While trying to explain poker to my father, I thought of an interesting problem which I cannot solve. I'm sure the more mathematically inclined people here can answer this, though. There is probably a name for it, and a dozen threads in another forum, but I'll call it "Gambler's Dilemma."

Someone asks me if I'd flip a coin for a dollar. I figure why not, and take the bet. I win.

Then he says, "Let's flip again. If you win, you get $4. If I win, we are even."

I know that is a great bet for me so I take it. I win again.

Now he says, "I will keep flipping with you. As long as you agree to each flip and win, you win four times the previous amount. The first time you lose, we are even, and the game is over."

How should I play this? Mathematically, it is always right to take the bet for each flip, but I am bound to lose everything eventually.

What you described seems to be a Martingale strategy. If you were to collect on each wager, as the wager increases exponentially, your opponent would eventually be unable to pay you unless he has an infinite bankroll. In other words, requiring a payout after each wager could end the game early. However, if your wins are on paper only, eventually you will lose your profits and the net result will be the loss of your initial wager. If you stick to the terms of the game and do not or cannot walk away at any point, you are guaranteed to lose that dollar -- no more, no less -- but you could get some excitement for your dollar.

For the math behind it, check out the Wikipedia entry on Martingale (probability theory).

In statistics, anything under 5% is rare occurrence and "doesn't happen" unless the premise of the probability is wrong. The probably of guessing 4 consecutive coin toss correctly is 6.25% and 5 is 3.13%. Personally, I'd stop after 2 (25%).

This does seem to play into it, but does not solve theh dilemma itself, does it? Assuming it is done on paper, if you make each decision mathematically, you would keep playing, I believe. Like you wrote, yes, it would be fun, but ruin is inevitable.

But they are independent events. On the third toss, your odds of winning are still 1:2 and your bet odds are 1:4. Wouldn't it be a mistake to refuse the third bet?

Each coin toss is an independent event (50% probability assuming a fair coin) but the successive outcomes are based on the previous results. Getting the third toss right is based on getting the previous two right (because there is no third toss if you get the second wrong).

The reason it isn't a good idea to continue, I think, is because once you lose, the game is over. There is opportunity cost involved. If he would keep playing it with you whenever you wanted forever, then this would be a good way to make a few bucks because when you lost, you could just start again (unless you had to offer him the same deal whenever you lost the first flip). the fact that the game ends when you lose and you can never play it again changes the game.

Perhaps I misunderstood the OP in thinking that once you accept toss #2, you can't quit while you're ahead. If that's not the case, play on as long as you wish, until you decide that the bird in your hand is worth keeping.

Although the outcome of each individual event is independent of previous events, as others have noted ITT, as the number of tosses increases, the more likely it is that the distribution of outcomes will approach the norm. In layman's terms, the further you push it, the more likely you are to lose.

That's funny because the chance of a straight flush in 7 card stud is 0.03% and I just saw one today. Not only that, but I have seen them before, too! Where are you getting this from?

In a strict mathematical sense you should of course take each bet (except the initial one which is nuetral).

But in a practical sense there comes a point where you should take your winnings because the utility of having those winnings exceeds the risk of losing them. Where that point is will be a personal decision.

Let's see if I understand this correctly . . .

Round 1: You bet $1. If you win, you get paid $2, which is a profit of $1.

Round 2: You bet $2. If you win, you get paid $4 (4 times your previous bet of $1). Bet odds are still 1:2 ($4:$2). Cumulative profit is $3 from the original dollar bet.

Round 3: You bet $4. If you win, you get paid $8 (4 times your previous bet of $2). Bet odds are still 1:2 ($4:$8). Cumulative profit is $7 from the original dollar bet. The chances of getting to this point are .875 or 7:1 (or 8 for 1), so it's a fair payout.

I suspect that the clause "you win four times the previous amount" might be the source of the gambler's dilemma, leading him to think that he's getting paid 4:1 odds on an even-money bet. Or am I not understanding the prop correctly?

To make sure I understand, this is the progression of the amounts you'd stand to win or lose on each bet, plus the EV of each bet:

Flip #1: $1 (–$1), EV = $0
Flip #2: $4 (–$1), EV = $1.5
Flip #3: $16 (–$5), EV = $5.5
Flip #4: $64 (–$21), EV = $21.5
Flip #5: $256 (–$85), EV = $85.5
Flip #6: $1,024 (–$341), EV = $341.5

Assuming I've interpreted this correctly, after the even-money bet, this silly person is repeatedly offering you a bet with an EV of just over 100% of the amount you're risking on each flip (actually starting at +150% and approaching +100%, which it will never quite reach). You will seldom see a bet this good anywhere.

Bankroll management is not even a consideration because he's funding your entire bankroll except the first dollar, and each successive bet is the same portion of that bankroll (all of it). The only consideration in deciding when to stop is that the game ends on your first loss.

Really, this decision is entirely personal. It's going to come down to how much you're willing to risk. A bet with +100% EV is obviously always a good bet, but it may not be worth taking if losing would be too devastating relative to your "real life" situation. So just pick a power of 4 that represents your risk threshold, and take it off the table if and when you get there.

Of course, if this silly person would offer this bet again in the future, that would change things significantly, but it does not seem to be the case.

Yes, you interpret it correctly. Sorry for any confusion about how this works.

You can stop at any time. He can only stop when he wins the first time.

So, is this a failure of mathematics as it relates to gambling, or is it a mathematical problem for which all of the information has not been provided? The missing information might include net worth, some value for risk aversion (like 1% of net worth, perhaps). Anything else?

Oh, and this isn't a Martingale strategy. A Martingale would size the bets so that Villain ekes out a fixed win (usually his initial bet amount) after the "inevitable" flip that goes his way. In a coin-flip situation with 0 EV, Martingale would be a really high-variance break-even strategy.

In this case, because he's skewed the payout odds of the flip, it's still high-variance, but breaking even is the best he can do after the first flip.

The math has this gambling exercise covered well. All of the information necessary to analyze the bet is there. The theoretical, mathematical solution is to keep betting forever and ever. It's an extremely good bet that never gets bad.

The only missing information is the subjective question of what the money means to you. There's no one answer for that, and how you should analyze that angle is a matter of personal preference.

i'm sure there are mathematical ways to approximate the utility of money to each individual, based on the kinds of things you said, net worth, income potential, risk aversion, etc. but there is no definite you should stop here calculation.

it's also a misnomer looking at it the other way - your friend won't inevitably win. it's not exactly a martingale, but that's the same illusion that martingaling relies on - hey it's inevitable that i'm going to win i'll just keep betting. but for any finite amount of money, it isn't.

Seriously? I completely dismissed this possibility as being, um, more than a little crazy. At the time the prop is made, "he" caps his profit at $1 in exchange for unlimited losses. Who in his right mind would do that?

Mathematically, play forever.

In real life, the only stopping conditions would be how much the money means to you vs how likely you are to get paid. If you're playing with a millionaire, maybe he'll pay off the $268 mill on Flip 15, but not the $1 billion on Flip 16.

Or if I go home to my wife and say I could've won $1 million after 11 flips, but flipped a coin and lost it all, I'll probably have to sleep on the couch.

Whomever it is, his ancestor is probably one of the 5 logical pirates.

Well, I'm not asking because I think I will run into it some day.

But who?...someone who pegs his opponent as so math driven, he cannot lose.

It seems crazy that the simple gambler's math we all use to play poker says you should keep playing indefinitely, until you definitely lose.

I understand that at some point you should opt out, but you would be doing it in defiance of the math we all use to play poker...

Well, we should also be using good bankroll management, and if you are ahead $100K, making a $100K bet would be bad bankroll management for most of us.

I guess that is the heart of the matter. When the amount you stand to lose (have won) becomes a bet larger than your bankroll management permits, you quit and take the cash.

For the flip of a coin, however, you're likely to never reach that point, starting at $1.

Clearly, he is a fellow of the finest character, a pleasure all-around, and a person whom we should never, ever cause to doubt himself about anything.

Bankroll management, for sure, both poker roll and life roll. There's also the long run aspect of poker math, which is missing in this prop.

Interesting exercise.

Seems similar to:

http://en.wikipedia.org/wiki/St._Petersburg_paradox

just to reiterate, you won't definitely lose. in fact if the person offering you this wager offers it enough, he will definitely go broke! this is exactly the same as an individual bettor who is doing a martingale vs the house, except your friend is the martingaler trying to claw back his 1 bet and you're the house where it is correct to take as much action as you can because it's +EV and you should only stop when the bets are so big you risk going under.

If the math you're using for poker doesn't include prudent bankroll management, you're doing it wrong. There are numerous analogs in poker to this situation, but the broader stroke is in the form of an adage:

While the example that leads to this adage has to do with actual bets on one day and the following day, its application in this spot is more theoretical.

EV-wise, you stand to double your money each flip. So when you reach the point where having that amount in your hand is more valuable to you than the mere possibility of having twice that much, you should stop.

It's a little like the question of whether you'd take the annuity or the lesser lump payment if you were to win a big lottery payout.