Prime

To put a finer point, if two siblings are two cells (or particles) resulting from splitting of the mother cell into two, then none is older.

I like happy endings. And that uncomplicated trajectory is nice too.

I thought lions could pounce. But even if that is not allowed, the lion still must eat.

Barring complex numbers and limiting ourselves to integers...

Why should a hungry lion devise the best strategy of escape for a tasty tamer?

I be the Lion. There. Someone solve it for me. BTW, it was an Ogre last time, not a dog.

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; your overall chance of ending up ahead looks like 48.78%; 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.

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.

What makes these topics so interesting to me, is my lack of education. Math education in particular. Perusing your argument from the post 24, imagine a casino offered you a choice between two games, 1000 dice roll each: 1) Stake $1 with a 5% chance of winning $1,000,000 or more; OR 2) Deposit $300 with a 50% chance of winning something between $16 and $50, while in the extremely unlikely cases losing your entire $300 or winning $500. Which game would you play? (The percentages here are illustration -- not an actual calculation.) I think, what we are arguing here is called Normal Distribution, or Gaussian Distribution, or Bell Curve, or some other such name. In view of the above, I think, the wining likelihood of a 1000-roll ride is being underestimated.

But I have run a bunch of simulations. And some numeric analysis as well. Enough to convince myself. See my posts 12, 20, 22, and 33 inside the spoilers. Riding your winnings for 1000 rolls, seems like a good choice. That won millions in my simulations in very short order of time. At this point I feel that for further discussion, we need to solve analytically what is probability of winning X $$ or more while riding for N consecutive rolls. E.g., $1,000,000 or more after 1000 rolls. For that one must find exact number of 1000-roll variations yielding more than 1,000,000 and divide that number by 61000. I don't feel up to the task at the moment. But I can provide any such data for up to 12 rolls.

When playing indefinitely, you cannot cash in your winnings and spend the money on this side of Infinity. And on that other side, who will care how much you've won, or what fraction of $1 you have remaining? Strange things happen at the infinity: 1. You are expected to end up with nothing, if you ride your winnings infinite number of times. 2. That is because the most likely outcome is the one where each of the six possible roll values will have appeared the same number of times. The probability of that most likely outcome is zero. (That requires a proof though. See the spoiler.) 3. Your average payoff after infinite number of rolls should be infinite, if you bet just $1 on every roll. 4. Your average payoff if you ride your winnings is going to be even more infinite. In fact, it is going to be infinitely greater than that from the previous point (3). That is because when riding your winnings, the infinite variable is an exponent whereas when betting a fixed amount on each roll the infinite variable is a multiplier. 5. To see your average payoff, you must play infinite number of sets of infinite number of rolls each. You need not live forever to go on that gambling binge. Just roll an infinite number of dice every time and roll them infinitely fast. Coming back from the Infinity, as I said here before, for all practical purposes, it is impossible to lose in this game. And you must ride your winnings, if you want to win big. The Expected Value (average payoff) for a 1000 consecutive rolls is (1.0333...)1000 = 1.74*1014. You cannot run enough experiments to confirm it empirically. Nonetheless, the winnings are good. In order to help its patrons to win large sums of money, the casino added a small modification to the game. Namely, the minimum bet requirement of $1. So while rolling the dice and riding your winnings, if your total falls below $1, you must add enough cents to make your next bet at least $1. To sum up: In this game having a bankroll of $20, and in reasonable amount of gambling time (few days) 1) If you ride your winnings, you most likely end up with the winnings of millions of $$. 2). When betting $1 at a time, in the same amount of time, you'll most certainly walk away with few hundred of $$. (Frankly, not fun and not worth your time.)

+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. 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.

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

Yes, there it is -- Expected Value vs. Expected Outcome. I fancy myself as an avid gambler. And in practice I would choose the mix of the two methods, as Rainman has suggested here. (Playing infinite number of times is rather impractical.) I have run an experiment playing 422 consecutive rolls 10 times with the following results: 4 times I had less than $0.01 left of my initial $1 bankroll. 3 times I had between $0.02 and $0.35 left. 3 times I won with the largest win of $83.62. That left me very much ahead (more than $80, while putting at risk just $10). The new question here is, what is the optimal string of consecutive rolls? 3 times I It seems your post was cut off unfinished. Expected outcome is the true value for any gambler who seeks to eliminate the "gambling" part. The probability is 1 that your own outcome approaches the expected outcome as the number of games approaches infinity. So in the long run, you are practically guaranteed to win if your expected outcome is positive. The same is not true for expected value. Expected value is too influenced by the extremely high payoffs in the extremely unlikely variations. In this case, if you play for example 100 times, the extremely lucky variation where you hit 1.5 every time would yield a net payoff of roughly +$406,561,177,535,215,236. On the other hand, the extremely unlucky variation where you hit 0.7 every time would yield a net payoff of roughly -$1. The average net payoff, or expected value, would be (31/30)100-1, or roughly $27. So the variance (actual payoff - average payoff) is 406,561,177,535,215,209 for the luckiest variation and only -28 for the unluckiest variation. As follows, the expected value is extremely tilted by the high variance from impossibly lucky scenarios. You would need to be insanely lucky just to get anywhere close to the EV. Your own experiments illustrate this perfectly. The average net payoff for running 422 consecutive games 10 times is 10*(31/30)422-1, or roughly $10,220,338. Your actual net payoff was just more than $80, falling way short of the EV. Had you kept your entire bankroll running for all 4220 games, you would have lost almost everything. This was not a case of bad luck, but rather quite expected. Had you instead bet half your bankroll every time, with a starting bankroll of the same $10, for 4220 games, your expected outcome would have been ~10*1.049244220/6, or +$4,800,000,000,000,000. Feel free to simulate it, you will end up somewhere around that number. Your EV would of course be even higher, 10*(61/60)4220-1 or roughly +$2,000,000,000,000,000,000,000,000,000,000. I am not disagreeing with the Expected Outcome concept. (And I have described the same in the post #12 under “PERSPECTIVE 1”, before I read your post #8. Actually, I should have read your post first. Now I invite everyone to read the spoiler in the post #12, where I derive the formula for the EV with a proof.) However, in view of the question in the OP “Can you win in this game?” the Average/Expected Value seems very relevant for practical gambling. I also like your system where you bet half of your entire bankroll on each turn. Although, I can't say I follow the 1.049N/6 formula for this method. I'd have to think about it. Finding “optimal” system in terms of available time, initial bankroll, Expected Outcome, and Expected Value seems like more serious mathematical research. Another interesting and, perhaps, simpler thing to solve would be: What are the chances of ending up ahead after N rolls when staking your entire bankroll? (I could tell it's 17/36, or better than 47% for 2 consecutive rolls and 105/236, or better than 44% for 3 consecutive rolls.) I played few more games against my computer and it owes me a lot of money.

I was not implying any actual "pulls" in my illustration. Nor did I say anything about withdrawing or adding any amounts to the stake. I did not mean it as an actual trial in the game. I meant it as the enumeration of all possible variations with their respective probabilities.

Yes, there it is -- Expected Value vs. Expected Outcome. I fancy myself as an avid gambler. And in practice I would choose the mix of the two methods, as Rainman has suggested here. (Playing infinite number of times is rather impractical.) I have run an experiment playing 422 consecutive rolls 10 times with the following results: 4 times I had less than $0.01 left of my initial $1 bankroll. 3 times I had between $0.02 and $0.35 left. 3 times I won with the largest win of $83.62. That left me very much ahead (more than $80, while putting at risk just $10). The new question here is, what is the optimal string of consecutive rolls? 3 times I

The six equally likely payoffs are .7 .8 .9 1.1 1.2 1.5. Their arithmetic mean AM is 1.033333 ... Their geometric mean GM is 0.9996530325. In Method 1 you pay $1.00 for each pull and remove your winnings. In Method 2 you pay $1.00 once and keep your stake at risk. Thus, AM (>1) gives you the expected return per dollar per pull for Method 1. (Winnings are added.) GM (<1) gives you the expected return per dollar per pull for Method 2. (Winnings are multiplied.) Post 3 paid $1.00, pulled the handle twice and then removed the winnings. This was done 36 times with the results averaged. This is not Method 2. That's not what I said in Post 3. It's actually Method 1 for a new set of payoffs: the 36 pairwise products of the original six payoffs. No, it is not Method 1. On the second turn the entire bankroll is staked -- not the original betting amount. This is a winning game, with an average return on a dollar bet of AM2. AM2 is in fact the arithmetic mean (AM) of the new set of payoffs. So this is one more example of a winning Method 1 result. To test Method 2 for 72 pulls of the handle, you bet $1.00 once and keep the result at risk for all 72 pulls. That final result is the just product of the 36 pairwise products, which is 0.9753235719 = GM72. And that is the expected result for Method 2: each pull multiplies your stake by GM. Again, why does Method 1 win and Method 2 lose? Because AM applies to added winnings and AM>1; GM applies to multiplied winnings and GM<1. Simulation: Pull the handle 100,000 times with random results: Method 1: Average return on each $1.00 bet is 1.032955 <--> AM = 1.0333 ... Method 2: Stake goes from $1.00 to $1.414863x10-31 = GM100000. The 100,000th root of 1.414863x10-31 is 0.9992899. <--> GM = 0.9996530325. There is an interesting thought provoking point there, presenting a kind of pseudo-paradox in this problem. I stand firmly by my solution in the post 3, though I am beginning to have some doubts about its straightforwardness property. The illustration I gave in support of the solution was not a proof. And since bonanova has misinterpreted my illustration, it must have been unintelligible. (It happens to me from time to time. In my head a statement I make seems precise and clear, but other people can't make out the ends of it.) So in the interest of befuddlement, perplexity, and creative thought... For anyone who is still not convinced, I am ready to play the game with a fair die, staking my entire bankroll on each roll of the die. (To speed up the process, we could roll 12 dice at a time.)

Change it back. The proof is coming.

The corners of the board are dots. George and Lennie have square bottoms exactly equal in size to a board square. They move from one square to another by sliding in a straight line until the entire square is covered with their bottom. This way whenever they are moving into the same square, the collision is unavoidable.

Since my last post has not been recognized as the answer, there must be something else to this problem. I suppose, it could be that different coins are encountered with different frequency. E.g., a penny is a lot more common than a half-dollar coin. I don’t have any data on how many of each coin type there are in circulation. However, I doubt an adjustment for each individual coin probability would change the final answer to the OP. I believe a 15-coin collection would still have a higher probability of spotting a dime.

To solve an unequal speed case, I feel a need for some disambiguation of the OP. 1). Moving 3 times faster does not mean moving in 3-square straight segments at a time. George still can zigzag in 1-squre steps. 2). Collision occurs whenever George and Lennie are found anywhere inside the same square at the same time.

.... It suddenly downed on me, what you meant. The (10, 9, 8) and (10, 9, 7) are not the actual cards drawn, but rather representations of two different strategies. So, (10, 9, 8) means: stay on 10 and up on the first card, on 9 and up – on the second, and on 8 - on the third.