What is the better bet?

[ynot]

[FONT=Verdana, sans-serif]I didn't want to derail the other current Lotto thread so have started this thread to ask . . . What is the better bet? [/FONT]


[FONT=Verdana, sans-serif]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?[/FONT]


[FONT=Verdana, sans-serif]Or are they both the same?[/FONT]

 

[Ziggurat]

[FONT=Verdana, sans-serif]I didn't want to derail the other current Lotto thread so have started this thread to ask . . . What is the better bet? [/FONT]


[FONT=Verdana, sans-serif]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?[/FONT]


[FONT=Verdana, sans-serif]Or are they both the same?[/FONT]

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.

 

[ynot]

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.

[FONT=Verdana, sans-serif]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?[/FONT]

 

[Ziggurat]

[FONT=Verdana, sans-serif]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?[/FONT]

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.

 

[ynot]

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.

[FONT=Verdana, sans-serif]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?[/FONT]

 

[Dan O.]

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?

 

[ynot]

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 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.

 

[Ziggurat]

[FONT=Verdana, sans-serif]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?[/FONT]

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.

 

[ynot]

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).

 

[Dan O.]

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.

 

[Ziggurat]

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).

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.

 

[ynot]

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 “average outcomes” really applicable in practice when the odds of ever winning are so incredibly unlikely?

 

[Ziggurat]

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.

 

[Ziggurat]

But are “average outcomes” really applicable in practice when the odds of ever winning are so incredibly unlikely?

Why not, ynot?

 

[ynot]

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).

 

[ynot]

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"

 

[Ziggurat]

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?

 

[Almo]

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.

 

[ynot]

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?

I’m 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.

 

[Dan O.]

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?