Preface: I've only high-school knowledge in Maths, so this is just an experiment I've run on my pc and I can't motivate the result I get. I'll be wrong in some points for sure so please be nice :)

Consider 2 different skilled people betting on sport events. When they bet on a given event, their probability of winning are $P_1$ and $P_2$.

We suppose that player 1 is more skilled and has an advantage on player 2 (player 2 is not skilled at all and bet randomly, so his probability of success is the same of the probability of the event).

Let's suppose $P_1 = P_2 + kP_2$ where $k$ represent the advantage of player 1 for example $k = 0.05$.

In a given amount of trials $n$ there's a chanche that player 2 has made more correct bet then player 1 (they're betting on events with the same probability of course). I've set $n = 10$.

So I'm calculating $P(X_2 > X_1)$ where $X_1$ and $X_2$ are "success counters" variables.

I expected that $P(X_2 > X_1)$ was maximum where $P(X) = 0.5$, where the variance is maximum, and then decrease simmetrically on both sides. Instead my experiment showed that the maximum was between $ 0.3 < P(X) < 0.4 $.

Why this happen? Is there a function that links $n$,$P(X)$ and $k$?

Is it correct to state that: "In a short-term period we can recognize more easily a more skilled bettor if he's betting on high variance events (those with $P(X) = 0.5$) instead if he's betting on low variance events"?

Is it correct to suppose that: "if we have no competence or if we have to take a negative expected value bet we have better probability of result short-term winner by taking bets with high variance $P=0.5$?"

  • $\begingroup$ Strictly speaking you have not said what $P(X)$ means. Perhaps is is the same as $P_2$? $\endgroup$ – Henry Aug 31 '12 at 21:31
  • $\begingroup$ You might also get closer to symmetry if you looked at $P(X_2 \gt X_1) + \frac12 P(X_2 = X_1)$ rather than just $P(X_2 \gt X_1)$. Making $k$ smaller might also help. $\endgroup$ – Henry Aug 31 '12 at 21:33
  • $\begingroup$ Yeah, not know what $P(X)$ means, since $X$ is not defined, and if it is a random variable, then $P(X)$ doesn't make sense - it only make sense to write $P(\mathit{expr})$ where $\mathit{expr}$ is a true/false expression. $\endgroup$ – Thomas Andrews Aug 31 '12 at 21:43
  • $\begingroup$ I think my answer is okay and it ignores the mysterious P(X). The question can be addressed without considering what this P(X) is. I think he may have meant when p$_1$=p$_2$. which is the minimum over all cases with p$_1$>=p$_2$. $\endgroup$ – Michael Chernick Aug 31 '12 at 21:46
  • $\begingroup$ Sorry if I didn't declare it. Where I say $P(X)$ I mean 'the probability of winning betting on a determined event'. Coin flip's $P(X) = 0.5$. So the experiment shows that there are better chanches for the less skilled player to be ahead of the other one when he bet on events that have a probability between 0.3 and 0.4. $\endgroup$ – KingBOB Sep 1 '12 at 8:03

When there is no advantage (i.e. p$_1$=p$_2$ where p$_1$ and p$_2$ are player 1 and player 2s respectively success probabilities) P(X$_1$>X$_2$)=P(X$_2$>X$_1$)=1/2 - P(X$_1$=X$_2$). But if X$_1$ has a larger p for success then P(X$_1$>X$_2$) will increase and become greater than 1/2 when the difference between the two success probabilities p$_1$ and p$_2$ becomes sufficiently large.


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