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Tags gambling , lotteries , math puzzles , probability

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Old 3rd March 2012, 01:02 AM   #1
ynot
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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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Old 3rd March 2012, 01:22 AM   #2
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Originally Posted by ynot View Post
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?
You actually need to be more specific about your setup. The latter has a slightly higher chance of never winning. But it also has a very small chance of winning more than once. So if winning twice is twice as good as winning once, then the two are equivalent, but if winning twice isn't twice as good, then the former is better than the latter.
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Old 3rd March 2012, 01:36 AM   #3
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Originally Posted by Ziggurat View Post
You actually need to be more specific about your setup. The latter has a slightly higher chance of never winning. But it also has a very small chance of winning more than once. So if winning twice is twice as good as winning once, then the two are equivalent, but if winning twice isn't twice as good, then the former is better than the latter.
If the ten attempts were each in different draws and each attempt could win “their draw”, would this be better or worse than the ten attempts being in the same draw and only one attempt could win?
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Old 3rd March 2012, 01:42 AM   #4
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Originally Posted by ynot View Post
If the ten attempts were each in different draws and each attempt could win ďtheir drawĒ, would this be better or worse than the ten attempts being in the same draw and only one attempt could win?
That question is exactly equivalent to your first question, with the ten attempts in one draw being the same as one attempt in one draw with 10x the chance.
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Old 3rd March 2012, 01:57 AM   #5
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Originally Posted by Ziggurat View Post
That question is exactly equivalent to your first question, with the ten attempts in one draw being the same as one attempt in one draw with 10x the chance.
Are you saying that having 38,383,800 attempts in 1 draw (guaranteeing a win) is equivelent to 1 attempt in 38,383,800 different draws?
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Old 3rd March 2012, 05:40 AM   #6
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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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Old 3rd March 2012, 01:03 PM   #7
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Originally Posted by Dan O. View Post
People with no money are willing forced to work all day to earn a minimum wage of about $50.
FTFY

I fail to see what this debate has to do with the value of money. I donít think anyone is seriously debating that Lotto is ever a good investment regardless of how itĎs 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 isnít. 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 ďpredictsĒ that a coin toss is 50/50 heads or tails but this doesnít mean that the actual results of coin tosses will be evenly distributed in practice. Itís more ďlikelyĒ 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 doesnít 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 donít have.
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Old 3rd March 2012, 01:16 PM   #8
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Originally Posted by ynot View Post
Are you saying that having 38,383,800 attempts in 1 draw (guaranteeing a win) is equivelent to 1 attempt in 38,383,800 different draws?
The average payout is the same. The distribution is not (you'll end up with some cases of not winning anything, but some cases of winning more than once). So it depends on what you care about, which was my point.
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Old 3rd March 2012, 01:48 PM   #9
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Originally Posted by Ziggurat View Post
The average payout is the same. The distribution is not (you'll end up with some cases of not winning anything, but some cases of winning more than once). So it depends on what you care about, which was my point.
I understand your point but wonder if it’s actually applicable in practice when the odds of any individual attempt ever winning are so incredibly unlikely. Would 1 attempt in 100 billion draws really make it any more likely you would ever win against 100 billion to 1 odds?

No amount of previous draws has any effect on current or future draws (the balls don’t have a memory).
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Old 3rd March 2012, 02:17 PM   #10
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Originally Posted by ynot View Post
I fail to see what this debate has to do with the value of money.

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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Old 3rd March 2012, 02:22 PM   #11
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Originally Posted by ynot View Post
I understand your point but wonder if itís actually applicable in practice when the odds of any individual attempt ever winning are so incredibly unlikely. Would 1 attempt in 100 billion draws really make it any more likely you would ever win against 100 billion to 1 odds?

No amount of previous draws has any effect on current or future draws (the balls donít have a memory).
Let's do a simple example. Our contest is to pick a number between 1 and 100. If I pick once, my odds of winning are 0.01. If I pick two different numbers for the same contest, my odds of winning are 0.02, and my odds of losing are 0.98.

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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Old 3rd March 2012, 02:24 PM   #12
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Originally Posted by Dan O. View Post
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.
But are ďaverage outcomesĒ really applicable in practice when the odds of ever winning are so incredibly unlikely?
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Old 3rd March 2012, 02:29 PM   #13
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Originally Posted by ynot View Post
I fail to see what this debate has to do with the value of money.
Would you rather get $1 million for sure, or flip a coin and get $2 million if you win, nothing if you lose? The average gain is the same either way. The distribution is not. Many people would pick the former because $2 million isn't twice as useful as $1 million, so the extra risk isn't worth the possible benefit. This is often referred to as marginal value: the million and first dollar is not as valuable as the first dollar. Declining marginal value is typical for most resources.

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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Old 3rd March 2012, 02:30 PM   #14
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Originally Posted by ynot View Post
But are ďaverage outcomesĒ really applicable in practice when the odds of ever winning are so incredibly unlikely?
Why not, ynot?
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Old 3rd March 2012, 02:41 PM   #15
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Originally Posted by Ziggurat View Post
Let's do a simple example. Our contest is to pick a number between 1 and 100. If I pick once, my odds of winning are 0.01. If I pick two different numbers for the same contest, my odds of winning are 0.02, and my odds of losing are 0.98.

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.
To demonstrate the point I think I have I would prefer a contest to pick a number between 1 and 100,000,000,000 rather than 1 and 100 as the likelihood of ever picking the correct number from 1 in 100 is far greater than from 1 in 100,000,000,000 (even thought the math is essentially the same).
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Old 3rd March 2012, 02:43 PM   #16
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Originally Posted by Ziggurat View Post
Why not, ynot?
As I just said - "the likelihood of ever picking the correct number from 1 in 100 is far greater than from 1 in 100,000,000,000"
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Old 3rd March 2012, 02:51 PM   #17
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Originally Posted by ynot View Post
To demonstrate the point I think I have I would prefer a contest to pick a number between 1 and 100,000,000,000 rather than 1 and 100 as the likelihood of ever picking the correct number from 1 in 100 is far greater than from 1 in 100,000,000,000 (even thought the math is essentially the same).
Well, now that you've seen me do the math in one case, you can go ahead and do it for whatever example you want, can't you?
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Old 3rd March 2012, 03:37 PM   #18
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Originally Posted by Dan O. View Post
To look at it another way, if you had just won $10 million, would you take an even odds bet for double or nothing?
No, I would not.
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Old 3rd March 2012, 04:28 PM   #19
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Originally Posted by Ziggurat View Post
Well, now that you've seen me do the math in one case, you can go ahead and do it for whatever example you want, can't you?
Iím not sure I need to provide an example when I only question the application of your example to a scenario that is highly unlikely to actually occur. It seems pointless to apply math to a scenario that is practically all-but a fantasy.

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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Old 3rd March 2012, 04:44 PM   #20
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Originally Posted by Almo View Post
No, I would not.

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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Old 3rd March 2012, 05:01 PM   #21
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Originally Posted by Dan O. View Post
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?
Why does this remind me of one of noworneverís Zen koans?
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Old 3rd March 2012, 05:04 PM   #22
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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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Old 3rd March 2012, 05:49 PM   #23
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Originally Posted by Ziggurat View Post
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.
Sorry I didnít mean to be vague and confusing. I appreciate the effort you and others have given.

I have no academic qualifications and am pretty much self-taught in what I ďknowď. I donít 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 isnít coming up with an answer itís knowing whether the answer is likely to be correct or not. I would rather know Iím wrong than believe Iím right.

Essentially Iím asking if there is some point at which odds become so unlikely they arenít 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? Itís not just that 1 in 3,838,380 is more likely, itís 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 Iím ďthinking with my gutĒ it seems to do so.
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Old 3rd March 2012, 06:11 PM   #24
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ynot, please read http://en.wikipedia.org/wiki/Decimal_mark. My brain is confused by your numbers.
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Old 3rd March 2012, 06:29 PM   #25
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Originally Posted by Dan O. View Post
ynot, please read http://en.wikipedia.org/wiki/Decimal_mark. My brain is confused by your numbers.
Sorry about that (fixed) - Itís not my fault that someone thought putting , and . next to each other on keyboards was a good idea.
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Old 3rd March 2012, 07:31 PM   #26
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Originally Posted by ynot View Post
Essentially Iím asking if there is some point at which odds become so unlikely they arenít worthy of serious consideration to be applied to the possibility of actual events.
There is no objective answer to that. One can calculate the odds, one can calculate the frequency of events, etc, etc. But whether or not it's worth considering, well, that's up to you. I am fine ignoring a 1 in 1000 chance of dropping an egg when I take it out of the refrigerator, but if there was a 1 in 1000 chance that I would die in a car crash on my way to work, that would worry me.
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Old 3rd March 2012, 07:52 PM   #27
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Originally Posted by Ziggurat View Post
There is no objective answer to that. One can calculate the odds, one can calculate the frequency of events, etc, etc. But whether or not it's worth considering, well, that's up to you. I am fine ignoring a 1 in 1000 chance of dropping an egg when I take it out of the refrigerator, but if there was a 1 in 1000 chance that I would die in a car crash on my way to work, that would worry me.
Would you rather have a 1 in 1,000 chance that you would die in a car crash on your way to work during one trip, or a 1 in 10,000 chance during each of ten trips?

Or do you think they are both the same?
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Old 3rd March 2012, 08:05 PM   #28
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Originally Posted by ynot View Post
Would you rather have a 1 in 1,000 chance that you would die in a car crash on your way to work during one trip, or a 1 in 10,000 chance during each of ten trips?

Or do you think they are both the same?
Since I can't die twice, the latter is better than the former, though the difference is small.
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Old 3rd March 2012, 09:19 PM   #29
WhatRoughBeast
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Originally Posted by ynot View Post
Or do you think they are both the same?
ynot -

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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Old 3rd March 2012, 09:58 PM   #30
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Originally Posted by Ziggurat View Post
You actually need to be more specific about your setup. The latter has a slightly higher chance of never winning. But it also has a very small chance of winning more than once. So if winning twice is twice as good as winning once, then the two are equivalent, but if winning twice isn't twice as good, then the former is better than the latter.
This.

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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Last edited by Beerina; 3rd March 2012 at 09:59 PM.
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Old 4th March 2012, 05:02 AM   #31
Roboramma
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Originally Posted by Ziggurat View Post
Since I can't die twice, the latter is better than the former, though the difference is small.
It also depends on the timing of the trips: if I have a 1/1000 chance of dying on my trip today, that's worse than a 1/10000 chance of dying every day from today-ten days from now.

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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