Today's math problem

[pokerfan]

involves rolling a pair of dice.

The total number of combinations is 36 (six times six). There is only one way to roll a specific double such as double sixes (AKA boxcars), therefore the odds against rolling a double six is 35:1.

So the question is: How many consecutive rolls on average will it take before you roll double sixes?

[mainegreen]

1. Just off the cuff....

18?
Or is it 18+1?

[pokerfan]

2. It is greater than 19

It is also not an integer.

[mainegreen]

6. is it closer to 35.009089?

Just curious.
(I just rolled 36 10,000,000 times, if I coded this correctly and got an average number of rolls).

No idea how to do the math.

:shrug:

[pokerfan]

11. It's less than 35

Another hint: The probability of NOT rolling double six is 35/36 = 0.972222222. The probability of NOT rolling double sixes in TWO rolls is (35/36)*(35/36) = 0.945216049.

[mainegreen]

15. This problem is driving me nuts.

Using that logic, I get the answer no problem.
But me, being me, I've got to actually test it.

I now no longer trust my sanity.

I'm rolling two electronic dice over and over in runs until I get the magic twelve. Total the number of rolls, increment the runs, and repeat.

No matter what I do, the more runs I do, the closer I seem to get to 36.

ARRRRG!

What the hell don't I get here?

[pokerfan]

17. You must have made a coding error

You are with me that the probability of NOT rolling double sixes in TWO rolls is (35/36)*(35/36) = 0.945216049?

Therefore the probability of NOT rolling double sixes n time in a row is (35/36)^n.

(35/36)^24 = 0.508596124
(35/36)^25 = 0.494468454

So the answer has to be greater than 24 but less than 25.

Setting (35/36)^n = 0.5 and solving for n, I get 24.60509772 (not the 24.255 that was given in the article I linked.

If you have nothing to do for the rest of the day, you could start tunning trials with real dice.

[mainegreen]

19. Thanks for the help!

I figured out my problem.
The code for rolling the dice and calculating runs was right, but my final analysis was wrong.

The way I was analyzing it, it was picking out the denominator of the probability of not getting the number on a single roll.

Duh.

That was fun!
Thanks for the mental challenge.

[pokerfan]

20. Thank you for making me double check as well

and discovering that the 24.255 in the article was wrong. Nothing like doing it yourself and proving it by checking your work to really "grok" something.

[EnviroBat]

3. Is beef a number?

[laconicsax]

4. (all three)

[Rabrrrrrr]

5. Somewhere around 21.6, I wager

[pokerfan]

7. Higher

[Canuckistanian]

32. You're on.

Who else wants in on the action?

[Lil Missy]

8. 36

[Orsino]

21. Has to be, unless the dice are loaded.

1 in 36 rolls will be boxcars.

[pokerfan]

22. That's not exactly saying the same thing

as asking for the average number of rolls required before you hit double sixes.

For it to take 36 rolls, you would have had to have missed double-sixes thirty-five times in a roll.

(35/36)^35 = 0.3730

Only 37.3% of the times will you miss double-six on 35 consecutive rolls.

[Orsino]

23. On the average, 1 in 36 rolls will be boxcars...

...with 35 non-boxcars in between. Quite often, more than 35 misses will occur between boxcars.

[pokerfan]

24. English is not well suited for explaining probability problems

It lacks the rigor of mathematics and it doesn't help that I have only a tenuous grasp of my only language. All I can say is that I trust my math and mainegreen's 10,000,000 trials which lend corroboration.

It's an old problem dating back to the seventeenth century and quite famous actually. Googling 'Chevalier de Mere' and 'Blaise Pascal' yields 625 pages on the subject. One of the better papers is http://www.probabilitytheory.info/topics/periodic_events.htm which explains it quite well, I think.

From reading that page it's interesting that Abraham de Moivre came up with an answer of 24.9516 in 1716. Which was too high.

John Scarne came up with 24.2585 in 1961. Which is too low.

Interesting to note that page comes up with the same formula as I did: n = log 2/(log 36 -log 35) but then evaluates it incorrectly. They say that n has a value of 24.6721? I keep getting 24.6051.

[Orsino]

25. That's a different problem from the one posed in the OP.

The ~25 figure is the number of rolls needed to make boxcars a better than even chance.

If you're rolling boxcars once in 25 throws on average, you've got loaded dice.

[pokerfan]

26. I think I found the problem

I think the source of the confusion in the question was the use of the word, 'average' to imply median instead of mean.

The median number of rolls require before throwing double-six is 24.605. In other words, in the long run, there will be just as many cases where boxcars are hit before that number of rolls as there are case that exceed that number.

[Orsino]

27. That's it exactly, I think. n/t

[flvegan]

9. On average? 25

Or so says Blaise Pascal and Pierre de Fermat.

[pokerfan]

12. Winner (24.255 to be precise)

But I don't know how to perform a 0.255 roll.

http://homepage.ntlworld.com/dice-play/Odds.htm

Pascal concluded that the true odds were the odds against winning in one roll multiplied by the colog of the hyperbolic log of two (0.693). This works out as 35 x 0.693 which equals 24.255 rolls for an even money game. He needed at least 25 rolls of the dice to get an edge and 26 or 27 would have been more like it.

[Dr. Strange]

13. I was going to say that...I even had all the calculations done, ready to be included.

But the message space isn't big enough to include them all. So you'll just have to take my word for it.

[flvegan]

14. Of course you did, Pierre.

Of course.

[Omphaloskepsis]

10. 2+6=Cat

[Dr. Strange]

16. You forgot to carry the doorknob.

Try again.

[LynzM]

18. LMAO

This was my first true LOL of the day. Thank you!! :rofl:

[Bake]

28. With MY dice it only takes one roll. Every time.

Whazzup with that?

Bake

[pokerfan]

29. You are the villian from Octopussy?

He could roll double-sixes whenever he wanted.

[Bake]

33. And I look GOOD! in a tux, too!

:rofl:

Bake

[BlueJazz]

30. 15

[pokerfan]

31. I'm actually surprised no one guessed 42

knowing this group.