For the discussion of math. Duh.

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WarDaft
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So, you're dying, and the reaper shows up early to play a game with you.

The game is, the reaper will flip a biased coin. There is a 55% chance of heads, which will double your remaining earthly time, and a 45% chance of tails, which will reduce your remaining earthly time by 70%. The reaper will play this with you as often as you want, but you must state ahead of time how many games you will play. The games will be resolved instantly and exactly according to these rules with no cheating.

So, if you have 10 days left, and you play, you have a 55% chance of ending up with 20 days left, and a 45% chance of ending up with 3 days left. Doing the math, this has an expected value of 12.35 days. Awesome! Each game increases your expected remaining time by over 20%, and the most probable outcome is an increase in time remaining! What could go wrong?

So you declare you'll play the game with the reaper 1000 times. That's not living forever, but it ought to be pretty close! At first it was great, you got three heads in a row and you have months extra! But by the time you're done, your time left on this plane is best measured in Plank units! The reaper offers to reset the game, and play once again with 1000 tries if you feel like you were unlucky. You get even worse results! Indeed, this is a loosing game!

----

This is a losing game, but I can't reason my way to why it doesn't make sense even once. It's not gamblers ruin, because you are neither strictly increasing the amount wagered (you clearly decrease the wager on a loss) nor is the expected value zero of any flip zero or negative.

Help me XKCD, you're my only hope.
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jaap
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### Re: I don't seem to understand my own brain teaser, please h

It is not really valid to take the (weighted arithmetic) average of two multiplication factors to get the expected multiplication factor. A simpler example would be a 50:50 chance of doubling or halving, which should be neutral in the long run, and not have an average muliplication factor of (2+.5)/2 = 1.25

Expected values are additive, but the setup of this problem is muliplicative. You can turn it into an additive problem by taking logs, and then you do get a negative expectation for the change of log(remaining time):
0.55*log(2)+0.45*log(0.3) < 0

quantropy
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### Re: I don't seem to understand my own brain teaser, please h

The expected value is still 10*(1.235)^1000 days. However, this is made up of a very small chance of living a very long time, and a very large chance of living a very short time.

If you could persuade the Reaper to allow you to keep some of your days left on each turn and just gamble with the rest of them then you could use the Kelly Criterion to exend your days with high probability.

PeteP
What the peck?
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### Re: I don't seem to understand my own brain teaser, please h

(10*2^1000)*(0.55^1000) (winning 1000 times in a row ) alone contributes a number with 42 zeroes to the expected value while you can at most lose 10 days. If you lose once you lose 7 days if you lose 1000 times an a row you still have only lost slightly less than 10 days.

Expected value alone just isn't a good guide if you have to bet everything. It doesn't warn you when you trade a huge possible gain for a lower victory chance.

WarDaft
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### Re: I don't seem to understand my own brain teaser, please h

Okay, that all makes sense for the multi play case.

But what about the single play case? Does it make sense to play the game once, or no times at all?
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elasto
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### Re: I don't seem to understand my own brain teaser, please h

The answer to that will be subjective: Seems to me there's a lot you could do in 10 days in terms of visiting and spending time with friends and family and such, that an extra 10 days wouldn't help with - but going down to 3 days could really screw you over.

It similar to how an extra \$10k is nice but no biggie to someone with \$10m in the bank, but would be huge for someone with only \$1k in the bank; Every dollar is not equal in value to every other, and nor is every day equal in value: In both cases, the more you have, the less value each additional one has.

Your own personal 'value curve' has to be employed to judge if going from 10 days to 20 days is worth enough to risk dropping from 10 days to only 3...

mfb
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### Re: I don't seem to understand my own brain teaser, please h

By the Kelly criterion, you should only bet 1/10 of your time each time (if you get that option). Betting everything is clearly far away from optimal.
If you get this option, you can play 10^10 times, or as long as you like (but ask for a suicide option in case 1 million years don't turn out to be so nice).

BedderDanu
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### Re: I don't seem to understand my own brain teaser, please h

elasto wrote:The answer to that will be subjective: Seems to me there's a lot you could do in 10 days in terms of visiting and spending time with friends and family and such, that an extra 10 days wouldn't help with - but going down to 3 days could really screw you over.

It similar to how an extra \$10k is nice but no biggie to someone with \$10m in the bank, but would be huge for someone with only \$1k in the bank; Every dollar is not equal in value to every other, and nor is every day equal in value: In both cases, the more you have, the less value each additional one has.

Your own personal 'value curve' has to be employed to judge if going from 10 days to 20 days is worth enough to risk dropping from 10 days to only 3...

Using something like the geometric mean to come up with an expected value, I get

(3^.45)*(20^.55) = 8.5167 < 10

Which is about a 20% loss. Over 1000 games, this amounts to 10^-96 days, or according to wolfram alpha 10^-48 planck times.

Bloopy
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### Re: I don't seem to understand my own brain teaser, please h

elasto wrote:The answer to that will be subjective: Seems to me there's a lot you could do in 10 days in terms of visiting and spending time with friends and family and such, that an extra 10 days wouldn't help with - but going down to 3 days could really screw you over.

Winning a bunch of time could really screw you over as well if it turns out you have to spend it dying slowly in the hospital bed.

Personally I wouldn't put a whole lot more value on 10 days than on dying the same day. 4 weeks or more might be a better amount of time for finishing things I started or travelling to places I always wanted to go. With that in mind, some of the probabilities look quite appealing considering the circumstances. 7 plays is the best chance of getting 4 weeks or more, and also adds the outside chance of gaining a few years. Some of the riskier options at least leave a reasonable chance of retaining a week or more:

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### Re: I don't seem to understand my own brain teaser, please h

quantropy wrote:The expected value is still 10*(1.235)^1000 days. However, this is made up of a very small chance of living a very long time, and a very large chance of living a very short time.

Wait, isn't this wrong? If I win 55 times and lose 45 times I end up behind. Isn't the only valid way to do this with logs as jaap said, so you'd expect the log of your multiple to be about -0.0697, which means you should multiply your days remaining by exp(-0.0697) = .9327 per game?
Let's have a fervent argument, mostly over semantics, where we all claim the burden of proof is on the other side!

Coyne
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### Re: I don't seem to understand my own brain teaser, please h

Okay, 55/45 is actually 11-of-20 vs. 9-of-20; of 20 turns, 11 heads and 9 tails. So let's take 20 turns as an example. Since multiplication is commutative, , it doesn't matter what order they come in, alternation or all heads first/tails first. So the results of the 20 terms can be written:

new_life = prev_life * 211 * .39 = prev_life * 2048 * 0.000019683 = prev_life * .040311

So on average, for every 20 plays, you lose a bit more than 96% of your life, if the distribution is truly random according to the given distribution. Therefore, for your 1000 term game, I would expect your remaining life to be 1.86665x10-68% of your starting life.

Death's game sucks: would not play. Now if he'd just change the game so I keep 70% of my life for tails, well, then, 1,000 turns and a look at the end of the universe, please.
In all fairness...

>-)
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### Re: I don't seem to understand my own brain teaser, please h

quantropy wrote:The expected value is still 10*(1.235)^1000 days. However, this is made up of a very small chance of living a very long time, and a very large chance of living a very short time.

Wait, isn't this wrong? If I win 55 times and lose 45 times I end up behind. Isn't the only valid way to do this with logs as jaap said, so you'd expect the log of your multiple to be about -0.0697, which means you should multiply your days remaining by exp(-0.0697) = .9327 per game?

No. If you win 55 times and lose 45 you end up behind, but this is made up for by the chance of living a very long time. The fact that your wins and losses are exactly proportional to their probabilities doesn't demonstrate anything in this case. I think the thing with the logs will give the expected geometric value, which is different from the expected value. This doesn't mean jaap's post is wrong, because he was talking about the average multiplicative factor, which does decrease.

Coyne's post makes the same error. Just because a run which perfectly matches up with the expected proportions, 9/20 and 11/20, result in a loss, that doesn't prove that on average the expected value decreases.

Coyne
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### Re: I don't seem to understand my own brain teaser, please h

>-) wrote:Coyne's post makes the same error. Just because a run which perfectly matches up with the expected proportions, 9/20 and 11/20, result in a loss, that doesn't prove that on average the expected value decreases.

You stated what the average case was, when you said, "There is a 55% chance of heads, which will double your remaining earthly time, and a 45% chance of tails, which will reduce your remaining earthly time by 70%." This means that for every flip there is a 55%/45% chance of heads versus tails; and that also, over a long series of flips, on average 55% of those will be heads and 45% will be tails. You can't lay down a stated probability and then expect the common case for the result to be something else.

Let's stop for a moment to consider the case of two flips:

• Heads/heads: probability .55*.55 = 30.25%; new_life=prev_life * 2 * 2 ; won
• Heads/tails: probability .55*.45= 24.75%; new_life=prev_life * 2 * .3 ; lost
• Tails/heads: probability .45*.55= 24.75%; new_life=prev_life * .3 * 2 ; lost
• Tails/tails: probability .45*.45= 20.25%; new_life=prev_life * .3 * .3 ; lost

Total outcome probabability covered is 100%, chances of winning are 30.25%; chances of losing are 69.75%. So right off the bat, the two-flip game is a bad risk.

As I noted, math is commutative, so it doesn't matter what order the flips come in. For four flips, all of these have the same outcome, in terms of your new_life:

• new_life = prev_life * 2 * 2 * .3 * .3 ... or ...
• new_life = prev_life * .3 * .3 * 2 * 2 ... or ...
• new_life = prev_life * .3 * 2 * 2 * .3 ...

This means that for any given game, we can simply count the heads as x and tails as y, and this equation gives the change to life:

• new_life = prev_life * 2x * .3y

For our 10,000 flip game, break-even can be determined from a pair of linear equations:

• 2x * 0.3y ≈ 1
• x + y = 1,000

Now I freely admit I don't know how to solve those equations directly, so I did it in Excel by starting with 1,000 heads 0 tails and heading toward 0 heads, 1,000 tails. I found that, if you get fewer than 635 heads (365 tails) you lose. Since in your given probability suggests the likelihood is 550 heads and 450 tails, you are quite heavily weighted to lose for 1,000 flips.

I don't know how to compute the actual chances of getting 635 heads, but you're way out at 63.5% heads and 26.5% tails, which is not what Death promised.

1. flip: 55% -- better than even
2. flips: 30.25%
3. flips: 57.48% -- better than even...surprising to me that it's a bit better than 1 flip
4. flips: 39.1%
5. flips: 25.6%
6. and from here on, the odds of winning get worse and worse.

So if you're the gambling sort, you might take a chance on 3 flips or 1; anything else is a fool's game.
In all fairness...

Cauchy
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### Re: I don't seem to understand my own brain teaser, please h

Coyne, I think you're misunderstanding expected value.

Consider the game where you roll a six-sided die. On a 1-5, you lose \$1. On a 6, you gain \$6. The expected value for this game is \$1/6, even though you're more likely to lose than to win. The small chance of a large payoff skews the expected value off of the "common value".

What I think you want to find is the median value, that is, the life expectancy that you exceed with probability 50%. This will certainly be less than your starting life expectancy.
(∫|p|2)(∫|q|2) ≥ (∫|pq|)2
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>-)
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### Re: I don't seem to understand my own brain teaser, please h

Coyne wrote:the average case was, when you said, "There is a 55% chance of heads, which will double your remaining earthly time, and a 45% chance of tails, which will reduce your remaining earthly time by 70%." This means that for every flip there is a 55%/45% chance of heads versus tails; and that also, over a long series of flips, on average 55% of those will be heads and 45% will be tails. You can't lay down a stated probability and then expect the common case for the result to be something else.

Yes, the average is 55% heads and 45% tails. Denote the time after an arbitrary but fixed number of flips with proportion of heads p to be T(p).
It is true that p = 0.55 on average.
It is also true that T(0.55) is less than 10 days, your starting value.
However, this does not mean the expected value of T, E[T(p)], is also less than 10 days.
E[T(p)] = ∫01 T(p) dp, and p happens to be a binomial distribution.
This does not necessarily come out to less than 10 days (and in fact, is more).

Coyne
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### Re: I don't seem to understand my own brain teaser, please h

>-) wrote:
Coyne wrote:the average case was, when you said, "There is a 55% chance of heads, which will double your remaining earthly time, and a 45% chance of tails, which will reduce your remaining earthly time by 70%." This means that for every flip there is a 55%/45% chance of heads versus tails; and that also, over a long series of flips, on average 55% of those will be heads and 45% will be tails. You can't lay down a stated probability and then expect the common case for the result to be something else.

Yes, the average is 55% heads and 45% tails. Denote the time after an arbitrary but fixed number of flips with proportion of heads p to be T(p).
It is true that p = 0.55 on average.
It is also true that T(0.55) is less than 10 days, your starting value.
However, this does not mean the expected value of T, E[T(p)], is also less than 10 days.
E[T(p)] = ∫01 T(p) dp, and p happens to be a binomial distribution.
This does not necessarily come out to less than 10 days (and in fact, is more).

Sorry, maybe I'm dense, but I don't see how the probability of occurrence of a head has anything to do with 10 days or 100 days or 33 years.

Starting with 33 years: You flip the coin: You have a 55% chance that you get heads, in which case your remaining life becomes 66 years; or 45% chance of tails, in which case your remaining life become 9.9 years.

Then you flip again. If your first flip is heads and your second flip is heads, your life increases to 132 years; if the second flip was tails, it shrinks to 19.8 years. If your first flip was tails and your second flip is heads, your life increases to 19.8 years; for tails, 2.97 years.

So yes, T(p) as you put it is the overall probability. Let's say that by some quirk you come up with T(p) = .589 after 1000 flips. That merely means you flipped heads 589 times, your life was doubled 589 times; and tails 411 times, your life was multiplied by 30%, 411 times. Which is...

new_life = previous_life * 2589 * 0.3411

Under that, your new life is 2.53223 x 10-38 of your starting 33 year life; about 2.63707x10-29 seconds.

I think you're confusing the probability of the event, with the consequence of the individual realized events upon your remaining life span.
In all fairness...

Cauchy
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### Re: I don't seem to understand my own brain teaser, please h

Coyne wrote:Starting with 33 years: You flip the coin: You have a 55% chance that you get heads, in which case your remaining life becomes 66 years; or 45% chance of tails, in which case your remaining life become 9.9 years.

Then you flip again. If your first flip is heads and your second flip is heads, your life increases to 132 years; if the second flip was tails, it shrinks to 19.8 years. If your first flip was tails and your second flip is heads, your life increases to 19.8 years; for tails, 2.97 years.

Right. The chance of the first outcome happening is (.55)^2 = .3025; the chance of the second and third are each (.55)(.45) = .2475; and the chance of the last is (.45)^2 = .2025. The expected value is (.3025)(132)+(.2475)(19.8)+(.2475)(19.8)+(.2025)(2.97) = 50.332425 = 33(1.235)^2.

Expected value is a technical term whose meaning is not really up for debate. The expected value of the game is greater than the initial wager. Indeed, the expected value grows every time you play. The key is to realize that in this game, unlike in many games that can be played repeatedly, the expected value is a very poor indicator of whether it is a good deal to play the game repeatedly.
(∫|p|2)(∫|q|2) ≥ (∫|pq|)2
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PeteP
What the peck?
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### Re: I don't seem to understand my own brain teaser, please h

Coyne for expected value you multiply the chance of an outcome with it's value and then add all outcomes together. Say you play 100 times. The chance that you win 100 times in a row is about 0.25% 1e-26 however the outcome is 10*2^100 is about 1,2e31 multiply those together and you get about 138000. Even if the chance for losing all 10 is 1-(1e-26) that wouldn't be enough to compensate for this one huge if extremely unlikely win case because it's at most -10. (And of course there are more likely combinations with positive results than winning every time.)

The expected value is positive because you can never lose more than 10 days but if you win you can win high. Just put 700 wins 300 loses in your formula to see it for yourself. Unlikely sure, but it's also ridiculously huge. (With 1000 games you start to get positive outcomes with 365 or less wins)

jaap
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### Re: I don't seem to understand my own brain teaser, please h

PeteP wrote:The expected value is positive because you can never lose more than 10 days but if you win you can win high.

Exactly. Even just playing twice, you have:

0.55^2 = 30.25% probability of getting 4 times your initial amount
2*0.55*0.45 = 49.5% probability of getting 0.6 times your initial amount
0.45^2 = 20.25% probability of getting 0.09 times your initial amount

The expected value is 0.3025*4+0.495*0.6+0.2025*0.09 = 1.525225

When playing twice, most of the time (69.75%) you lose out but the large gain of a double win offsets those losses. This is why using expected value here is misleading - it is correct but does not necessarily give you the information you are looking for.

How you decide should also depend on Utility. How much is a loss worth to you compared to a gain? If you use the Expected Value, you are implicitly assuming that the value of gaining 10 days is exactly equal to the value lost when losing 10 days. Clearly this is not true in the dire situation described in the OP.

One particular utility function is the use of logs as I mentioned in my first reply. In that case you are assuming that doubling has the same value to you regardless of your current number of days, e.g. doubling 10 days has the same worth to you as doubling 20 days. This is a reasonable choice, essentially measuring values relative to the amount you already have. In that case, playing even once is a losing proposition. You may however feel that this is not the right utility function and decide differently.

MOJr
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### Re: I don't seem to understand my own brain teaser, please h

I don't know it seams simple to me.

Your expected life after X trials is 10*(2^(0.55*x))*(0.3^(0.45*x)). It is easy to show that as X->infinity expectancy->0.

The expected value calculations you use are for binomial distribution. And this example is not a binomial distribution, since you are multiplying the results and not adding them.

If you could always bet a fixed number (lets say a 1 day) then you could use binomial distribution EV and live forever (barring some really unlucky rolls in the beginning).

Cauchy
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### Re: I don't seem to understand my own brain teaser, please h

MOJr wrote:I don't know it seams simple to me.

Your expected life after X trials is 10*(2^(0.55*x))*(0.3^(0.45*x)). It is easy to show that as X->infinity expectancy->0.

This is not your expected life. This is the exponential of the expected log of your life, which is not the same thing. Alternately, this is the average value of your life if you average geometrically instead of arithmetically. Which is fine to do and all, but it doesn't produce the expected value, so labeling it as such is incorrect.
(∫|p|2)(∫|q|2) ≥ (∫|pq|)2
Thanks, skeptical scientist, for knowing symbols and giving them to me.

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### Re: I don't seem to understand my own brain teaser, please h

Well based on the discussion from this thread, you shouldn't play the game at all.
Why would you trust the reaper? For all you know, all your extra time could be a state of unbearable agony. How certain are you that what is promised will be done?
This is a block of text that can be added to posts you make. There is a 300 character limit.

MOJr
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### Re: I don't seem to understand my own brain teaser, please h

Cauchy wrote:
MOJr wrote:I don't know it seams simple to me.

Your expected life after X trials is 10*(2^(0.55*x))*(0.3^(0.45*x)). It is easy to show that as X->infinity expectancy->0.

This is not your expected life. This is the exponential of the expected log of your life, which is not the same thing. Alternately, this is the average value of your life if you average geometrically instead of arithmetically. Which is fine to do and all, but it doesn't produce the expected value, so labeling it as such is incorrect.

I was technicali correct with wrong words . It is median expected life after x flips.

And yhat is the whole story of the game, the difference between EV (average, mean) and median.
Avarage (EV):
With increasing number of tosses the average goes to infinity - this means that if large enough number of people play this game then some will win high enough, so on average you will end up with more days than the initial investment. (if you could share the days, then each will have more than 10 days)

Median:
But the median goes to zero so the life that "average person" gains with increasing number of games. So a person who is not extremly lucky will die...

Also EV is only usefull when you can repeat the same experiment a large number of times. If you could bet for example only a day at a time, then you should and you could live forewer.

The problem is to break even you need to win around 63% of flips. (If I win y flips out of x, then I get 200% y times and get 30% (x-y) times). But due to binomial distribution properties, there will be more and more possible outcomes in the <63% range as number of flips increases.