
#1
Nov1911, 12:37 PM

P: 56

This interesting thought experiment was brought up by my philosophy TA last week and I thought I'd pass it along in hopes for a lively discussion.
The game is as follows. You flip a coin. H TH TTH TTTH .......... $2 $4 $8 $16 If you flip heads, the game is over and you get the coressponding prize money. The question is, how much would you be willing to pay to play this game? Seeing that winning $2 is equally as likely as winning 4$, and winning 4$ is equally as likely as winning $8, ad infintium. Any thoughts? 



#3
Nov1911, 07:01 PM

Mentor
P: 14,432





#5
Nov1911, 09:47 PM

P: 1,256

If head is non first toss you lose. If tail is on first toss and tail is on second toss, you lose. I head is on second toss you win Wouldn't that be 1 out of 3 to win? For an $8 bet. You lose if H TH TTT You win if TTH chance of winning is 1/4 etc. 



#6
Nov1911, 10:39 PM

Mentor
P: 14,432

Here's how the game works. You say something like "I'll pay $10 to play that game". Suppose the person running the game accepts. You flip a coin. Get heads on the first toss and you get $2 back, for a net loss of $8. Heads on the second toss represents a net loss of $6, on the third, $2. Live past the third toss and you are making a net profit. Live past the 20th toss and you will have won over a million, minimum. Suppose the person running the game says "Not enough" to your paltry offer. You offer more and more, but the answer always remains "Not enough." At what point should you claim "Too much" and walk away? (Answer is further down.) Here's how the probabilities work. First off, the probabilities for each of the distinct outcomes must sum to one. Look at what you have: 1/2+1/3+1/4+1/5+... That is the divergent geometric series. The sum is not one. So right off the bat something is wrong. What you did wrong was to erroneously apply the principle of indifference. For example, One way to determine how the amount one should pay to play a game is to compute the expected payoff from the game. For a discrete game such as this, the expected return is the sum of the payoffs weighted by their corresponding probabilities. For this game, the expected return is [tex]\sum_{n=1}^{\infty} $1 \cdot 2^n\cdot\frac 1 {2^n} = \sum_{n=1}^{\infty} $1[/tex] You should be willing to pay any finite amount to play in this game! However, there is no gambling house in the world that could cover the potential payoff for this game. In the real world there would be some upper limit beyond which the game would end because you just won the house. So, to make the game a bit more realistic, lets say that you win $2 if you first flip heads on the first toss, $4 for the second, $8 for the third, and so on. Flip 20 tails in a row and you are paid $1,048,576; game over. Now how much would you be willing to pay to play? 



#7
Nov2011, 08:52 AM

Sci Advisor
PF Gold
P: 4,859

In a PM to Evo I raised that the cute feature of this game is the infinite expectation combined with that fact that really relying on this in the real world would be absurd. Besides DH reasoning (which would say pay no more than $20 to participate in the modified game), a general feature of pure reliance on expectation is asymmetry of consequence. The consequence of losing all of your money is often much worse than the real gain in your life of some large amount of money. This depends on your circumstances, but expectation simply ignores asymmetry of consequence.




#8
Nov2011, 08:56 AM

P: 15,325





#9
Nov2011, 09:20 AM

Mentor
P: 5,340

I would offer £4, just to make the game interesting for me but also not to waste money.
Can I twist the question slightly: How much would you be willing to accept from a gambler to play the game if you represent the house? 



#10
Nov2011, 10:09 AM

Sci Advisor
PF Gold
P: 4,859

There is, conceptually, a way to formalize asymmetry of consequence. That is simply that you should be computing expectation of f($), where f codifies your personal posititve/negative consequence. A very generic feature of f(), for example, is diminishing returns  a million on top of a billion is enormously less consequential than a million on top of zero. On top of this, the shape for negative $ will not match that for positive dollars. The utility of expectation for 'every day' bets is then formally understood as nothing more than that linear approximation of a sufficiently small region of any smooth function works well.




#11
Nov2011, 10:15 AM

Mentor
P: 14,432

There is a flip side to this: People can be riskaffine when the perceived risk is small. A good number of people would be more than willing to pay more than $20 to play my modified version of the game. They'd line up by the droves. Evidence: Modify my game a bit more and you get the governmentrun lotteries. 



#12
Nov2011, 10:22 AM

P: 94

Love this type of question that is supposedly math but in reality coming across such a game:
The person offering the game and the "chance" to win money, is out to take your money. I know this: so would keep my money in me pocket. 



#13
Nov2011, 10:27 AM

PF Gold
P: 3,173

I'd pay 0.5 cent so I can't pay (cash at least) if I lose, though I can win some money.




#14
Nov2011, 10:36 AM

Mentor
P: 14,432

Now if I could make one slight modification to the game, say one flip per day, a billion or so should do it. That gives me almost a month to find people who, for a modest set of fee (and with a billion dollars on hand, even a million is a modest fee), would supply me a bunch of completely new and relatively untraceable identities, buy me some cool hideouts around the globe, and develop an escape plan should the mark manage to flip tails on each of the first 29 days. Nobody said I had to follow the rules. 



#15
Nov2011, 12:29 PM

Engineering
Sci Advisor
HW Helper
Thanks
P: 6,339

My answer to the question would depend how many times I expected to play the game (for both the original or D.H.'s modified/truncated version). If I knew I was playing only once, the answer would be $2, because that is the maximum amount where I can't possibly lose anything. I offered less than $2 I would expect the other player to walk away rather than play. I wouldn't offer more than $2 because I'm not a gambler by nature. If I was playing repeated games, it would be worth a bigger stake each time, to keep the person was running the game interested in playing longer by letting him/her make lots of small wins and occasional big losses. It might even be worth raising the stakes as the game progressed, to give him/her a "chance" (against the odds!) to even the score. 



#16
Nov2011, 06:20 PM

Emeritus
Sci Advisor
PF Gold
P: 12,257

I know I'm risk aversive when it comes to gambling. I'd pay $2. I can't really see any reason to pay more if you can win all the higher prizes regardless of how much you spent on the game.
Perhaps you meant the question to be, "How much could someone ask to pay to play and you'd still be willing to try?" I probably still wouldn't play for more than $2, but you might get more people to play if charged $4 or $8 considering they play lottery tickets for that much. 



#18
Nov2011, 06:42 PM

Sci Advisor
PF Gold
P: 4,859

What is amusing is that the same theory that correctly says all casino gambling and all lottery games (unless the winning amount rolls unclaimed a number times) are foolish, claims that you should be willing to bid any amount to play this game once. 


Register to reply 