## Welcome to BrainDen.com - Brain Teasers Forum

 Welcome to BrainDen.com - Brain Teasers Forum. Like most online communities you must register to post in our community, but don't worry this is a simple free process. To be a part of BrainDen Forums you may create a new account or sign in if you already have an account. As a member you could start new topics, reply to others, subscribe to topics/forums to get automatic updates, get your own profile and make new friends. Of course, you can also enjoy our collection of amazing optical illusions and cool math games. If you like our site, you may support us by simply clicking Google "+1" or Facebook "Like" buttons at the top. If you have a website, we would appreciate a little link to BrainDen. Thanks and enjoy the Den :-)
Guest Message by DevFuse

# Comparing gambling systems

49 replies to this topic

### #21 plasmid

plasmid

Senior Lolcat

• VIP
• 1484 posts
• Gender:Male

Posted 21 November 2013 - 06:46 AM

Would it help to rephrase the problem by saying "You've heard of double-or-nothing games. You have a chance to play a triple-or-nothing game with 50/50 odds, but you have to bet your entire holdings each time you play. Should you take it?" It's in the same spirit as the OP but the math becomes trivially simple and doesn't call for simulations.

If you start with \$1 and play for N rounds, you have a 1/2N chance of winning \$3N, and a (2N-1)/2N chance of losing your wager. The average outcome is 3N/2N so it's a clearly winning game from that perspective, the most likely outcome is you lose everything so it's a losing game from that perspective.

I'd say play the game a few times. It's not really different from, say, rolling a six sided die and getting \$10 if you roll a 1 vs lose \$1 if you roll anything else.
• 0

### #22 Prime

Prime

Senior Member

• Members
• 872 posts
• Gender:Male
• Location:Illinois, US

Posted 22 November 2013 - 03:08 AM

Upon further reflection, I am leaning back to my first post(#3) and the solution therein. The Expected Outcome and Geometric Mean have no bearing on the winning scheme you must adopt in this game. In particular, the Geometric Mean formula does not offer any tangible means of predicting an outcome of a gambling binge, whereas the Expected Value formula works just fine in practice.

Spoiler for You cannot lose in this game

Once again, those who don't believe it's a winning game, can play the Casino side.

Edited by Prime, 22 November 2013 - 03:17 AM.

• 0

Past prime, actually.

### #23 Rainman

Rainman

• Members
• 162 posts

Posted 22 November 2013 - 07:10 AM

Upon further reflection, I am leaning back to my first post(#3) and the solution therein. The Expected Outcome and Geometric Mean have no bearing on the winning scheme you must adopt in this game. In particular, the Geometric Mean formula does not offer any tangible means of predicting an outcome of a gambling binge, whereas the Expected Value formula works just fine in practice.

Spoiler for You cannot lose in this game

Once again, those who don't believe it's a winning game, can play the Casino side.

I played few more games against my computer and it owes me a lot of money.

Spoiler for COMPUTER SIMULATION

I agree that the EV is (31/30)N for N rolls. But the problem is, as N grows larger, that value will become distributed over a relatively smaller set of variations. For a large enough N, everybody in the world could play the game and everyone would lose. You asked if the ratio of wins to overall variations approaches 0 as N approaches infinity. It does, and I will prove it as you asked. The proof is at the end of this post.

As for your simulations with 10,000 rolls each: for some perspective, consider what the EV is for 10,000 rolls. (31/30)10000 ~ 3*10142. The expected outcome is about 3*10-2. Your largest result was about 9*1011, followed by 4*106, 1*102, 2*101, and six results of something times 10-something small. Now imagine if you were the expected value, sitting around 142 at the logarithmic scale, looking down at those ten simulations, knowing that the expected outcome is sitting around -2. You might think "hey, those ten results are all packed pretty neatly around the expected outcome, while I'm all alone up here". It should become clear that it's unreasonable to expect to get the expected value. If you ran 7 billion more simulations, one for each person in the world, I can't imagine that a single one of us could be expected to win nearly that much. 7 billion simulations is still a very small sample from the 610000 possible variations. Also, if you had combined those ten simulations into one 100000-roll game, you would have lost that game (based on your remark that the losses gave way below a penny in return).

The statement that you practically can't lose with method 2 is just wrong. In the long run, you practically can't win (proof still coming up). The only reason you can still win after 10000 rolls is that the expected outcome per roll is so close to 1. The expected outcome for six rolls is 0.99792, which means 0.997921/6 ~ 0.99965 per roll.

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.

• 0

### #24 plasmid

plasmid

Senior Lolcat

• VIP
• 1484 posts
• Gender:Male

Posted 22 November 2013 - 03:05 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?
• 0

### #25 bonanova

bonanova

bonanova

• Moderator
• 6161 posts
• Gender:Male
• Location:New York

Posted 22 November 2013 - 05:09 PM

Add 0 and 10 to the set of payoffs: 0, 0.7, 0.8, 0.9, 1.1, 1.2, 1.5, 10.

The arithmetic mean is now greater than 2 and the comparison of the methods is clear.

• 0

Vidi vici veni.

### #26 Rainman

Rainman

• Members
• 162 posts

Posted 22 November 2013 - 06:16 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?

I'm saying that if \$1 was my entire bankroll, I definitely wouldn't play that lottery. Sure, I could get lucky and win the big jackpot, and sure the EV is positive, but protecting your bankroll is just as important as finding favorable games. You can win in the short run playing unfavorable games as well, but in the long run you are going to lose. And if you don't protect your bankroll, you're going to lose in the long run.

If you're a recreational gambler with a steady job, and you spend a small part of your weekly salary on some weekend bets, hoping to cash in big so you can have a nice long holiday or early retirement, then bankroll management is not really an issue for you. Your bankroll is refilled on a regular basis, and you're not risking anything. However, I believe the appropriate definition of recreational gambling is that you're doing it for fun. If you're doing it to win money, if you're looking to make your living as a gambler, constantly looking for bets with a small edge to grind, knowing that if you do it right your profits will increase exponentially over time, until one day you're rich enough to retire, then you simply must know how to protect your bankroll. Otherwise you'll end up an addict, putting yourself in debt because you "know you can beat the game, you've just been unlucky so far".

• 0

### #27 Prime

Prime

Senior Member

• Members
• 872 posts
• Gender:Male
• Location:Illinois, US

Posted 22 November 2013 - 07:20 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.

• 0

Past prime, actually.

### #28 Rainman

Rainman

• Members
• 162 posts

Posted 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.

• 0

### #29 bonanova

bonanova

bonanova

• Moderator
• 6161 posts
• Gender:Male
• Location:New York

Posted 22 November 2013 - 10:27 PM

This post refers to OP and the implications, if any, that stem from the fact that pulling twice for the 36 possible outcomes gives you a positive payoff.

OP asks to compare two methods: M1 and M2.
OP gives the payoffs as .7 .8 .9 1.1 1.2 1.5. and assures us they are not biased.
Over time there is no preference.

AM=1.0333 ...
GM=0.9993061854

M1: Bet \$1. (Keep your winnings; bet a new \$1.) Repeat (.) Repeat means: do not stop after n pulls.
M2: Bet \$1. (Bet your winnings.) Repeat(.)

Over time, M1 wins. (AM>1.) I will take player's side on this game.
Over time, M2 loses. (GM<1.) I will take the house's side on this game.

Variation:

M3: Bet \$1. (Pull twice. Keep your winnings. Bet a new \$1.) Repeat(.)
To be precise, pull (twice) only for each of the 36 outcomes then stop and average the outcomes.
Over time this difference goes away.

For the 36 outcomes (and over time as well,) M3 wins. I will take player's side on this game.

There seem to be differences only about M2.

There is a conjecture that (M3 wins) ==> (M2 wins).

Let's revise the payoffs:
.49 .56 .56 .63. 63. 64. .72 .72 .77 .77 .81 .84 .84 .88 .88 .96 .96 .99 .99 1.05 1.05 1.08 1.08 1.2 1.2 1.21 1.32 1.32 1.35 1.35 1.44 1.65 1.65 1.8 1.8 2.25

AM=1.0677777...
GM=0.9993061854

Over time, M1 wins. (AM>1). I'll be the player here.
Over time, M2 loses (GM<1). I'll be the house here.

The second set of payoffs comprises the pairwise products of the first set.

M2 (win or lose) has the same outcome in the two cases.

It differs from M2 in the first case only because we look at results after even-numbered pulls.

M1 still wins, and now it winds faster: AM is larger.
M1 in the second case must have the same outcome over time as M3: the same actions are taken.

M3 thus does not relate to M2 at all. Nor does not depend on GM.

M3 is equivalent to M1 with better payoffs. It wins, therefore, only because AM>1.

Conclusion: (M3 wins) =/=> (M2 wins).

• 0

Vidi vici veni.

### #30 bonanova

bonanova

bonanova

• Moderator
• 6161 posts
• Gender:Male
• Location:New York

Posted 23 November 2013 - 06:56 AM

Can a Method-2 strategy win using the payoffs given in the OP?

Using the {.7 .8 .9 1.1 1.2 1.5} payoffs of the OP, I took 2,000,000 random selections (pulls) and put them into 20 sets of 100,000 each. From this I calculated {twenty products}, and they ranged in value from 1.38 x 10-72 to a whopping 1.55 x 1072! Six of the twenty were >1; the median value was about 10-18. All payoffs were positive, of course, so the average was at least 1/20 of the highest value.

The AM, in fact, was a whopping 7.74x1070.

In the context of comparing Methods 1 and 2 from the OP, what can we say from these {twenty numbers}?

1. We might assert that if we play Method 2 multiple times, say n times, where n is large enough, we'll get at least one whopping payoff which, when we add the results, will more than offset the \$(n-1) we may have lost on the other games. Viola! Therefore we conclude that Method 2 is a winning strategy that pays off handsomely. In this case, where n was a meager 20 games, using Method 2, \$20 became \$1072. Wow. We will play that game any day of the week!

2. Or we could consider the {20 numbers} to be a {set of representative payoffs} that apply to the 100,000-pull game. Let's play that game twenty times, using the strategy of Method 1. Thus, we bet \$1. (Pull 100,000 times, Take our winnings. Bet another \$1) Repeat (20 times). We win big. \$20 became \$1072. Wow. We will play that game any day of the week, too!

Notice these stories describe the same actions. Not surprisingly they give the same results. The only difference is that first we say we are using Method 2. Then we say we are using Method 1.

But in Method 2 there is no provision for periodically starting over with a fresh \$1 and then at the end adding the results. That process is the heart of Method 1. To rightly apply a Method-2 strategy to these {20 numbers} we must multiply them. And that makes an amazing difference.  Their sum is 1.55x1072; but their product (the result of betting \$1, then pulling 2 million times) is a very disappointing 6.88x10-244. That is to say that our initial stake could have been as high as \$10244, and we'd end up with enough money for a McDonald's #7 meal with a medium Coke. My favorite.

Not surprisingly, the 2-millionth root of \$6.88 x 10-244 is 0.9997200883, the GM of {.7 .8 .9 1.1 1.2 1.5}.

• 0

Vidi vici veni.

#### 0 user(s) are reading this topic

0 members, 0 guests, 0 anonymous users