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Prime

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Everything posted by Prime

  1. 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.
  2. I like happy endings. And that uncomplicated trajectory is nice too.
  3. I thought lions could pounce. But even if that is not allowed, the lion still must eat.
  4. Barring complex numbers and limiting ourselves to integers...
  5. Why should a hungry lion devise the best strategy of escape for a tasty tamer?
  6. I be the Lion. There. Someone solve it for me. BTW, it was an Ogre last time, not a dog.
  7. 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 som
  8. 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.
  9. 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 call
  10. 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 num
  11. 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. You
  12. +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
  13. 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.
  14. 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
  15. 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.
  16. 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
  17. 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 averag
  18. Change it back. The proof is coming.
  19. 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.
  20. 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.
  21. 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.
  22. .... 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.
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