mersenneforum.org Dueling Pennies ... my Son found this
 Register FAQ Search Today's Posts Mark Forums Read

 2014-11-27, 19:56 #1 petrw1 1976 Toyota Corona years forever!     "Wayne" Nov 2006 Saskatchewan, Canada 3·5·349 Posts Dueling Pennies ... my Son found this http://www.futilitycloset.com/2014/1...eling-pennies/ 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 n-tosser 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?”
2014-11-27, 20:34   #2
retina
Undefined

"The unspeakable one"
Jun 2006
My evil lair

147618 Posts

Quote:
 Originally Posted by petrw1 http://www.futilitycloset.com/2014/1...eling-pennies/ 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, ...
I stopped reading when I got to here because anything after this must be based upon erroneous computations.

2014-11-28, 00:35   #3
Mini-Geek
Account Deleted

"Tim Sorbera"
Aug 2006
San Antonio, TX USA

11×389 Posts

Quote:
 Originally Posted by retina I stopped reading when I got to here because anything after this must be based upon erroneous computations.
The source shows 1/2[I]n[/I]. There are other superscripts lost in his post. This should be closer:

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 $8^{n-1}$ dollars; if he’s tossed $n + 1$ times, the house pays him $8^n$ dollars; and in all other cases the payoff is zero.

The probability of tossing the coin exactly $n$ times is $1/2^n$, so the customer’s expected winnings are $8^n/2^{n + 1} - 8^{n - 1}/2^{n - 1} = 4^{n - 1}$ 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 n-tosser pays the $(n + 1)$-tosser $8^n$ 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 Mini-Geek on 2014-11-28 at 00:37

 2014-11-28, 02:21 #4 TheMawn     May 2013 East. Always East. 110101111112 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 n-1 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.  2014-11-28, 04:18 #5 TheMawn May 2013 East. Always East. 11×157 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 2014-11-28 at 04:19
 2014-11-28, 05:33 #6 LaurV Romulan Interpreter     "name field" Jun 2011 Thailand 101000000001012 Posts The game is not symmetric, the player always wins, and there is a fallacy in David Gale's argument. The conclusion "the n-tosser pays the (n+1)-tosser 8^n dollars" should be restated as "the n-tosser pays the (n+1)-tosser 8^n dollars if..." Last fiddled with by LaurV on 2014-11-28 at 05:34 Reason: s/strike/spoiler/ grrrrr

 Similar Threads Thread Thread Starter Forum Replies Last Post tha Data 65 2020-08-05 21:11 jord161 PrimeNet 10 2018-06-22 21:57 science_man_88 Twin Prime Search 27 2010-05-05 16:24 em99010pepe ElevenSmooth 8 2006-01-02 10:12 Unregistered Data 1 2005-02-19 18:20

All times are UTC. The time now is 06:48.

Sat Dec 3 06:48:00 UTC 2022 up 107 days, 4:16, 0 users, load averages: 0.86, 0.99, 1.05