Bold and cautious play (probability theory)

Statement of the problem
A man in Las Vegas needs $40 but only has $20. He does not want to wire his wife for the money and decides to play roulette (having a negative attitude towards the game) according to one of two strategies:
▪ Bet all of your $20 on even and end the game immediately if he wins or loses.
▪ Bet one dollar on "even" until he wins or loses $20.
Which of these two strategies is better?

Solution to the problem
Another problem on random walks with absorption. Its solution is not given in full, but as a continuation of previously considered problems.

A "bold play", i.e. a bet of $20 at once, gives the player a winning probability of 18/38 ≈ 0.474. Calculating the winning probability when playing cautiously at a dollar per game comes down to the problem of ruining a player with the following parameters:

initial capital of the 1st player: m = 20;
initial capital of the 1st player: n = 20;
probability of a successful outcome in a single game: p = 18/38;
the probability of an unsuccessful outcome (loss) in a single game: q = 1 - 18/38 = 20/38.

Substituting these values into the formula for the probability of winning M, we obtain
P = [1 - (p/q)^m] / [1 - (p/q)^m+n] = [1 - (20/18)²⁰] / [1 - (20/18)⁴⁰] = [1 - 8.23] / [1 - 67.65] ≈ 0.108.
So, "playing carefully" reduces the player's chances of winning by four times compared to "playing boldly".



What random walks lead to: a decrease in the probability of winning when splitting bets

The intuitive explanation for this phenomenon is that old play is also fast play.

Playing fast reduces the time spent playing against the house, which is not harmless.

However, we have seen that intuition based on averages does not always lead to correct conclusions about probabilities. Dubins and Savage note that they are unaware of a proof based on such intuitive reasoning. However, in our special case of doubling the initial sum in the game of red and black, Savage's explanations below are based on this idea.

In preparing these explanations of casino play, Savage deliberately omitted some mathematical subtleties concerning the case of equalities in inequalities for probabilities.

“Golden Paradise”

Any harmless game can be played in Golden Paradise, provided the player has sufficient initial capital. A player entering Golden Paradise with x dollars and desiring a return of y dollars can achieve his goal with probability x/(x + y) by staking his entire fortune x on the only chance of winning y dollars with probability x/(x + y), which is obviously a harmless game. As is well known, no strategy gives a greater probability of winning, and the probability of winning is maximum if and only if the player is known to either lose x or win y dollars.

"Lesser Paradise"

"Lesser Paradise" is similar to "Golden Paradise", but with the significant difference that when leaving the gambling hall, the player must pay a tax of t (0

The example is about American roulette with "0" and "00". Therefore, the probabilities are calculated as 1/38.