Median vs Mean

Would you provide the Median number of rolls before a 7-out in craps?
Mean of 8.5 is skewed by the very long roll counts and median better represents where 1/2 of the shooters get to before the 7-out.
Also, as a request, what would the various percentile rank rolls be... 80%tile, 90%tile?
Thank you for your help.


The ShadowMaster
"What evil lurks in the hearts of men... the Shadow knows!"

I am trying to follow my personal rule of no math after 3 am but I can conduct simulations in the morning for various percentiles

I failed to follow that rule

About half the time you get more than 4 rolls and the other half 4 or less.

About 20% of the time you get 10 or more rolls

A little under 10% of the time you get 14 or more rolls

A little over 1% of the time you get 26 or more rolls

0.1% for 39 or more

0.01% for 52 or more

I get just below 6 as the median. 6 rolls or fewer should happen about 50.28% of the time.

There's a difference between the median number of rolls for just the dice before a 7 appears and the median of rolls in Craps before 7-out. To calculate the latter is not easy because you have to include the come-out rolls. The 7 can show on the come-out and not 7-out. I believe 6 rolls or fewer—I trust ThatDonGuy---is the median for Craps.

 So how many rolls of the dice does it take for a 7, not just 7-out, to have an even chance of showing? This is not only easier to calculate, but more practical as guideline if you're making place bets and want to take your bet down. 

The probability of 7 appearing in 1 roll is 1-(5/6)= 1/6. And then you continue by multiplying 5/6.

In 2 rolls:   1-(5/6)^2 = .3056
     3 rolls:    1-(5/6)^3 = .4214
     4 rolls:    1-(5/6)^4 = .5179

By the 4th roll, the 7 will show 51% in 4 rolls.

To get the exact median number, let's X equals the number of rolls it takes for the 7 to have an even chance of showing:
X / 4 = 50 / 51.79
51.79X = 200
X = 3.86

I'm not a math-person. Do I have this right?

Quadriga I believe we did basically the same math.

I did not understand the terminology difference between 7-out and simply rolling a 7, I just calculated how long it would take to roll a 7 on average.

Thank you for the quick response!!

I was feeling the median was lower than the reported mean. This makes a big difference when evaluating various strategies that employ "press & collect." The low median also confirms data showing "for the most part" (i.e. most typical shooters <10 rolls) that 'same bet' collect generally gives more profit than "press & collect."

Thanks again for the responses!

The ShadowMaster
"What evil lurks in the hearts of men, the Shadow knows."

Quote: harris

… I did not understand the terminology difference between 7-out and simply rolling a 7 ...
link to original post

You don’t play enough craps.

A 7 rolled before a point is established (the puck says OFF), is a come out 7 winner.

Once a point is established (the puck says ON), a 7 is a loser and is called a 7-out. This is also when the Don’t bets win, all other bets lose, the layout is cleared, and the dice move to the next player.

If the point gets hit, and then a 7 is rolled before a new point is established, it’s a mid-roll come-out-7. And it wins just like an initial bet. The only bets that lose or come down are the come bets (while don’t come wins).