What is the better bet?
I didn't want to derail the other current Lotto thread so have started this thread to ask . . . What is the better bet?
One attempt at winning with 3,838,380 to one odds for the attempt, or ten attempts at winning with 38,383,800 to one odds for each attempt? Or are they both the same? |
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The value of money is not linear. People with no money are willing to work all day to earn a minimum wage of about $50. Give those same people $10 million and see how many of them don't give a damn about that $50.
The difference between buying 38 million tickets in one draw and buying 1 ticket in each of 38 million draws is in the latter case there is a 38% chance of not winning (actually 1/e or about 36.79%). Although it averages out with a probability of winning multiple times, your value of the multiple jackpots will probably not cover the loss. To look at it another way, if you had just won $10 million, would you take an even odds bet for double or nothing? |
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I fail to see what this debate has to do with the value of money. I don’t think anyone is seriously debating that Lotto is ever a good investment regardless of how it‘s played. That some people play Lotto with disposable income and others play with food and rent money is a different issue. The significant difference between having 38,383,800 attempts in 1 draw and 1 attempt in 38,383,800 different draws is that the former is an absolutely guaranteed result and the latter isn’t. The significant difference between having 0 attempt compared to having 1 attempt is the difference between impossible (0 in 38,383,800) and possible (1 in 38,383,800). Math “predicts” that a coin toss is 50/50 heads or tails but this doesn’t mean that the actual results of coin tosses will be evenly distributed in practice. It’s more “likely” than not in practice that any series of coin tosses will not be exactly evenly distributed. Regardless of how many attempts you have in the same draw the odds of any individual attempt winning are 1 in 38,383,800. Having multiple attempts only makes it more likely that one of those attempts will be the winning one. If you have 38,383,800 attempts in the same draw the likelihood that one of those attempts will be the winning one is 100% guaranteed. Having 10 attempts at 1 in 38,383,800 in the same draw doesn’t make it significantly more likely that you have the winning attempt than if you only had 1 attempt. Having 10 attempts still leaves 38,383,790 attempts you don’t have. |
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No amount of previous draws has any effect on current or future draws (the balls don’t have a memory). |
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The math says that the average outcomes are identical so the relative values you place on the possible outcomes makes the only difference. I suspect that many people over value the lottery jackpot. |
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Now let's say I play the game twice, picking one number each time. The odds that I'll win any one of those contests is 0.01, and the odds I lose any one of those contest is 0.99. That means that the odds I win twice are 0.0001 (0.01*0.01). I could also win the first time and lose the second time with a probability of 0.0099 (0.01*0.99), or lose the first time and win the second time with a probability of 0.0099 (0.99*0.01), or never win with probability 0.9801 (0.99*0.99). So in this scenario, I have an 0.0198 chance of winning once (add the two different ways of doing it together), an 0.0001 chance of winning twice, and an 0.9801 chance of never winning. That's close to the odds of winning with two picks in one draw, but it's not quite the same. But the smaller my overall odds are, the less difference it will make how I arrange things. But does it make a difference? Yes, it makes a difference. Maybe not enough for you to care, but that's a different issue. |
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The original post was ambiguous about what the questioner wanted. It presented alternate scenarios where the average payout was the same (assuming multiple payouts were possible), but like this extreme example, the distributions of payouts was not. If the prize for winning was something with decreasing marginal value, then the value of the winnings would not be equal for the two scenarios, even though the strict average would be. |
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I would rather entertain the possibility that I might ever win from 1 in 3,838.380 odds rather than from 1 in 38.383,800 odds regardless of how many attempts at the latter there are. |
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Yet we can probably come up with a scenario wher you would. Let's say you are out for a walk on a rather warm day and you happen to find a quarter on the ground. A little later you come across a street vender that will sell you an icecreme bar for 50 cents. Having left your wallet, change and a full pint of Ben&Jerry's Cherry Garcia at home, the quarter is all you have. The vender then offers you a deal of a lifetime (or at least the immediate moment) and will give you double or nothing on one flip of that quarter. Will you take the bet? Will you walk away with the 50 cents if you win? |
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ynot, at this point, I really don't even understant what you want to know. I gave you an answer to what I thought you meant by your initial post. You don't seem satisfied by the answer I gave, and now you appear to want to know something else, but I really can't tell what. I don't even have a clue as to why you can't answer whatever it is you want to know yourself.
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I have no academic qualifications and am pretty much self-taught in what I “know“. I don’t assume that everything I know is actually correct so I like to test it against people that might know better than I do. The problem isn’t coming up with an answer it’s knowing whether the answer is likely to be correct or not. I would rather know I’m wrong than believe I’m right. Essentially I’m asking if there is some point at which odds become so unlikely they aren’t worthy of serious consideration to be applied to the possibility of actual events. In this case is 1 in 38,383,800 at such a point? It’s not just that 1 in 3,838,380 is more likely, it’s that 1 in 38,383,800 is more unlikely. Does the math of 1 in 38,383,800 alone pretty much negate all other math to do with multiple attempts and draws? When I’m “thinking with my gut” it seems to do so. |
ynot, please read http://en.wikipedia.org/wiki/Decimal_mark. My brain is confused by your numbers.
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Or do you think they are both the same? ;) |
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That's the question, and the phrase "do you think" is at the core of it. The "mathematical" answer is yes (pretty close, but how do you deal with the fact that you can't lose more than once?). For most folks, however, there exists an ill-defined point where the perception of risk and reward change. This is another version of Dan O.'s "The value of money is not linear." It also pertinent to your question about behavior in the face of long odds. As the saying goes, "A bird in the hand is worth two in the bush." And the sort of question you posed in your OP, with a few changes, is instructive. Let's make the choice between a lottery with 10,000 to 1 odds (and a $10,000 payoff) and one with 1,000,000 to 1 odds and a $1,000,000 payoff. The average return on a bet is the same in each case. However, this assumes an infinite (or at least very large) number of tries. And none of us will live that long. Just for grins, let's assume you'll buy one ticket per week for 50 years, with no purchase during your two-week vacations. Then a more reasonable judgment is: what are your chances of winning something during your lifetime? In round numbers, you have a 1 in 4 chance of winning for the first lottery, and a 1 in 400 chance for the second. So which is better? That depends. Why would you play the lottery in the first place? For some folks, the first choice wins. 1 in 4 is not completely unreachable. But for many others, the longer odds are better, since the bigger payoff is more attractive. After all, you can't quit your job and retire on 10 grand, but you miight do so with a million. |
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Given most lottos split the winnings between multiple winners, winning 2x or more in the second lotto gains you nothing. You're just splitting in half the exact same pot you'd win with 1 winning ticket anyway. Hence you are slightly better off with the one bet on the first lotto. The math isn't all that hard, either. To calculate it, you invert the calculation. Consider two lotteries, one with 1/1000 and the other, 1/10,000 odds. The chance of winning is 1 minus the chance of not winning. In the first case, the chance of not winning is 999/1000. In the second, it's 9999/10,000, but that's with just 1 ticket. Now why calculate the inverse? Because in the second case, you are buying 10 tickets. And the chance of winning at least once is one minus the chance of winning none at all. And the chance of winning none at all is the chance of not winning on the first time, 9999/10,000, times the chance of not winning with the second ticket, 9999/10,000, and so on. That's 9999/10,000 to the 10th power, or 0.99900044988002099748020998800045, which is a whisker above the 0.999 chance of the 1 ticket lottery. So you're slightly more likely to not win anything with the 10 ticket lottery, the first number. |
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Why? Because not only can I not die twice, but I'd rather die tomorrow than today. On the other hand, if we took the trips in the reverse order (so that the ten trips remain the same, but the 1/1000 chance of dying were to fall ten days from now), then I might make the opposite choice. |
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