Probability Question For The Wiz - Craps!

Hey Wiz,

I have a friend that told me he was on the craps table last week and had a nice score. He proceeded to tell me how the roll unfolded.

A guy rolled a 10 , 10, 4, 4, 6, hard 4, hard 4, hard 10, 5, hard 4, 10, 4... then 7 out. My budy was only playing the 4 or 10 and kept pressing up the bets. He was on the hard 4 for the three times it hit in a row while pressing it up. Regardless of how much he made, I told him to enjoy that win because a roll like that occurs once in maybe millions of rolls.. probably hundreds of millions of rolls. I have played craps for 20 years an have never seen anything like that considering the hard fours conseculatively.

Is there a way to have a round about idea of what the probabilities of that roll happening? If anyone can figure it out, it is you. I welcome other math guru's as well.

Thanks!!

Well there's a couple of things to look at. First the chance of hitting a single hard 4 is 1/9. The chance of hitting three before losing would be (1/9)^3 or about a 1.37% chance.

The chance of hitting ANY four before a seven is 1/3 so hitting six would be (1/3)^6 which is the same as hard 4's, 1.37%. The chance of hitting any 4 OR 10 before a seven is 1/2, so ten of those before a seven would be (1/2)^10 r 0.0977%.

This doesn't factor in how many total rolls it would take to get these wins, as in there could be more sixes and eights or five and nines in between, that doesn't change the math. So the hardways aren't really that crazy, it's ALL the fours and tens that make it an incredible roll. Probability would indicate it should happen once every 1024 shooters.

Yes, I think that was an unusual roll. Almost two years ago I posted here about my experiences in a loop through Mississippi. In the next-to-last paragraph I commented on an interesting string of 6s that I hit but had not bet on.

Oh, one can dream about a roll like that.... lol The probability is way over one in one thousand rolls. Heck just to have 3 hard 4s come up before an easy 4 or a 7 is close to that. We are talking ten fours and tens with a string of 3 hard 4s consecutively all within one roll.

I personally never stick around long enough in a shoot to see such a thing because I like to hit and run... but amen, that was a shoot!

This SPECIFIC set of rolls is much more than 1/1000, that's not what I figured out. I just looked at certain aspects of the roll. For instance, you might roll a hard 4 on the comeout but have turned the bet off. That's makes it seem that it's more improbable to people, but if the bet wasn't on, then it's like that roll never happened.

Forgetting the aspect of the game and turning bets on or off, the numbers are right. I suppose one thing that I didn't analyze was all the hard 4's on the SAME roll as all the other 4's and 10's. You might be right in that that's quite improbable.

This EXACT sequence is :

1/18 * 1/18 * 1/18 * 1/18 * 1/9 * 1/36 * 1/36 * 1/36 * 1/9 * 1/36 * 1/18 * 1/18 * 1/6 (assuming soft 6, not hard 6) :

1 in 1,542,441,841,901,568. But most 13 rolls sequences of the dice are kinda unusual. Though for comparison, 13 straight 7's or 13 straight doubles would be 1 in 13,060,694,016, or 118,000 times more likely (that's 1 in 6 to the power of 13).

But 3 hard fours before a 7 out is about 1 in 1024.

One can also ask the probability that in 13 rolls one will have exactly 10 - 4s and 10s. This is a Binomial Dist problem.

One can also ask since betting on the 4 and 10 would have won 10 times in a row before the 7.

Slightly different questions.
Given 13 rolls the probability of 10 - 4s and 10s B4 a 7.
The word "given" is the clue.

or given 10 - 4s and 10s the probability of exactly 3 hard 4s.

or given 6 - 4s in 13 rolls the probability of exactly 3 hard 4s.
on and on
This can help for those that want to do the math
bayes-rule-calculator