# which of these fair gambles gives your the greatest chance of winning?

I came to know about a gambling game with following conditions:

a> Betting option for 3/4 winning chance.

1> You will have to bet 1 of 4 numbers.

2> If the out come of game doesn't match with your bet, You WON 100 coins.

3> If the out come of game matched with your bet, You LOSS 300 coins.

b> Betting option for 1/4 winning chance.

1> you will have to bet 1 of 4 numbers.

2> If your bet matched out come of game, you got 3 times. i.e., if you bet 100 coins, you get 300 coins if you won.

3> In case it you loss, you pay to dealer 100 coins.

c> Betting option for 2/4 winning chance.

1> For same game, You will have to bet any two 2 out of 4 numbers.

2> If you won you get double.

In case a , your probability of winning is 3/4 and loss 1/4. Ratio of of money you get when win to money you pay when loss is 1:3.

Here is my questions:

1> I think in big picture, the player have more chance to win because odd of player winning is higher. How do dealer make money at all?

2> If player triple his stake when player loss the bet, next time player win the bet, he should got his money back. The probability of loosing 3 game consecutively is very low (0.25*0.25*0.25 = 0.015). So there is more chance for dealer to went bankrupt. Again How do dealer make money at all?

3> Do you think there is another way to win this game or increase probability of winning the game?

P.S. : This question is truly for academic purpose. I am not trying promoting gambling here.

-

## migrated from mathematica.stackexchange.comMay 23 '13 at 9:48

This question came from our site for users of Mathematica.

Try this simple simulation and see who wins at the end.

play = function(asset){
outcome = sample(c(1,2,3,4),1)
yourpick = sample(c(1,2,3,4),1)

if(outcome != yourpick)
asset = asset + 100

else
asset = asset - 300

return (asset)
}

asset = 5000
times = 1

while (times < 100 & asset > 0){
asset = play(asset)
print(asset)
times = times + 1
}


It is generally true that the dealer won't make any money if the winning chance is greater than 50% (the dealer, however, can charge a fee for playing the game and make money that way).

You can play around with my code and test your strategy in part 2).

-
Thanks Linda. I tried writ a code (in c++) to simulate the situation. but its look like there is some pattern in random number i generated. in the simulation, this game will cause player bankrupt and on the other hand, the player playing roulette (33.33% winning chance) is winner. So i was looking for mathematical proof that this game give player advantage. –  someone_ smiley May 27 '13 at 3:11
appreciate your answer, thanks :) –  someone_ smiley May 27 '13 at 3:12

I am very surprised that no one has answered this before!

The answer to this question is that, the 3 formats of the game:

a) pays out $1:3$ with probability $3/4$

b) pays out $1:1$ with probability $1/2$

c) pays out $3:1$ with probability $1/4$

Consider a strategy like this: you start with $0<x<100$ dollars, you bet anything you want, until you either reach $100$ or $0$, playing any of these games. say, if you are at 95 dollars, you can bet 2 dollars on (a) or (b) but you cannot do so on (c), because it would take you over 100 dollars, but you may bet 1 dollar or 50 cent, should you wish.

Here is a fun fact:

No matter which strategy you use, the probability you win is $x/100$. This follows from something known as optional sampling theorem.

I am sure you hear of this strategy:

bet on head, if you win stop, if you lose double your bet. For this strategy, eventually, you will win, so what is the catch?

The answer is that you might be down by an arbitrarily large amount before you make your 1 dollar. You may be down $1023$ bucks if it went tails 9 times consecutively or maybe $2^{100}−1$ if it went tails first $99$ times consecutively. Basically you are trading a big chance of winning a small amount of money (1 dollar) for a small chance of losing a vast a mount of money, so in the end it is all fair. Notice for this strategy, there is no cap, comparing to the previous example.

MORAL: YOU CAN GUARANTEE TO MAKE MONEY ONLY WHEN YOUR POCKET IS INFINITELY DEEP

I have previously written a post on gambling on unfair games. See Can you make money on coin tosses when the odds is against you?

Notice that when the odds is even, everything I said on the other post still remains true, because a martingale is also a supermartingale.

-

The definition of "fair game" is quite shaky.

Let's imagine a trivial two-player game - repeatedly a coin is flipped; if it's heads then player A gives a dollar to player B; if it's tails, then player B gives a dollar to player A; the game ends when someone runs out of money.

Nonetheless, the game is assymetric and 'unfair' if the starting amount of money is different. If A starts with 1 dollar and B starts with 9 dollars, then someone will end up with \$10, but A will get it less than the expected 10% times and B will get it more than 90% times; thus the game benefits B.

-