20141127, 19:56  #1 
1976 Toyota Corona years forever!
"Wayne"
Nov 2006
Saskatchewan, Canada
1000100100011_{2} Posts 
Dueling Pennies ... my Son found this
http://www.futilitycloset.com/2014/1...elingpennies/
Dueling Pennies A certain strange casino offers only one game. The casino posts a positive integer n on the wall, and the customer flips a fair coin repeatedly until it falls tails. If he has tossed n – 1 times, he pays the house 8n – 1 dollars; if he’s tossed n + 1 times, the house pays him 8n dollars; and in all other cases the payoff is zero. The probability of tossing the coin exactly n times is 1/2n, so the customer’s expected winnings are 8n/2n + 1 – 8n – 1/2n – 1 = 4n – 1 for n > 1, and 2 for n = 1. So his expected gain is positive. But suppose it turns out that the casino arrived at the number n by tossing the same fair coin and counting the tosses, up to and including the first tails. This presents a puzzle: “You and the house are behaving in a completely symmetric manner,” writes David Gale in Tracking the Automatic ANT (1998). “Each of you tosses the coin, and if the number of tosses happens to be the consecutive integers n and n + 1, then the ntosser pays the (n + 1)tosser 8n dollars. But we have just seen that the game is to your advantage as measured by expectation no matter what number the house announces. How can there be this asymmetry in a completely symmetric game?” 
20141127, 20:34  #2  
Undefined
"The unspeakable one"
Jun 2006
My evil lair
3^{2}×7^{2}×13 Posts 
Quote:


20141128, 00:35  #3  
Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
10AB_{16} Posts 
Quote:
A certain strange casino offers only one game. The casino posts a positive integer on the wall, and the customer flips a fair coin repeatedly until it falls tails. If he has tossed times, he pays the house dollars; if he’s tossed times, the house pays him dollars; and in all other cases the payoff is zero. The probability of tossing the coin exactly times is , so the customer’s expected winnings are for , and for . So his expected gain is positive. But suppose it turns out that the casino arrived at the number n by tossing the same fair coin and counting the tosses, up to and including the first tails. This presents a puzzle: “You and the house are behaving in a completely symmetric manner,” writes David Gale in Tracking the Automatic ANT (1998). “Each of you tosses the coin, and if the number of tosses happens to be the consecutive integers and , then the ntosser pays the tosser dollars. But we have just seen that the game is to your advantage as measured by expectation no matter what number the house announces. How can there be this asymmetry in a completely symmetric game?” Last fiddled with by MiniGeek on 20141128 at 00:37 

20141128, 02:21  #4 
May 2013
East. Always East.
11·157 Posts 
This isn't the kind of question where something in the question presented as TRUE is actually FALSE.
You can work through the math yourself to find that your winnings are in fact positive. If the toss numbers are consecutive, then whether you are the one with more or fewer tosses is a 50% chance (read: GIVEN n and n+1, being n or n+1 is completely random). This makes sense because as it stands, there is nothing to distinguish between the n tosser and the n1 tosser. I made an excel spreadsheet for this and it confirms, the number of "victories" is even. For example After 1000 attempts, YOU won 166 times and THEY won 162 times, yet your net money is $2,434,520 for some reason. One element of the game is NOT symmetric. The casino determines n, not you. Whoever determines n is at a disadvantage. On average, your winnings are greater because your wins are with larger n's. For example, in my simulation, with n = 1, you win 0 times and the Casino wins 127 times. With n = 2, you win 121 times and the Casino wins 35 times With n = 3, you win 35 times and the Casino wins 9 times With n = 4, you win 7 times and the Casino wins 1 time You CANNOT win with n = 1 because it would require that the casino tosses 0 times before getting tails. This is the reason for the "offset" in win numbers at different n's. With the exponential nature of the winnings AMOUNT, you are at a massive advantage. 
20141128, 04:18  #5 
May 2013
East. Always East.
11010111111_{2} Posts 
If this problem had a trap, it's one that a lot of people will fall into when learning basic probability. The whole "symmetric problem" where "the n tosser pays the n+1 tosser" makes it look like the Casino and Customer just win the same amount of times. Both players DO win the same amount of the time in general but NOT for a particular n.
For example, say n is 5. This means the Casino tossed a tail on the fifth try. You WIN if you toss a tail on the fourth, and lose if you toss a tail on the sixth. There is no winner if a tail is first tossed on any other attempt, so the only times that a winner is even declared are the ones where either the fourth OR sixth are the first tail. There is a 1/16 chance that the flip order is HHHT and a 1/64 chance that the flip order is HHHHHT, so your chances of winning are actually 80% ASSUMING there is a winner at all (your chances of winning are only 6.25% in general). The exception to this is where the Casino tosses a tails on the first toss. There is a 50% chance of this happening, and a 100% chance of you losing ASSUMING there is a winner at all (your chance of losing in general is only 25% because you must toss HT). You can't win because it would require that you flip a tail on the 0th toss. If we go back to my numbers, we see the 80/20 win rate on every other n except n=1. 121 / 35 ~ 4. 35 / 9 ~ 4. Your 4:1 win rate occurs on every n except n = 1 where the winnings are the lowest. So while you have a 50/50 chance of winning IN GENERAL, 80% of the times you lose are where the winnings are the lowest (8[SUP]1[/SUP] = $8) and none of the times you win are for $8 so on average, each time you win, you win more money. Last fiddled with by TheMawn on 20141128 at 04:19 
20141128, 05:33  #6 
Romulan Interpreter
Jun 2011
Thailand
2·3·31·47 Posts 
The game is not symmetric, the player always wins, and there is a fallacy in David Gale's argument. The conclusion "the ntosser pays the (n+1)tosser 8^n dollars" should be restated as "the ntosser pays the (n+1)tosser 8^n dollars if..."
Last fiddled with by LaurV on 20141128 at 05:34 Reason: s/strike/spoiler/ grrrrr 
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