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Best Answer superprismatic, 23 January 2013 - 05:40 PM

Spoiler for I get

Go to the full post


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18 replies to this topic

#11 superprismatic

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Posted 26 January 2013 - 03:39 AM

Spoiler for had thought...

Please explain further.  I can't parse this in any sensible way.


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#12 phil1882

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Posted 26 January 2013 - 05:16 AM

Spoiler for

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#13 Prime

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Posted 26 January 2013 - 05:23 AM

Spoiler for had thought...

 

 

Yes. Simple concept that cuts thru all the computational complexity.

This concept, does not apply to the question in the OP. The problem called for finding the minimum number of rolls (or dice) to get a winning bet. A winning bet is one, where your probability of winning is better than 0.5. It is not the same concept as average number of rolls (dice) required to get to the objective.

A simple example:

If you roll repeatedly a fair dice, then any given number, say "1", will occur on average once in six rolls. However, if I bet on "1" coming up just after 4 rolls or sooner, I am going to come out ahead. (Same as rolling 4 dice and betting that at least one of them will roll "1".)

P = 1 - (5/6)4 = 0.5177

 

The answer to the problem as stated is like Sp and myself have found. I have another equation, which is not recursive, but encompasses all variations. It yields the same result. I'll make a separate post for it.

 

But of course you still get a winning bet with 14.7 rolls. Your winning chance is better than 60%, but it is not the minimum number of rolls.


Edited by Prime, 26 January 2013 - 05:31 AM.

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Past prime, actually.


#14 Prime

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Posted 26 January 2013 - 07:04 AM

The bottom line, the ultimate truth for calculation of a probability is the number of all variations that meet your criteria divided by the number of all possible variations.

 

Spoiler for TOTAL VARIATIONS


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Past prime, actually.


#15 bonanova

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Posted 26 January 2013 - 09:42 AM

Prime, that analysis is convincing. Consider the following as well.

I wonder, in light of your betting on "1" appearing in just four die rolls

means we should be looking at median outcomes rather than average.

 

Spoiler for Translating the OP into math

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The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
- Bertrand Russell

#16 Prime

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Posted 26 January 2013 - 10:28 AM

You throw m fair six-sided dice and bet that all numbers from 1 to 6 will show at least once.
What is the smallest value of m that gives you a winning bet on average?

That's the OP. A computer, or real life simulation of this simply and economically stated probability problem must be simple and straightforward.

I look at it as a gambling game, where I can bet $1 on the outcome.

 

Spoiler for the game


Edited by Prime, 26 January 2013 - 10:34 AM.

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Past prime, actually.


#17 bushindo

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Posted 26 January 2013 - 05:00 PM

Prime, that analysis is convincing. Consider the following as well.
I wonder, in light of your betting on "1" appearing in just four die rolls
means we should be looking at median outcomes rather than average.

Spoiler for Translating the OP into math


I agree with the note about median. I think the problem here is that the OP and post 14 (see blue-colored part in spoiler above) are conflating the median and the average (or mean) in the interpretation of the winning condition

Spoiler for


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#18 Prime

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Posted 26 January 2013 - 10:23 PM

The statement of the problem is unambiguous. It requires finding the smallest number of dice for the winning bet  on average (not the average).

Bonanova is a Grandmaster of stating problems. If he wanted us to find the average number of rolls before you get all 6 numbers, he would say so (in a better way.)

Spoiler for On the other hand...


Edited by Prime, 26 January 2013 - 10:26 PM.

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Past prime, actually.


#19 bonanova

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Posted 27 January 2013 - 06:44 AM

I brought up the idea of median, since the distribution is skewed and I needed a way for both 13 and 14.7 to have significance. Bushindo stated exactly, and clearly, why it suggested itself. Prime explained well why and how wait times and <p> = .5 are different conditions. I've thought about it for a day now and I haven't decided for sure how a puzzle like this one might be worded in order to require finding a skewed median. but i think it could make an interesting puzzle. For sure, the wording would need to be precise.

For this puzzle, I had in mind the cute 14.7 calculation. But I didn't word the OP to ask for it. I was surprised to confirm by simulation that 13 is the answer, as I worded it. Prime is way too kind to say a mistake on my part is anything close to a noteworthy event. But it is clear that (using his example) since betting on a "1" in four rolls wins, wait time (six rolls) isn't the answer for an even bet.

Well guys, I enjoyed this puzzle too. Thanks for teaching me something. ;)

Footnote: sorry plainglazed, the solution post goes now to SP! :)


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The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
- Bertrand Russell




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