BrainDen.com - Brain Teasers

# Prime

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1. ## Half as old as my brother

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.

3. ## A lion and its tamer

I like happy endings. And that uncomplicated trajectory is nice too.
4. ## A lion and its tamer

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

Barring complex numbers and limiting ourselves to integers...
6. ## A lion and its tamer

Why should a hungry lion devise the best strategy of escape for a tasty tamer?
7. ## A lion and its tamer

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

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.
9. ## Comparing gambling systems

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.
10. ## Comparing gambling systems

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.
11. ## Comparing gambling systems

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.

13. ## Comparing gambling systems

+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.
14. ## Comparing gambling systems

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.

16. ## Comparing gambling systems

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.
17. ## Comparing gambling systems

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
18. ## Comparing gambling systems

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.)
19. ## Comparing gambling systems

Change it back. The proof is coming.
20. ## Let's meet up

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.

22. ## Simple money probability

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.
23. ## Let's meet up

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.
24. ## A shady game?!

.... 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.
25. ## Having sisters

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