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First of all, I am not sure if this is the right place to ask this question, I wanted to post it at the mathematics community but also wasn't too sure. I read somewhere that when dice are fair, the odds are always with the casino. Shouldn't it be the other way around? Which also makes me question is a fair die harder to control than an unfair die?

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closed as off-topic by Deusovi Aug 19 at 1:46

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If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Welcome to Puzzling! This doesn't seem to have anything to do with puzzles, so I have closed it as off-topic. If you'd like to learn about what types of questions are appropriate for this site, I recommend taking a look at the Help Center. $\endgroup$ – Deusovi Aug 19 at 1:46
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Do you mean craps? The casino wins without cheating because the mathematical odds gives them an edge – based on fair dice. So they don't need to cheat, and if caught could lose their license. Only the punter uses a loaded dice, to try to swing the odds in their favour, because as said, the odds are against them.

Please see Loaded Dice – What Is It & How Does It Work?

The reason casinos exist is because the games they offer are always loaded in their favour mathematically (for example by the 0 and 00 in roulette). Sometimes casinos do contrive to load the games artificially in their favour, but not usually.

For the punter, it's not so much the dice they control, but the way they place their bets (unless there is a way to throw the dice to take further mechanical advantage of the loading).

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In games where players win money from other players, the house wins by shaving a percentage off of every bet, meaning the fairness of the dice is irrelevant to them. In games where players win money from the casino, they win by ensuring that a player has a less than 100% return on average. In Roulette, this is achieved with the zero where no one wins. They also win by the nature of gambling itself: people get addicted and tend to scale their bets to their current wealth. If you do this, you are guaranteed to lose. The only way to even nearly break even with a casino is to keep betting a flat amount.

Example of how scaling your bet to your current wealth results in losing money: Let's say you start with $100. Let's say you are in a non-existent casino where the win rate is exactly 50% and you win as much as you bet. Let's consider a loss-win and a win-loss where you keep betting 50% of your current wealth:

START: 100 LOSE: 50 WIN: 75

START: 100 WIN: 150 LOSE: 75

As you can see, you would end up with less either way.

Games can also be structured in a way that encourages or forces players to put more money in to hold onto a chance of winning. Human psychology has a number of vulnerabilities, such as the Sunk Costs fallacy.

Of course, if someone was to come in with loaded dice and play a game that used dice, they'd turn the odds in their favour, just because they'd know that certain numbers would show up more than usual. You'd have to plant the dice and risk getting in a fair amount of trouble.

The casino is unlikely to use loaded dice because it would be an unnecessary risk when they are already making a sizeable profit, and with a game where the dice could be exploited in favour of the players, it is safe to assume it would be.

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    $\begingroup$ Your scaling example (in a fair game with 50% win/loss probability) is faulty as you cannot just consider win-loss and loss-win. You also need to allow for loss-loss and win-win. In the former case you end up with 25 and in the latter case you end up with 225. Each possibility (L-L, L-W, W-L, W-W) has the same probability so your expected capital at end of play, if you bet half your capital each time and play two rounds, is (25 + 75 + 75 + 225)/4 = 100. Which is exactly what you started with. i.e. a 'fair' game gives no advantage or disadvantage to the player. $\endgroup$ – Penguino Aug 18 at 23:09
  • $\begingroup$ Of course someone can walk away after a win and effectively "beat" the house, but you're missing the point; an addict will keep playing and their win/losses will average out. If they scale their bets to their current wealth, they constantly lose most their money. The money is going somewhere of course - the mechanics of gambling essentially concentrates wealth in fewer and fewer hands. And so there will always be someone who seems to have a lot of money due to a recent win, but if they keep playing and keep scaling to their current wealth, they will also lose it. $\endgroup$ – Joshua Bizley Aug 19 at 0:38
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    $\begingroup$ That was not my point, For any 'fair' 50% game if the player follows your described strategy,f always playing half their capital (starting from some initial value C) the game is balanced with no advantage or disadvantage to the player. It only becomes a disadvantage to the player if a) there is a limiting stake beyond which the casino won't accept a bet, or b) if the player has a set goal for capital (either in terms of winning or losing) beyond which they won't make any more bets. If play runs for a pre-chosen number of games or an arbitrary time limit then the expected return is exactly C. $\endgroup$ – Penguino Aug 19 at 3:00
  • $\begingroup$ "The expected return is exactly C". Except most future versions of you lose, and in any given set of people using the strategy, the vast majority of participants will lose. Anyway I should stop debating on a closed question. $\endgroup$ – Joshua Bizley Aug 19 at 3:50

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