. (In the sense that it is justifiable to say that an individual \$1 billion lottery ticket that has a one-in-a-million chance of winning is almost certainly a losing ticket, and that the statement becomes stronger if both the winnings and odds are escalated by a factor 1000.)

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Vidi vici veni.

### #49 plasmid

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Posted 01 December 2013 - 04:28 PM

#5 is factually true but I disagree with its implications, mainly because the OP doesn't say you have to play a Large number of times. Simulations show that the odds of winning after 1000 plays are not too far from 50/50 (roughly 45%), adding another couple of lines to the code showed the odds of winning >\$1000 are roughly 17%, and you have at least a 1/1000 chance of winning a bankroll that has to be expressed in scientific notation. I'd consider that favorable for a \$1 wager.
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Posted 01 December 2013 - 10:30 PM

I have no idea who to give credit to for this one. team victory maybe??
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