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# All numbers showing

### #11

Posted 26 January 2013 - 03:39 AM

Spoiler for had thought...

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

### #12

Posted 26 January 2013 - 05:16 AM

### #13

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.**

Past prime, actually.

### #14

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__.

Past prime, actually.

### #15

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.

*Vidi vici veni.*

### #16

Posted 26 January 2013 - 10:28 AM

You throw

mfair six-sided dice and bet that all numbers from 1 to 6 will show at least once.

What is the smallest value ofmthat 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.

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

Past prime, actually.

### #17

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

### #18

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.)

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

Past prime, actually.

### #19

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!

*Vidi vici veni.*

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