Poker Return on 100 Hands Just looking for a bit of validation of a theoretical problem I’ve been working on.
It may be completely unoriginal, but take 100 hands of poker. For arguments sake, say the difference between the odds of your cards winning and the pot odds is 5%. Also, say you have a starting balance of £100, and only ever risk £2 on one hand (remember, this is only theoretical). Could you say the following?
Starting Balance: £100
Number of Hands: 100
Odds Difference: 5%
Risk Level: 2% (or £2)
Total Return: £10

Answers on a postcard please!!
 A: Assume the pot size is $a$ (all in GBP), of which you have already contributed $b$. You are facing a decision to either fold or or to add another $c$ to the pot. After this the betting ends and your winning probability is $p$.
In this situation, if you call, your expected win is $$\tag1p\cdot (a-b)-(1-p)\cdot(b+c)$$ and if you fold it is $-b$. 
Hence calling seems to be an advantage as soon as $(a+c)p>c$.
This is precisely covered by the notion of  pot odds, which are $c:a$, or converted to a probability: $\frac{c}{a+c}$. I read your assumptions as $p=\frac c{a+c}+\delta$ with $\delta=\frac1{20}$, hence you do call and your expected win (computeed by plugging this $p$ into $(1)$) is $\delta\cdot(a+c)$. 
However, this cannot easily be related to your 2GBP beting limit (i.e. the constraint $b+c\le 2$). If there is a total of $n$ players (and none of the others have folded), we know that $a+c=n(b+c)$, i.e. after your call the pot would contain equal contributions from all players. If $n>2$, some may have folded after havng contributed at least something to the pot (and be it an ante or  blind), and then we only know that $a+c\le n(b+c)$. That makes your win per round $\delta(a+c)\le n\delta(b+c)$ which equals $\frac n{10}$GBP with the numbers you provided, so after $100$ rounds this makes $\le 10n$GBP (or $20$GBP if you have only one opponent).
