That’s interesting, but Clothahump didn’t get into that, making no mention of either the lump sum or the annuity. All he mentioned was the taxman’s take.
Or you’re underthinking it! 
The calculation you refer to calculates the casino’s average win (or player’s average loss) on total action (*), but (although this peripheral to the main point) there’s a lot more total action when you’re dropping dollar coins in a slot machine than when you’re playing $10 Keno tickets during lunch.
The detailed example I gave involved mutliplying your money by a large factor with even-money bets. You can reverse this for a similar conclusion: doubling your money with long-odds bets.
For example, suppose you have $70 and want to double your money at the roulette wheel. Your best first play is not $70 on Red, but $2 on any single number. If you win, you walk away, “forfeiting” the vigorish on just one $2 bet.
I don’t know what your math or programming skills are, but there is a clear mathematically-defined question/answer here. If you’d like we could … er … bet on the answer!
(* - that means “total action” for the player. One way to help understand the paradox is that casino’s “action” is different than player’s action unless bets are specifically even-money.)
It’s a sucker’s bet if you play with any kind of real hope or expectation of winning (i.e. had a bad week at work, buy a stack of lotto tickets so you won’t have to go back on Monday), or if you play using a stake you can’t afford just to throw away (i.e. your children will go without shoes because you bought lotto tickets).
Sadly, both/either of those scenarios are not uncommon.
I’ll work a more detailed example to clarify my claim and its correctness. (The claim, briefly, is that very-high-payoff high-vigorish bets are better for the player than low-payoff low-vigorish bets.)
Suppose the casino offers two roulette wheels.
Wheel A has only a single 0 but allows only color bets (18 winning numbers) or column bets (12 winning numbers), with the column bets allowed only immediately after winning color bets. The house vigorish is a fixed 2.70% on all bets. (The peculiar betting restriction is not essential to the underlying idea; it just gives us very well-defined bets for the example.)
Wheel B has 0 and double-0 but allows bets on six numbers at a time. The house vigorish is a fixed 5.26% on all bets.
You have $X and want to win an additional $Y. For simplicity suppose X = 100 and Y = 500. At which wheel should you play? Wheel A, with its 2.70% house advantage or Wheel B with its 5.26% house advantage?
Answer: Your best chance is Wheel B, despite its higher vigorish!
Proof: Bet $100 on a 6-number set and win $500 with chance 6/38 = .1579. If instead you play at Wheel B, you’ll bet a color, hope to win, then let-it-ride on a column bet. Your success chance is (18/37)(12/37) = .1578.
This is why, if you’ve set yourself a specific win target, a Keno ticket may be a much better bet than an even-money roulette bet despite its much higher raw vigorish.
And why Lotto isn’t as bad as the raw vigorish figure would imply. (But the Lotto analysis will be complicated by different payoffs, the non-cash payoff, and taxes.)
What you say boils down to the fact that you want a scheme that involves a single bet, to avoid allowing the house to take multiple rakeoffs.
That’s just ridiculous.
First, your talk about losing vigorish on your bet makes no sense. The casino looks at vigorish as a percentage of all its bets. But for you as a player making a single bet, the concept is meaningless. You can’t lose 2.7% of your bet - you either win and lose nothing or lose and lose 100%.
Second, your strategy for winning also doesn’t add up. Sure, it’s true that the money you’ll win on a $70 color bet is the same as the money you’ll win on a $2 number bet. But you can’t blithely say “if you win” and pretend the two bets are the same. You have a 47.37% chance of winning a bet on red and doubling your money. And you have a 2.63% chance of winning a bet on thirty-six and doubling your money. So the two strategies are not equivalent chances for doubling your money on a single bet.
Look at it this way, suppose two hundred people all go into the casino with $70 and a desire to double their money on a single bet. A hundred of them bet everything on red and a hundred of them bet two dollars on thirty-six. In the first group, fifty-three lose their seventy dollars and forty-seven double their money. In the second group, ninety-seven lose two dollars and leave with sixty-eight and three double their money. So betting on red is a much better strategy if your goal is to double your money in a single bet.
Let’s not quibble about the semantics, or the meaning of “vigorish” or trying to look at the casino’s point-of-view. I’ve defined a very straightforward mathematical problem. You can approach it directly with mathematics or computer programming. I even worked a specific example in detail.
If you have $70 and want to double it at the roulette wheel then (assuming for math simplicity that the casino will allow bets of any fractional amount) your best chance is to start by betting $2 on a 35-1 bet (the max payoff on the table); a bet on any lower payoff would yield a lower net probability that you will achieve your goal.
This is a simple mathematical fact. Casino reasoning or psychology is irrelevant.
I think we could find some mathematicians or computer scientists among the Straight Dopers we could agree to use as arbiters! Would you like to … er … bet on the question? 
I’ll give odds!!
I’m just having a hard time understanding what you’re claiming.
A roulette bet on red pays 1-1. You bet $70, you collect $70 if you win. That doubles your money.
A roulette bet on a single number pays 35-1. You bet $2, you collect $70 if you win. That also doubles your money.
So I agree that both bets will double your money if you win.
But how can you say the two dollar bet is “your best chance”? It just is not. Your chance of winning the color bet is 47.37% - your chance of winning the number bet is 2.63%. A 47.37% chance is a better chance than a 2.63% chance.
I didn’t realize explaining this would be so hard. :dubious:
If you bet $70 and lose, you’re out of money. Game over. Go home.
If you bet $2 and lose, you still have $68 left. You now bet $2.057 on a 35-1 proposition. If you win that bet you’ve got your $140 total. Go home with success.
If you lose the second bet also, bet $2.116. Again, victory sends you home successfully. If instead you lose all three of the bets, you still have $63.827 and place a 4th bet. Get the idea?
I’ve made the claim that the final probability of achieving the $70 win goal is greater following the procedure just described than the 47.37% chance you correctly calculate for the $70-on-red bet. (I’m not going to type in the detailed tedious arithmetic for this! I did work the details for a simple case upthread.)
I (and Xema, partially) have indicated that one way to understand why this works is that you may avoid house “rakeoff” (“vigorish”). For example, if the $2 bet wins, the house has “raked off” only from $2.
There are various other ways to understand the principle intuitively or to describe the relevant math more succinctly, but I won’t go there with you: you’ll find someway to confuse yourself about my terminology and call me “ridiculous”!
P.S. Rereadingmy earlier post I see “your best chance is to START by betting $2”. Apparently I should have added the emphasis then.
It sounds like you’re trying to reinvent the Martingale.
I don’t know if you think a smug pointer to Wikipedia refutes simple arithmetic.
Look, suppose you have $70 and need $140 quickly for an insulin shot and that a roulette wheel is your only option. There’s a best strategy and there are inferior strategies. Smug cliches about Martingales and casino motives have no effect on simple arithmetic facts.
The simple fact I’ve outlined about payoffs and probabilities explains why Keno (or Lotto) tickets are not as bad of bets as often portrayed. Note that I wrote “not as bad”, I didn’t write “good.”
(I’m afraid I have become exasperated in this thread. For what it’s worth, septimus is not my real name; whatever my weaknesses I do know about gambling, Martingales, and mathematics.)
I do see what you’re saying now. Your earlier posts were implying that your system was for a single bet - either a seventy dollar bet on a color or a two dollar bet on a number. I now understand you’re talking about the possibility of a series of bets on a number.
But you’re still wrong. Let’s look at the numbers.
Here’s the series of bets you’d be making. The second number is the amount you currently have left of your original seventy dollars. The third number is how much you’d have to win on this bet to walk away with $140. The last number is how much you have to bet to win that needed amount.
1 = 70 = 70 = 2
2 = 68 = 72 = 2.06
3 = 65.94 = 74.06 = 2.12
4 = 63.82 = 76.18 = 2.18
5 = 61.64 = 78.36 = 2.24
6 = 59.40 = 80.60 = 2.31
7 = 57.09 = 82.91 = 2.37
8 = 54.72 = 85.28 = 2.44
9 = 52.28 = 87.72 = 2.51
10 = 49.77 = 90.23 = 2.58
11= 47.19 = 92.81 = 2.66
12 = 44.53 = 95.47 = 2.73
13 = 41.80 = 98.20 = 2.81
14 = 38.99 = 101.01 = 2.89
15 = 36.10 = 103.90 = 2.97
16 = 33.13 = 106.87 = 3.06
17 = 30.07 = 109.93 = 3.15
18 = 26.92 = 113.08 = 3.24
19 = 23.68 = 116.32 = 3.33
20 = 20.35 = 119.65 = 3.42
21 = 16.93 = 123.07 = 3.52
22 = 13.41 = 126.59 = 3.62
23 = 9.79 = 130.21 = 3.73
24 = 6.06 = 133.94 = 3.83
After that, you’re down to $2.23 so you don’t have enough left to make the bet you’d need.
So what you have are 24 chances to win. If you win on any one of them you’ll have $140. And what are the odds of winning one bet when each of them has a 2.63% chance of winning?
47.25%
And what are your chances of winning a single bet on a color?
47.37%
I’ve never had any problem with people who buy one ticket per drawing or less. You don’t even really need any sort of chance of financial gain to do that (although you’d have to think you did) - if simply having a plausible fantasy brings you enough happiness to be worth a buck, then it’s a fairly reasonable deal.
But… buying a second ticket doesn’t mean you can fantasize twice as hard. Really, it has no additional value at all but it doubles your (poor) investment. The first ticket is reasonable enough - it’s a sucker bet (and to quickly resolve the question in the OP - can you come up with a definition of sucker bet that doesn’t fit the lottery?), sure, but you may deem it to be worth a dollar based on the secondary benefits. However, given that there are no additional secondary benefits with the second ticket you buy, that’s when you become an idiot.
I worked as a gas station cashier years back - most of the people I encountered who played the lottery don’t fit in that group. They were mostly retards who spent way too much money on the lottery, and usually money they didn’t look like they had to blow.
My previous post was so tediously detailed as to almost be sarcastic, but apparently wasn’t tedious enough.
When you’re down to $2.23, do you go home? Think carefully before reading the spoiler.
You bet the $2.23 on a 35-1 shot. You won’t have $140 if you win, but you’ll have enough to continue play.
Now, repeat your calculations with this correction. If you still get a wrong answer I’ll help again. Or use this shortcut:
2.23/140 = 1.593%. (No, I won’t try to explain that trivial formula for you.) Adjusting for “rakeoff” reduces this to about 1.45% The situation occurred about 52.7% of the time. 1.45% * 52.7% + 47.25% = 48.01% roughly. That’s larger than 47.37%. (I’ve used your numbers without checking them.)
Hell, anybody who has ever played Keno in Vegas or at a Casino, knows it’s a sucker bet, one of the lowest on odds games in casinos. Same game as the lotto, except there’s thousands of drawings a day. Even the State lottery commissions are starting to get into Keno, too. Helluva shrewd move… but the people who sit at bars and Restaurants playing it religiously round after round are bigger suckers than the lotto folks.
One good thing about Keno, though… if you play it for any amount of time it will give you a better real-life, temporal, and mathematical idea of just how low your odds are of winning the lotto.