I guess a 100% if you pose the question like that right?

( 55%*2x against 45%*0 )/ 100% = 1.1x

In other words, Peter's expectation value is 110% of what he put in. The more he puts in, the higher his expected return?

100% is a bad idea, since he might lose it all in one bet. This is statistics question. If he play infinite amount of games (until he doubles his money, since this is the stopping condition), he will get to 2000 for 100%. The question is how sure does he want to be in this process. The more times he bets (the smaller the bet is), the more likely it is he will get to 2000, without loosing it all, since the chances are in his favor (55% vs 45%).

"This is statistics question."

What have I been doing, nuclear physics?

I'll give a brief example. Suppose he wants to fence off the chance that he will lose 4 consequent times (4 bets of $500 each). The chance for losing in this case is 0.45^4=4.1%
So, if this is not enough for him, he should bet smaller bets. Of course, this is only partial solution, because he can win and lose in a lot of possibilities, and for that we need to use a more formalized statistical approach.

redhouse: probability.

@redhouse1938
It wasn't meant as a reply to you.

P.S. Being a nuclear engineer, I can safely say that you were not doing nuclear physics there.

He should stake 55% of his bankroll every night.

There is no correct answer. It depends on how much risk he wants to take. If he wants to be almost certain that he won't go broke then he should stake as little as possible. If he wants to make as much as possible then he should stake as much as possible, but will have a higher risk. You need to specify what chance he is prepared to risk going broke.

(Although, redhouse, it kind of is probability. So yeah).

What FreeThing said is correct.
An answer can be given in terms of "With n% probability of going broke in k years." n and k should be specified.

When I think of it, it is a more sofisticated question than I thought. Peter's goal is not to get to 2000 - but to maximize his return over an infinite amount of games. So, while it is safer to bet each night 1$, it does not mean that it is the strategy that will provide the maximum return. In fact, I think that the solution should not be as a number of $, but a percentage - like hellalt suggested.

Hellalt, can you explain why do you think 55% is the correct solution?

Baskineli, now you're making a very valid point. If the time at which you study the revenues is irrelevant, then you better put in low stakes to avoid the risk of losing everything.
However, if the guy is playing for a fixed number of nights, or you can "check" his revenues at any given time, I believe the precise answer to Maniac's question; the highest possible /expected return/ is the one I gave above.
You are all debating the /statistical spread/ around his earnings. That's interesting, but that's not Maniac's question, because he's asking to maximize revenue and not to minimize risk.

Yes, the question does say 'maximise his expected return' and a 100% stake would do that. Maybe the question needs rephrasing e.g. Given that Peter wants a less than 5% chance of going below £100 how much should he stake?

I only partially agree with you, redhouse.

Low stakes does not mean maximum return. I am only giving an example here, but if to bet each night 1$ will bring a revenue that might look like y=x^2, then it is possible that betting 5$ each night will bring a revenue that might be sharper, like y=x^3, which will grow faster than the first option. The question is what strategy will bring the fastest growing AVERAGE revenue, considering two parameters: 1. The daily bet. 2. The amount of risk Peter is willing to take to lose it all.

(And I am not even mentioning the fact that this is not a Poker game by description, but roullete - in Poker the probabilities change after each game, since the players learn to know each other).

Look at that I've been a little rude I guess... Didn't properly read all the commentaries. Sorry @Baskineli. Still sticking to my answer though, although proposed modifications to the question itself by Baskineli and @FreeThing are interesting. Back to work now :(

FreeThing, exactly.

He cannot go broke. Note that his investment in every game is a percentage of his bankroll, not a constant. So say he invests 1000 and loses it. Then in the next round, he will only put in 500, and so on.
Suppose he has money m, and invests t fraction of his money (i.e. tm) every time. Then there's a .55 chance of him doubling it, and ending up with m(1+t), and a 0.45 chance that he ends up with m(1-t).
Overall, in n rounds, his money will be multiplied by (1+t) 0.55n times and by (1-t) 0.45n times. So you want to choose t to maximise (1+t)^0.55n * (1-t)^0.45n, or equivalently, maximise 0.55log(1+t) + 0.45log(1-t). The derivative for this expression is 0.55/(1+t) - 0.45/(1-t). It is maximised when the derivative is 0, which is when 0.55/(1+t) = 0.45/(1-t). So t = 0.1. He should invest 10% of his money every round to maximise expected gain.

Wow - thanks guys, I did word the question poorly, I did want an answer that would work over infinate time and always avoid potential of going bust - which (I think) is the Manus has provided. Is the answer always the difference between the two percentages 55%-45% = 10% or is that just a coincidence?

It's a coincidence. You could do the Manas's calculation with variable pw which indicates the probability of winning as a number between 0 and 1, and then pl, the probability of loosing, would, of course, be (1-pw). In this case t = 2 pw -1, which is again number between 0 and 1 and for pw = 0.55 this gives Manas's result of t=0.1.

It also makes sense, if you probability of winning is 100% you should invest 100% of your money every time.

@Manas
Nice way to look at the problem. GG ;)

@Maniac
There is one problem, that you cannot bet on fractions of a cent, so you cannot eliminate the risk completely - there is always a chance that you will end up with 1 cent and lose it.

Actually I'm stupid and 2pw -1 is pf course nothing else than pw - pl, assuming pw > pl.

"he starts with a bankroll of £1000.00 and either wins double his money or loses everything. How much as a percentage of his bankroll should he stake each night in order to maximise his expected return. Assume that he wins 55% of the time and loses 45% of the time. "

1. if he won 100% of the time then he would have to stake 100% of his bankroll to maximize his return.
2. Now that he wins 55% of the time he has to stake 55% :P

@Baskineli - thanks for you comments

Am I right to assume that the formula holds true irrespective of number of players and frequency of results as long as the percentage winning rate is accurate. For example suppose the original question related to player against just 1 opponent. Now imagine that Peter players in large tournaments say 100 players and now he only wins 10% of the time but when he does win his prize is 55 times his stake - should he still stake 10% of his deminishing bankroll on each game?

As far as straight statistics go, its hard to take into account because somebody isnt going to lose that exact percentage of the time all the time. You have to take into account variance, and consecutive losing sessions. And sure, also consecutive winning sessions.

But as a poker player myself and somebody who has a career specialized in the poker industry, ideal bankroll management is to have roughly 20 buy ins for the stakes you're playing at. Also, consider that you may be bringing two buy ins each time you play. Yes, i know the example provided states one buy in, but I'm assuming you're probably asking for real life advice on the matter?

Consider that if you're playing 1/2 NL, a $10,000 bankroll gives you 50 buy ins. Your goal is to maximize your wins, and minimize your profits in each session. Therefore, by leaving after you have doubled up is unwise and a losing mentality. You are leaving good games too early and losing out on expected wins.

Basically, I would suggest playing with no little than 20 buy ins at whatever stake you choose. Try to win as much as you can, and only bring one buy in and stick to it. Being a poker dealer at the Bellagio, and conversing with many pros, I can actually ask them for their own opinion on the matter. That would probably give a better answer.

I think where I was going with the 1/2 NL $10,000 reference is that if you go to each game with two buy ins and intend to play both bullets, its kind of halving your bankroll. The same example would hold true with a 20 buy in bankroll.

@Jok3r - thanks for your comments and I think you are right with regards to cash/ring games and there does seem to be plenty of advice about managing a bankroll for such games. However, I am favouring single table tournaments at the moment and whilst I'm not crushing the game I'm doing OK. I just wanted to know if there was an optimum percentage of your bankroll to stake each session to maximise earnings without going broke.

What regulars do you deal to and are the tips good?

Well, if you're playing tournaments you have an even higher chance of variance because tournaments involve more luck. I am not sure what the recommended number for buy ins at a tournament would be but it definitely be higher as opposed to cash games. For tournaments, I would say no less than 30 buy ins? Maybe more. Variance is huge, and not going bust is the most important thing.

As far as regulars, the big TV professionals only come into town when we have tournaments going on or the world series. But I've dealt to nearly all of them. And yes, the money is excellent!

@Maniac

Please do not take the advices people gave you here into the real world - we were talking about pure statistical problem which was more similar to a roullete than poker.