Chance (quick question)

[Tricky]

If you could choose between one try of a lucky dice throw at a 60% chance of winning, or two tries of winning something at a 40% winning rate (with the added possibility of winning double if both tries succeed), which would be the option to choose?

I saw a billboard on my way to town this morning that displayed some kind of computer sale with this very action in mind. The question stuck in the back of my head, but I'm having a bit of a bad brain day. It seems there's a really crystal clear solution to this.

[fable]

One chance at 60%. You stand an above average chance on one throw, as opposed to a below average chance on each of two throws.

[Dottie]

Two chances of 40% is better. Even if you don't care about the double win the risk of not winning anything is only 36% with two chances at 40%.

[Tricky]

How do you get to 36% Dottie? It's an odd percentage..

Leaving out the percentages, outcomes for double drawings are false/false, false/true, true/false, true/true. A 1/4th chance of not getting any winnings at all. On the 40% game, wouldn't that mean 1/4th of the 60%? 15%?

[Dottie]

How do you get to 36% Dottie? It's an odd percentage..

To lose you need to fail two tests at 0.4 which is the same as two tests at 0.6 need to succed: 0.6 * 0.6 = 0.36

Leaving out the percentages, outcomes for double drawings are false/false, false/true, true/false, true/true. A 1/4th chance of not getting any winnings at all. On the 40% game, wouldn't that mean 1/4th of the 60%? 15%?

Those options are not of equal value, and the last operation is not understood by me at all.

The value of all the options should be as follows:

false/false 0.6*0.6 = 0.36
false/true 0.6*0.4 = 0.24
true/false 0.4*0.6 = 0.24
true/true 0.4*0.4 = 0.16

so, 16% chance of a double and 48% of a singel win.

[Tricky]

Wow. Somehow I feel I knew this stuff better way back in highschool. Math has never been my strong point, but that doesn't mean it fails to fascinate me. Thank you for explaining this.

[endboss]

Two at 40%. I hate probability. You either win or lose. It's always 50/50 no matter how statisticians try to gussy it up. Two chances are better than one, therefore take the second bet.

[Fljotsdale]

Well; being totally hopeless at figures, and not understanding - or even wishing to understand! - the explanations, I would go with one @ 60% chance of winning. I understand that in normal circumstances every throw of a die is 50/50, but in the given circumstance it's either one at 60% or two at 40% each. The one is above the average, the two are below the average, as Fable said, so you have a slightly better chance with the one 60-percenter, imo. But what do I know? I'm dumb with numbers.

[Tricky]

Hmm, I should've made this into a poll..

[endboss]

Hmm, I should've made this into a poll..

Then the statisticians will win!

[galraen]

Instinctively I side with Dottie, but can't be assed to work it out, but statistically I believe that two at 40% is the better option.

Of course that's with honest random numbers, start including computer generated pseudo-random numbers and conventional statistics go out the window.

[Tricky]

That depends on the computer. Pseudo generated random numbers are a major security leak. My old Pentium 3 had a small temperature measurement tool that measured the thermal vibrations of a small metal element and based its random calculations on that.

[Loki[D.d.G]]

That depends on the computer. Pseudo generated random numbers are a major security leak. My old Pentium 3 had a small temperature measurement tool that measured the thermal vibrations of a small metal element and based its random calculations on that.

Good point. It'll be pretty interesting to see just how random a random generated computer number is. After all, it is a machine coupled with a software that conforms to their own specific rules.

[Dottie]

It is true that random number generation isn't always as easy as it sounds. However, simulating the problem above is really difficult to mess up. The difference in probability is quite large, and all that is required for a two chance strategy to prevail over time is that the numbers produced have a close to uniform distribution.

[Tricky]

Good point. It'll be pretty interesting to see just how random a random generated computer number is. After all, it is a machine coupled with a software that conforms to their own specific rules.

You can actually literally 'see' that, if properly implemented:

Pseudo-Random vs. True Random - A Simple Visual Example

That site compares a pseudo random number generator with a true random number generator.

RANDOM.ORG - Statistical Analysis ~ more info and where I found that link.

[Vicsun]

Dottie is right about everything and you should all listen to Dottie.

That depends on the computer. Pseudo generated random numbers are a major security leak. My old Pentium 3 had a small temperature measurement tool that measured the thermal vibrations of a small metal element and based its random calculations on that.

Not really a 'security leak' unless you're using them for encryption. In general, depending on the quality of random numbers needed, and the quality of the algorithm generating them, pseudo-random numbers will do the trick just fine. It doesn't matter if after the millionth use some sort of pattern emerges (if you look very carefully), if all you're using them for is calculating damage in Baldur's Gate 2. Or settling an internet argument about dice.

I would go with one @ 60% chance of winning. I understand that in normal circumstances every throw of a die is 50/50, but in the given circumstance it's either one at 60% or two at 40% each. The one is above the average, the two are below the average, as Fable said, so you have a slightly better chance with the one 60-percenter, imo.

However, the chance of making either one of two 40% throws is above average, and also above 60%. As a matter of fact it's closer to 80% My inner mathematician really dislikes the acronym 'imo' when it comes to problems with a single, clear solution that couldn't be further away from a matter of opinion if they tried. It's sort of like saying "in my opinion, the world is a thousand years old, and global warming is a hoax". It's actually slightly worse, even.

In order to be more sciency, I thought I'd make a small python script to test the hypothesis (using, *gasp*, pseudo-random numbers):

Code: Select all

import random
winOne=0
winTwo=0
for i in range(0,1000000): #A MILLION DICE ROLLS
	if random.random() < 0.6:
		winOne+=1
	if random.random() < 0.4:
		winTwo+=1
	if random.random() < 0.4:
		winTwo+=1
print 'A 60% roll: ', winOne
print 'Two 40% rolls: ', winTwo

here are the results:

Code: Select all

vicsuns-macbook-pro-109 :D esktop Vicsun$ python diceroll.py 
A 60% roll:  600592
Two 40% rolls:  799387
vicsuns-macbook-pro-109 :D esktop Vicsun$ python diceroll.py 
A 60% roll:  600707
Two 40% rolls:  800127
vicsuns-macbook-pro-109 :D esktop Vicsun$ python diceroll.py 
A 60% roll:  600286
Two 40% rolls:  801798

Turns out that with two 40% rolls, you'd win 80% of the time if you count double-wins. Whodathunkit?

[fable]

However, the chance of making either one of two 40% throws is above average, and also above 60%. As a matter of fact it's closer to 80%

I'm not a numbers guy. I leave that to people who do algorithms in their head, like my wife. But...

You roll once, and your chance is 4 out of 10 of getting what you want. You roll the second time, and your chance is still 4 out of 10 of getting what you want. So how does that add up to an 8 out of 10 chance? Isn't it rather 2 attempts at a 4 out of 10 roll?

And what's the chance of making that 60% roll? 6 out of 10, correct?

[Vicsun]

You roll once, and your chance is 4 out of 10 of getting what you want. You roll the second time, and your chance is still 4 out of 10 of getting what you want. So how does that add up to an 8 out of 10 chance? Isn't it rather 2 attempts at a 4 out of 10 roll?

And what's the chance of making that 60% roll? 6 out of 10, correct?

Exactly, and if you are successful at either one of the two attempts, it counts as a win If you are successful at both it counts as two wins.

Think of it the following way: you are rolling a 1d10. In the first case, you 'win' if you roll 6 or higher. In the second case, you 'win' if you roll 4 or higher, but you can try to roll 4 or higher a hundred times. It's incredibly likely that at least one of those hundred rolls will be higher than a 4. As a matter of fact, you'd get, on average, 40 wins. In the first case, you'd only get 0.6 wins on average.

Tricky's question is the exact same situation, but instead of a hundred rolls, he only gets two. It so happens that two is enough.

[Dottie]

I would like to clarify though, that it's not the case that a chance of 8 out of 10 is equal to two chances of 4 out of 10 because 4+4=8. That it ends up that way is because the chance of a double reward. If you only counted one of the wins if both 4 out of 10 chances turned out to be lucky the final chance would not be 8 out of 10.

Edit: I also slightly object to calling the 8 out of 10 a chance to win, since it really isn't, it's more of a reward rate.

[fable]

Exactly, and if you are successful at either one of the two attempts, it counts as a win If you are successful at both it counts as two wins.

Think of it the following way: you are rolling a 1d10. In the first case, you 'win' if you roll 6 or higher. In the second case, you 'win' if you roll 4 or higher, but you can try to roll 4 or higher a hundred times. It's incredibly likely that at least one of those hundred rolls will be higher than a 4. As a matter of fact, you'd get, on average, 40 wins. In the first case, you'd only get 0.6 wins on average.

Something's wrong, there. You don't get 40 wins out of 100 rolls. You're assuming that this isn't a matter of random rolls, but of a fixed rate of success. It isn't, it's a 40% chance of success on each successive roll: less than half likely, on every roll. On the other hand, you have 1 roll at a 60% rate of success, more than half a chance at winning.

Can someone contact a statistician? Or a mathematician? Maybe I should just wait until my wife comes home Thursday night...