Why are higher risk-reward ratios better than lower ones in gambling?

Question

Assume a game where you can place bets of $1\$$. The game is played as long as you either win $4\$$ or lose your $1\$$. There're two ways to place your bets:

Option 1: With a high risk-reward ratio of $1:4$, which means you win $4\$$ for each $1\$$ you risk, however your odds of winning are only $1:4$.
Option 2: With a low risk-reward ratio of $1:2$, which means you win $2\$$ for each $1\$$ risked, but with odds that are twice as high ($1:2$).

So, the probability of winning $4\$$ with option 1 is $\frac{1}{5}$, however with option 2 only $\frac{1}{3}\cdot\frac{1}{3}=\frac{1}{9}$. Why is that? Shouldn't the chance of winning be equal given that the odds for option 2 are twice as high?

Answer 1

I think that there is a malaise (particularly in USA ?) to treat odds and probability as synonymous which they definitely aren't.

Further, the term "win" or "winnings" is also subject to misinterpretation as to whether it is net winnings or the payout including the stake.

The correct way to compare two situations is to use probability and expected value

So let us say that for situation one, the probability of winning is $\frac14$ , you bet $1$, if $E[x]$ is the expected value of the game,

$E[X] =\frac14\cdot4 - 1 = 0$

and for the second case with the probability of winning being $\frac12$,

$E[X] = \frac12\cdot2 - 1 = 0$

In both the cases, it is a fair game

Of course, the casino will never offer such schemes, they will skew the scheme, but they can skew it (if they want) so that whether you risk losing/winning large amounts or small amounts, the casino's expected value for each game is the same.

For a real life situation, you will have to work out the expected value for each game, because by and large, all games aren't pegged to give exactly the same % gain for the casino, of course !

There is one high reward scheme which attracts a lot of people, lotteries. There the psychology of the better is that the modest entry fee (even if repeated, say periodically) doesn't really pinch, but if you win, you get a HUGE amount that catapults you into a different level altogether !

Answer 2

As mentioned in the comments, your calculation for Option $2$ is incorrect, as you only consider the path where you win twice in a row, and of course there are other ways to win. You could, for instance, win on the first round, then lose twice (putting you back in the starting slot) and then win twice. And there are many more winning combinations.

To do the computation correctly, I suggest working by states. Index a state by the amount of money you have. There are $7$ possible states, as you might have anywhere from $\$0$ to $\$6$. Easy to work out the transitions and from there, to get the chances of winning.

And that also exposes a conceptual problem with your analysis. You are absolutely correct in thinking that any bet you build out of a sequence of bets each with expectation $0$ must itself have expectation $0$. You can not manufacture expected gains or losses out of fair bets. Thus, if it were possible to take option $2$ and design a bet that either resulted in going bust or ending with $\$5$ then that must have the same probabilities as Option $1$ has. But...you can't. Not as stated. A possible sequence of bets would be $$(1)\to (3)\to (2)\to (4)\to (6)$$

Here, of course, $(n)$ denotes the state in which you have exactly $\$n$.

Note that that outcome gives you a profit of $\$5$ which is impossible in Option $1$. This makes it hard to directly compare the two strategies.

Still, worth working out the probabilities of winning in Option $2$ correctly, just as an exercise.