Upon proving that the best betting strategy for “Gambler's Ruin” was to bet all on the first trial Any explanation please to make the meaning of the following theorem (stated in The Gambler's Ruin (1909) by J. L. Coolidge in Annals of Mathematics, Second Series, Vol. 10, No. 4 , pp. 181-192) clear.  
"The player's best chance of winning a stated sum at an unfavorable game is to stake the sum that will bring that return in one turn. If that be not allowed, he should stake at each turn the largest amount the banker will accept."
I found the aforementioned paper where this theorem originated from but could not follow it.
There is also the following quotation by Julian Coolidge in H Eves Return to Mathematical Circles (Boston 1988):
[Upon proving that the best betting strategy for "Gambler's Ruin" was to bet all on the first trial.]
It is true that a man who does this is a fool. I have only proved that a man who does anything else is an even bigger fool.
what does it mean? please.
Any reference about this specific question is appreciated.
 A: Assuming a 50/50 chance of winning and losing, any "strategy" you come up with will lose, or tie the simple bet all on first turn. One possible end is that you go so far into debt that the banker will not accept the amount required to get you out of debt. Another possible end is just playing the game over and over again until the law of large numbers takes over, and you end up having an equal chance of making or losing money. In real life however, your odds are typically lower than 50/50, say 49/51, so even the second strategy is worse. You are guaranteed to lose money over a long period of time, compared to having a 49% chance of gaining money by just betting all on your first turn. The man is a fool, because he is betting all of his money on a single chance, but any other strategy is a bigger fool, because it is expected to end worse.
A: An unfavourable game is one in which the expected gain to the player is always negative.  A player who continues to play such a game is certain to lose his entire initial bank, no matter what betting strategy he uses.  If his aim is simply to win a pre-specified amount, however, he can do so with positive probability—unless, of course, any bet always loses—as long as he quits as soon as he has achieved his aim.
The result referred to by the two authorities cited is that the player maximises the probability of achieving his goal if he adopts the strategy described—that is, always bet exactly the amount that will achieve his goal immediately, or the betting limit whenever it is smaller than the required amount, and quitting as soon as he has achieved his goal.
As a simple example, consider the case where the player's probability of winning at each turn is $\ p\ $, and the odds paid are $\ r:s\ $. The player's expected gain per unit bet is $\ pr-(1-p)s\ $, which is negative whenever $ \frac{r}{s} < \frac{1-p}{p}\ $. In these circumstances, a player whose aim is merely to win an amount $\ w\ $ should first bet $\ \frac{w}{r}\ $, and, thereafter, $\ \frac{\left(1+r\right)^nw}{r^{n+1}}\ $ after any series of $\ n\ $ consecutive losses.  After $\ n\ $ consecutive losses, the player will have lost a total of $\ \left(\frac{1+r}{r}\right)^nw-w\ $. If he wins on the next turn he will win $\ \left(\frac{1+r}{r}\right)^nw\ $ for a nett total gain of $\ w\ $, whereupon he has achieved his goal and quits. If the player's initial bank is $\ \left(\frac{1+r}{r}\right)^bw-w\ $, he will achieve his goal unless he suffers a run of $\ b\ $ consecutive losses, which occurs with probability $\ (1-p)^b\ $.  His probability of achieving his goal is therefore $\ 1-(1-p)^b\ $, and no other betting strategy can do better than this.
A good reference for this kind of problem is Dubins and Savage's classic How to Gamble if You Must.
