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May 9[edit]

Gambling[edit]

Say you're betting on something with 50/50 chance. Is there a way to put your stakes so that you make money in the long run? Money is tight (talk) 09:18, 9 May 2014 (UTC)[reply]

Be a bookmaker. Seriously - if the actual underlying probability of thing happening or not happening is 1/2, the only way to make money on it consistently is to charge other people to bet on it.

Simple proof: Because P and ~P have equivalent likelihoods, it doesn't matter which one you bet on. Let's assume you bet on P every time. Now, assuming you can find someone to make a fair bet with you, you'll win that bet exactly half the time and lose it exactly half the time. So your expectation is zero every time. While it's very likely you'd make either a small profit or a small loss in the long run, there's no way to ensure a profit instead of a loss, and no way to ensure a large profit rather than a small one. Therefore: be the bookie. AlexTiefling (talk) 10:14, 9 May 2014 (UTC)[reply]

What if one keeps increasing the bet (if one can find someone to play against) to cover and exceed previous losses? For instance, bet $1 and lose (balance: -$1); bet $2 and lose (balance: -$3); bet $4 and win; balance: +$5); bet $1 and lose (balance: +$4)... Hayttom 10:36, 9 May 2014 (UTC) — Preceding unsigned comment added by Hayttom (talk • contribs) [reply]

That's the Martingale betting system. It doesn't work in the long run (and often in the short run too). It just means you could be increasing your financial risk exponentially. If you're on a losing streak, you're risking more and more to win $1. VERY, VERY BAD. Clarityfiend (talk) 11:19, 9 May 2014 (UTC)[reply]

That's not quite right: bet $1 and lose (balance: -$1); bet $2 and lose (balance: -$3); bet $4 and win; balance: +$5: Should be bet $4 and win; balance: +$1 That's the problem with the Martingale, every eventual win only gives you $1 more than what you lost before. Using the Martingale system, and assuming minimum bet is $1 and you've got $2ⁿ-1, every time you play (until you win) your odds of losing everything are 1/2ⁿ, your odds of winning $1 are (2ⁿ-1)/2ⁿ. Ssscienccce (talk) 14:43, 9 May 2014 (UTC)[reply]

It works in theory - but in practice, it fails. The practical problem is that the amount you have to bet to get your $1 win rapidly grows to the point where the time taken to play the game multiplied by the interest rate you'd have gotten on your stake money exceeds the interest you could earn from putting it into a savings account. So although doubling your stake with every bet until you win does assure that you'll eventually win - that's only a theoretical matter. And in theory, it could take you an infinite amount of time and money to do that. Since nobody has infinite money (or time), you're taking a chance that you'll eventually be forced to give up right after losing your entire net worth - and the odds of that happening by far outweigh the benefits of a potential $1 win!

So on a true 50:50 chance bet, there is no way to do any better than to break-even...which is what you'll do eventually without any complicated 'system'.

An alternative to the Martingale system is to simply continue making $1 bets until you are "up" by $1, and then stop gambling...that eliminates the exponential need for cash - and replaces it with a linear increase in cash input over time. But it still suffers from the problem that there is still a 50/50 chance that you'll never get into the black before you've lost your entire net worth.

The answer should really be "NO!" there is no way to reliably win at 50:50 odds. SteveBaker (talk) 15:07, 9 May 2014 (UTC)[reply]

You actually can't win EVER in the long term, even if the odds are in your favor. That's because the house has an essentially infinite bankroll. Take a 50/50 game for example. You don't win every other game, there will be streaks where you win some, and streaks where you lose some. There's some significant chance that you'll eventually hit a losing streak that will equal your entire net worth. The casino also has that chance, but the difference in odds between you hitting your killer losing streak and the casino doing the same are going to be the difference between the size of your cash reserves and that of the casino. The longer you play, the better the odds get for the casino that you'll hit a streak bad enough to put you out of cash. The concept even has a name. The Gambler's ruin. --Jayron32 18:08, 9 May 2014 (UTC)[reply]

The easiest way to imagine this is to consider the events of a series of coin flips as a Random walk. A graph of your winnings looks exactly like a one-dimensional random walk...and in that case, the expected distance you get from the starting point is the square root of the number of steps you take. So if you start off with $10 and bet $1 on each coin flip - you should expect to either lose the entire $10 - or to have taken $10 from your opponent within about 100 coin flips. If your opponent has $100 and you have only $10, he can most probably withstand a game of 10,000 coin flips and you can only withstand 100...so in a long series of such "flip coins until bankruptcy" games, he's going to win about 100 times more often than you despite having just ten times as much money to lose. That really does highlight the reasons why you're not going to break the bank at a Casino anytime soon - if they have a million times more money than you stashed away then breaking the bank is a longer-than-the-life-of-the-universe scale of difficulty....even if the games are "zero sum" - which they aren't! SteveBaker (talk) 21:02, 9 May 2014 (UTC)[reply]

If the odds are in your favor, you definitely can win, as Gambler's ruin: #Unfair coin flipping shows. n1 and n2 are the amount of money, p and q the respective odds (sum is 1), P₁ is the chance of player two going broke (the article says player one, but on the talk page someone commented already in 2007 that the formulas were reversed).

P_{1}={\frac {1-({\frac {q}{p}})^{n_{1}}}{1-({\frac {q}{p}})^{n_{1}+n_{2}}}}

If q<p and n2 is infinite, the denominator equals 1 and the expression becomes:

P_{1}=1-({\frac {q}{p}})^{n_{1}}

With p = 0.51 and q = 0.49, and enough money for ten games (n1 = 10), you have a 32% chance of breaking the bank, and with n1=18 your odds are better than 50% ... Ssscienccce (talk) 01:24, 10 May 2014 (UTC)[reply]

Or get with The Mob and have the game rigged so that the supposed odds don't reflect reality. ←Baseball Bugs ^{What's up, Doc?} carrots→ 15:00, 9 May 2014 (UTC)[reply]

I don't think there is a way to win doing that either...The Mob doesn't come cheap! SteveBaker (talk) 15:07, 9 May 2014 (UTC)[reply]

Good point. The issue of winning or losing in the long run was explained in the movie Trading Places, in which the Duke brothers explain to Eddie Murphy's character how being a stockbroker works. One brother says, "Whether the stock price goes up or down, we still get our commission." Murphy says, "Sounds like you guys are a couple of bookies!" The second brother says to the first, "See, I told you he'd understand!" ←Baseball Bugs ^{What's up, Doc?} carrots→ 19:51, 9 May 2014 (UTC)[reply]

If the expected payoff of a single game is zero, then the expected payoff of any number of games is zero regardless of your betting strategy. But if your goal is not to maximize your expected profit, but, say, to maximize your chance of ending with more money than you started with, then some strategies are better than others. For example, if you start with $999, bet $1 per game, and quit when you reach $0 or $1000, then (as discussed in gambler's ruin) you have a 99.9% chance of coming out ahead, whereas if you just make 999 bets of $1 you have only a 50% chance. These are not really long term strategies, but you can turn them into long-term strategies. For example instead of quitting when you reach $1000 you can start making very tiny bets so that it will take longer to get down to $999 than you plan to spend playing. If bets can be arbitrarily small then you can arrange to never drop below $999 once you exceed it, by betting less than the difference each time. However, the zero expected payoff means that you can't expect to win more than $1, if you do win, regardless of strategy (still assuming $999 to start and a 99.9% chance of winning). -- BenRG (talk) 02:16, 10 May 2014 (UTC)[reply]

If they give you comps (free hotel room, meals, drinks) for gambling, then you could, in theory, come out ahead. However, casinos that do this obviously don't give you a 50/50 chance of winning, and the comps come out of the house's cut. But, if you could get to 50/50, say by counting cards, then the comps could put you over the top, at least until they figure out what you're doing and toss you out. StuRat (talk) 03:06, 10 May 2014 (UTC)[reply]

The casinos don't offer even money bets, with the exception of some slot machines. You could enjoy comps from slot play for "free" (see Jean Scott (author)), but that's about it. Clarityfiend (talk) 03:26, 10 May 2014 (UTC)[reply]

Why would they offer even money bets on anything, considering that once their overhead is figured in that would mean a loss for them ? To get people in the door, in the hopes that they will also play more profitable games ? (In other words, a loss leader.) StuRat (talk) 14:54, 10 May 2014 (UTC)[reply]

Let me be a bit more precise. There are machines that have an expectation value over 100%, and according to this article, they are indeed loss leaders. The article doesn't mention it, but from what I've read, in video poker, you could easily determine by the payout table if that particular machine pays out over 100% with the correct playing strategy. Clarityfiend (talk) 19:31, 10 May 2014 (UTC)[reply]

The article progressive jackpot discusses the situation. Some lotteries have those too. 70.36.142.114 (talk) 19:38, 10 May 2014 (UTC)[reply]

The short answer to your question, as others have already remarked, is "no". This is true both for the simple example that you have mentioned, and also for more complicated examples that people may try to persuade you to invest in. A good general rule is "if something seems too good to be true, it probably is".

As it happens, things are worse than you think. Suppose you have $100, and you are offered a 50-50 bet (the toss of a coin, say). If you win you'll have $200, but if you lose you'll have nothing. The drop from having $100 to having nothing at all is much worse than the gain you make by going from $100 to $200: if you've got $100 at least you can call home to get help, or buy that vital medicine your child needs, or just stay warm and get some food, but if you're down to nothing you're in real trouble. Economists describe this effect in terms of utility. Daniel Bernoulli, who first introduced the idea, presented it by writing "[The] utility resulting from any small increase in wealth will be inversely proportionate to the quantity of goods previously possessed." That is, what you might gain from your bet in terms of the value you place on your spending power is less than the value you might lose, even though the bet itself is "fair". Or to put it another way, you'd need better than 50-50 odds for the bet to be "fair" on a utility/value basis.

A good rule of thumb is "If your future personal finances rely on a lottery win, you should probably change your behaviour". RomanSpa (talk) 08:23, 10 May 2014 (UTC)[reply]

A quicker remembered rule of thumb is "Nothing ventured, nothing gained." "Time is money" and "thinkin' ain't drinkin'". That said, you make fine points. For those of you who want some "lucky" money, kicking expert Daron Cruickshank is +375 against the merely well-rounded Erik Koch tonight. You won't have to bet a hundred times to win here. InedibleHulk (talk) 19:54, 10 May 2014 (UTC)[reply]

Head kick at 3:21. Ka-ching! InedibleHulk (talk) 03:45, 11 May 2014 (UTC) [reply]