Can you explain what you mean by this? If you win or lose, you have to pay $1. Either way, you lose the voucher.
Here’s a very simple example.
Suppose you have $100 and all you want is another $300 so that you’ll have just enough to buy your next fix of [del]heroin[/del] [del]vampire blood[/del] to tip the hatcheck girl enough to retrieve your briefcase crammed with cash. :eek: Let’s consider just three possibilities:
[ul]
[li] Bet it all on Red; if you win, let it ride. To get your fix you need to win two of these even-money shots. 18/38 * 18/38 = 22.44%[/li][li] Bet it all on 1-9 which pays 3 to 1. Your chance of buying your fix is 9 out of 38 = 23.68%, significantly better than you’d get with the even-money chance.[/li][li] Bet $8.57 on #18. If it wins, walk away winner. If it loses, you’ll bet $8.82 on the next spin; and so on. The arithmetic details get tedious, but your chance of success is now 23.82%, better still.[/li][/ul]
Comparing options 1 and 2, for which the arithmetic is very simple, should be enough to demonstrate the principle. When the house offers two bets with the “same vigorish,” the one with higher payoff is generally better for gambling purposes. (Casinos understand this implicitly, as you can see checking the odds on most games. Roulette is an exception, due to the 0/00 tradition.)
My reasoning was the same as Riemann’s. In a normal casino game, if you lose, then you lose your original stake, and if you win, you get it back plus something more. And in most games, all the bets are structured around this assumption, such that they all have the same expected value. In this game, though, if you lose, it works just like a normal play, but winning is different: When you win, in addition to what normally happens, you lose your voucher. The fact that you also lose it when you lose is irrelevant, because that’s just like a normal game.
Since this is an actual casino OP is talking about, if they take the coupon away after a win, they also take it away after a loss. You lose the coupon every time, either way.
The EV is the same for every bet in roulette, on a 00 wheel, it’s -0.526
As other posters have alluded to, how you look at it is going to depend on how many coupons you’re going to have.
I think the answer below kind of covers it. I think this answer argues for the single number bet strategy (at least for some medium number of trials) because if you’re only playing the coupon, your real loss is zero:
Q:
Hi Wizard, you’ve got a great site. In double 0 roulette, I realize all the bets have the same high house edge, but I’m just finishing a stats course and it seems to me that not all the bets are quite the same due to their standard deviations. A $1 bet on Red, for instance, has an S.D. of 1.012019 while a $1 bet on a single number has an S.D. of 5.839971, according to my calculations. Thus, the expected probability of coming out ahead over 1, 100, and 10000 trials, respectively is 0.4793, 0.3015, and 0.0000 for an even-money bet, and 0.4964, 0.4641, and 0.1837 for a single number bet. Is my analysis correct? (I assumed normality) Thanks!
Mike from Toronto, Canada
A:
Thanks for the compliment. First of all the standard deviation on any even money bet is 0.998614 and on a single number is 5.762617. The probability of coming out ahead by flat betting even money bets over 1, 100, and 10000 spins is 0.473684, 0.265023, and 0.00000007 respectively. The probability of coming out ahead by flat betting single number bets over 1, 100, and 10000 spins is 0.0263158, 0.491567, and 0.18053280 respectively. It seems you are trying to argue that single number bets are better because of the higher probability of finishing ahead over multiple bets. This is true, however the probability of a substantial loss is also much greater. Over a session the expected results always fall somewhere on a bell curve. With low volatility bets like red or black that bell curve is sharp and doesn’t stray far from a small loss. With high volatility bets like single numbers the bell curve is wide, allowing for a much wider range of net results, both good and bad.
Edited: Damn, I’m sucking at reading and math today, I erased my question and I’m going home!
Oh, and about the longer-odds games being better for the player, consider the extreme example: Suppose that the casino also offered a bet for all-numbers-and-zeroes. If you make this play, then you put down your dollar, and if the ball lands on anything at all, then you walk away with 97 cents. This is actually a considerably better expected value than any of the standard bets. Would you take it? Why or why not?
What it comes down to is, if you’re in a casino, it’s because you like taking risks, and are willing to pay the house a small amount for the opportunity to take a risk. If you don’t like risk, then you wouldn’t play any casino game. And a longer-odds bet lets you buy more risk for the same price.
This is, at best, very misleading. It assumes that the size of the wager(s) would be the same in each case. See Post #22.
ok, i really am going home because i’ve become a moron. Sorry for wasting everyone’s time 
This analysis is wrong. In the OP’s scenario, losing your stake when you win is something unusual that happens in addition to the usual payout. If you lose, you normally lose your stake, so nothing unusual is happening.
Thus, if normal bets (with stake returned when you win and not when you lose) are structured with identical EV, then voucher bets in the OPs scenario - which add something unusual to the win payout but not the lose payout - do not have the same EV.
No. Forget variance, the expected return for a 1/36 chance voucher bet is almost double the expected return for an even-money voucher bet. However many vouchers you get, always always make long-odds bets.
Sorry I can’t explain it better.
Something that may simplify thinking about roulette is to focus on ONLY single-number bets. An $18 bet on Red is exactly the same as betting $1 on each of the 18 red numbers. Note that, after betting $1 on #18, a $1 bet on #19 is “betting against yourself.”
Carrying things to an extreme is very often the easiest way to demonstrate a combinatorial fact of this type. It may not be obvious that playing 18 numbers is worse than playing 1 number, but carry it to Chronos’ extreme and bet $1 on each of 38 numbers. Now you’re guaranteed to lose $2 every spin! That’s incompatible with any gambling objective.
Sorry to have apparently abandoned this thread, I went on holiday shortly after posting it. Thanks for all the replies, which have convinced me the long odds bet is much better. Quercia and Turble nailed it.
I’ve seen this at more than one UK casino (where you get half your initial stake back on an even money bet if the wheel lands on 0) but I don’t think it’s universal over here. We also only have one 0, improving the odds still further. None of that is particularly relevant to the original discussion, of course.
I don’t criticise anyone who takes this attitude as it does have some merit. But most people go to a casino hoping to leave with more money than they arrived with. As such it makes sense to try and work out the best chance of doing so (and obviously it will always be lower than 50%).
I’m pretty sure most UK casinos absolutely forbid their staff from playing in their own casino at all, probably across the whole national chain actually, and I believe it’s frowned upon for them to gamble in other casinos. In this particular case one of the conditions of use of the voucher is it has to be played by the recipient. You are correct that it would not be possible to cash in the voucher in any other way, such as using it to pay for drinks.
Incidentally, having lost my voucher on a single number bet, I discussed the odds with the croupier. They didn’t particularly commit one way or the other, but they did point out that for the single number strategy to pay off, it would take three years (as they generally send one voucher a month). That’s fine by me, given the much better return.
I can see why this would be the case now everyone has explains it, but not so in my casino, which is handy.
If your goal is to leave with more money than you arrived with, and you want to have the best chance of doing so, you can actually get that chance very high indeed (precisely how high depends on the size of your bankroll). That’s what a Martingale scheme does. The catch is that, in the unlikely event that you don’t leave with more, you end up losing so much that it’s still in the casino’s favor.
Absolutely- I’d plop my free voucher down on that one and walk out with my free 97 cents. Why? Highest EV. Better than the long odds roulette option.
Using my own money, I’d be equally likely to play that one as any of the pure gambling options with a house rake. 0=0.
This hypothetical bet has a 97c payoff including your stake. Since (with the voucher bet) you have to return your stake, it’s equivalent to just throwing your voucher away.
The statement of the hypothetical was that “if the ball lands on anything at all, then you walk away with 97 cents.” You appear to be fighting the hypothetical.
Your new, different hypothetical, in which you do not walk away with 97 cents, would be a worse bet.
It’s not a different hypothetical, you just haven’t read the thread carefully. Chronos was hypothesizing a regular (non-voucher) bet that returns 97c, including the stake. A voucher bet is like a regular bet except that the stake is not returned. So Chronos’s entire point was that you would never make that bet with a voucher, because it’s equivalent to throwing away your voucher.
Indeed. Personally, I’m not happy with that particular high risk/low reward strategy.
Yeah, the voucher version of that all-numbers bet would be that you put your voucher down, and then whatever result came up, you’d lose your voucher and also have to pay an additional three cents. That’s clearly terrible. But the point is that, even in the normal play version, it’s still a bad bet, because you’re still paying a house cut, but not getting any variance in exchange for it.
Here’s a slightly different (but equivalent) way of looking at things. We can break the behavior of the voucher into three simultaneous “games”:
- A completely ordinary roulette spin with $1 stake
- The casino pays you $1 unconditionally (the face value of the voucher)
- You pay the casino $1 on a win (you don’t win the stake)
But we can combine 2 and 3 into this simplified version:
- A completely ordinary roulette spin with $1 stake
2a) The casino pays you $1 on a loss
Clearly, we want to lose as much as possible! Assuming, that is, that the expected value across the different bets is even–which is the case for roulette. So we go for the long odds and get the casino to pay us $1 across most of the spins.
Forgot about this thread. Yeah, I figured it out on the drive home that day. It only seems to work since all bets are normally the same EV, and with the lower odds bets, the amount you have to give to play is a bigger percentage of your win.
thanks for the explanations.