A very interesting problem

Well, provided that it is pure probability where I'm at a disadvantage, I would most likely refrain myself from putting all money on a single bet. Without going too deep into why I'd do this or that, I'd probably start off by betting 10-max 20% of my total available money. However, if I go into a loss streak of 2 (I would probably prefer the number to be 3 actually), I will go with a all-in bet afterwards. Statistically it might be incorrect or inefficient, but I'm not looking at it from a statistical point-of-view, or at least not going too deep.

Curious what would be statistically efficient though.

The first question I would have is how do I ensure that 45% winning and 55% losing are the actual odds?

Assuming I was satisfied with the answer, I would bet $2,000 the first time. If I won, I would bet $4,000 the next time. I would bet $4,000 the third time and walk away, regardless of the outcome.

I would have a 20.25% chance of having $8,000 after my second roll. After my third roll with a $4,000 bet, I will end up with either $4,000 (11.1% cumulative probability) or $12,000 (9.1% cumulative probability), giving me an expected value of $1,539. Given that I started with $1,000 of my money; that's a decent % return. Also, I'm done after only three rolls, so it's not costing me a lot of opportunity cost in the form of time.

bet $1 if you lose bet double until you win

i.e. bet $1 lose (at 55% chance) bet$2 lose (at 55% chance) bet $4 lose (at 55% chance) bet $8 lose (at 55% chance) bet $16 win get $32 bucks cost to you was $31 profit is $1

when you win start back at betting $1 rinse and repeat until you made 10k of bets and clean house

depends on your utility function. Since you're at a disadvantage, definitely withdraw at the first opportunity. Your expected loss is .1(10,000) = 1,000 dollars. To minimize variance, bet house minimum until you've bet 10,000, but you're almost definitely going to leave a (moderate) loser. Any attempts to increase your probability of leaving a winner however will just increase your risk.

This is assuming that it's your own money and your interests are suitably aligned. Obviously on a Lehman trading desk, shoot for the moon and bet max every time.

Edit: actually the below is incorrect, i didnt read the condition that u get to exit if you lose all your money. that makes things different. but i'll leave the stuff below anyways

@dickfuld @yelloweat

no, polymorphic is correct, all ur approach does is increase variance. it's just basic probability.

for a simple illustration, compare betting $2 vs. betting $1 and $1. The first has an EV of 2.45 - 2.55 and variance of 3.96 while the second has a mean return of 2.45-2.55 and variance of 1.98.

basically you make a series of bets a_1...a_k, a_i>0 and sum(a_i) = 10,000.

your ev on this is sum( a_i.45 - a_i.55) = -.1 * sum(a_i) = -1000 your variance on this by linearity (since a_1 can be considered independent random variables) is

sum(var(a_i)) = sum(a_i^2 * .99) = .99 * sum(a_i^2).

to minimize the variance u need to minimize sum(a_i^2) subject to sum(a_i) = 10,000.

it can be proven that for each fixed k, we minimize the quadratic sum by taking a_1=a_2...=a_k (if any 2 are different you can "smooth" it so they are equal and reduce quadratic sum)

then sum(a_i^2) = ka^2 with ka = 10,000, so a = 10,000/k so sum(a_i^2) = 10,000^2/k

as we can see, taking k->infinity brings the variance to 0.

(too bored..)

@mikesswimn @illiniprogrammer

Please opine.

@DickFuld - Will do, sir.

From where I'm sitting, the best strategy here is utilizing the Kelly criterion. Now, the tricky part is getting it to work properly, since as the question states, you're always in a losing situation given the unfair coin, making the typical calculation unworkable since you can't "be the casino" or "bet against yourself". However, that little detail of the casino giving you an extra grand looks to be the key. The way I'd look at this is not by including the casino's money in the original bankroll, but instead by using it as an augment to the payout. For example, for every $1 of YOUR money that you put up, you have a 45% chance of winning $2, not $1, since the extra $1 put up in the bet isn't really yours, it's the casino's. So:

fraction of bankroll = [(probability of winning)(net odds + 1) - 1] / net odds fraction of bankroll = [(0.45)(2 + 1) - 1] / 2 = 0.175

So, using this information, I'd put up 17.5% of the account value for each bet (i.e. $350 for the first bet) and leave with more than $1,000 (my original bankroll).

@dickfuld

oh actually i didnt read a key part of the question, sorry: you get to exit if you lose all your money. in this case i think you're right. i'll leave my answer up tho since it works if u remove that condition.

@polymorphic123 @cauchymonkey - This part of the question is probably a red herring:

boib:

You are able to withdraw from this game, when you made bets on at least 10000 dollars

That shouldn't matter if you have a profitable strategy. It's probably in there to mess you up.

@yelloweat - What you're talking about is a martingale betting strategy. So, in this case, since q > 1/2, using this strategy your expected payout per round (a round being complete when the gambler wins or goes bust) is -1.85. Suggesting that utilizing a martingale strategy in this situation isn't profitable (read: you'll probably go bust).

So technically, ignoring statistics as I pointed out, my 'common sense' way of doing this is not exactly wrong, is it? Seems to be close enough to the Kelly criterion at least in terms of the size of the bets. @mikesswimn Though I do realize that in reality I would actually care enough to come up with a fail-proof way of doing this.

@mikesswimn

Is this math wrong? This is the way I would do this:

Here's the math (you can ignore the losing in the first two throws, because it goes to zero):

0.45 x 0.45 = .2025 chance of winning $8,000

One more try, where I bet $4,000: 0.45 x .2025 x $12,000 (comes from the doubling of $4,000 to $8,000 and the $4,000 I didn't bet) = $1,093.50 0.55 x .2025 x $4,000 (only kept the $4,000 I didn't bet) = 445.50

First bet was $2,000, second bet was $4,000, and the third bet was $4,000; so the total amount bet was $10,000.

$1,093.50 + $445.50 = $1,539 (expected value)

@mike

dickfuld is correct, you cannot have a profitable strategy placing -ev bets. the point is to invest as little money into the game as possible. i dont think it makes sense to use some sort of "risk management strategy" that makes you play the losing game longer... that's why you should bet $2000 right from teh start.

if you lose 2k, alright you only had to make 2000 of negative ev bets. if you win, then you are obviously in better shape.

it's still a negative ev strategy though but you can come out with more than 1k since the casino gave u 1k.

ur expected pnl from the strategy is -2000.55 + -2000.45.55 + 2000.45.45.55 + 10000.45.45*.45 = -461

so we r expected to have 2000-461 = 1539 as dick fuld got.

@DickFuld & @cauchymonkey

There's a big glaring issue with the "bet it all" strategy because you can't calculate the expectation in the traditional way as your strategy becomes path dependent. In other words, after the first round, you're not calculating a straight expectation, you're calculating a conditional expectation. Let me crank the math out and get back to you guys.

Besides, isn't expected value not really sufficient for this question? Don't you care about the variance of your profit, too?

I basically agree with Dick Fuld.

This game kind of works like a weird option.

It is impossible to get your money out using a low vol strategy. You expect to lose $0.10 every time you bet $1 (45% chance of winning $2= $0.90 expectation from game). So if you bet $1 10k times, your expected loss is $1K with low variance.

If you use a high vol strategy, and try to maximize the expectation of having unbet money short of $10k, that's basically your expected value from the account.

@DickFuld & @cauchymonkey

Okay, here's the problem with your math, you don't include the potential loss in the two prior periods in your expectation. Both of you are correct that the final two outcomes result in $1,093.5 and $445.5, but that's only if you get there in the first place. You need to include the prior probability that you'd lose all $4,000 (probability = .2475) and the initial probability that you'd lose all $2,000 (probability = .55) giving you an expected value of -$551.

edit: actually im not completely sure how the money accounting is done, the question states 1k is "your money" and 1k from the casino is put into the account. when you incur losses, is that taken directly from your 1k? so you could exit if you bet 1000 dollars and then lose? then illini's strategy would make sense.

@mikeswimn

the losses are incorporated in the calculation.

@euroazn

yeah i guess one should incorporate variance... for example using the very small bets you can guarantee a -1000 loss with very little variance. or you can have an ev -541 with dickfuld's strategy with greater variance.

Let me adjust my previous answer since I, like cauchy didn't take into account you can leave if you lose all your money. In this case, slightly counterintuitively, you want to MAXIMIZE your chance of going broke. This is because in a negative EV game, you want to minimize the amount of time you play it. So DickFuld's maxbet strategy is best.

btw, stop wasting your time trying to find a profitable strategy...you can't win money in a negative EV game. Best you can do is minimize your losses.

I'm going to count this as my first win since 2006!

I think Dick Fuld is right. Conceptually I'm not sure if I'm thinking about it correctly, but I'd look at it as a simple case of cutting off one end of your probability tail. If at any point in time I end up with nothing left, I simply leave. Increasing the variance allows me to cut off a longer tail, effectively increasing the expected value of what's left on the right side.

As soon as I'm no longer forced to go all-in I kill the variance (although this wouldn't affect the expected value) and leave.

Edit: hurr read the problem wrong

actually there may be an improvement on the strategy,

again, first bet 2k. if you win, you are at 4k. again, bet 4k. now if you win, you are at 8k and have to bet 4k more.

But at this point, it makes sense to do a ton of 1$ bets because no matter what you do, you are going to have to bet 4k more into the negative ev game. So at this point, making a ton of tiny bets will minimize the variance.

I think of this as a sophisticated version of you need to double your money or you'll get killed. The only game in the casino is the roulette table. The most efficient strategy is to bet it all on black (or red). Sure, you have a little more than a 50% chance of dying, but it beats all the other strategies.

For the Kelly Criterion to work either the odds (assuming 1:1) or probability would have to improve for this to be an attractive venture. I would not entertain this problem, even with the $1k kicker...

I think you use the Kelly Criterion to determine how much to bet on one game, not on a particular roll of the dice.

kelly criterion doesn't apply to -EV games...well it does, but it tells you not to bet anything.

This is a real-life problem as many sports gambling websites offer this type of deal...they match ur account on opening and you are required to bet some amount. I am quite rusty on stats but i think that the EV of putting 10k into play in this game is -1k, which is the value of the 1k comp you are getting so at best this is a 0 EV setup...there is no betting strategy that can overcome that so long as your odds of wining each flip are only 45%. However, there is an additional problem in that you are taken out of the game with a 1k loss if you hit 0 in your account (if you lose 2k)...ie you cant use the option to leave the game if you lose all your money before you put that 10k into play. My strategy would be to flat bet the minimum i was allowed in order to reduce the variance around that -1k EV and increase the odds that you make it to the point where you have put 10k into play. But in the long run you cannot win with those odds even with that 1k comp.

This is very serious problem and very good approach.i think this is very beneficial for every one. i wanna discus on this topic if any body is online so please reply me i m waiting. Thanks for sharing this information.