Multiple 4s/10s Before Big Red

I searched through a number of similar threads before posting here, as I wasn't able to find one that addressed this topic specifically. Hoping someone sees this and can help...

I came across a Craps strategy with a component that has you place a Lay bet on the 4 or 10 if a 4 or 10 has been rolled four times prior to a 7 having been rolled. The idea is to begin with a Lay of $40 and if the 10 is rolled a fifth time before the 7, increase the lay to $120, so forth and so on.

My question is, what is the probability that a 10 is rolled four times before a seven? Five times? Six times? The 10 and 4 are counted INDIVIDUALLY - so a 4 being rolled twice and a 10 being rolled twice a before a 7 appears doesn't trigger you to begin placing the Lay. Also, I'm not looking for the probability of four consecutive 4s or 10s before a 7 - just four 4s or 10s overall before a 7.

Thanks!

Remember when you sat in the back of the math class saying, "When will I use this in real life?" Well, this is when that thought bites you in the ass.

The math on this is actually kinda simple.

Since you're talking about independent bets, I'll only discuss one of them, and to avoid confusion, I'll pick the ten.

Of the 36 dice roll combinations, 3 will result in a ten, 6 are seven, and 27 are something else. You're ignoring those 27, which means out of the 9 rolls you're interested in, 3 will produce the ten.

Therefore the odds of a single ten before a seven is 3/9 = 1/3.
The odds of a two tens before a seven is (1/3) * (1/3) = (1/3)^2 = 1/9.
The odds of a three tens before a seven is (1/3)^3 = 1/27.
The odds of a four tens before a seven is (1/3)^4 = 1/81.
The odds of a five tens before a seven is (1/3)^5 = 1/243.
The odds of a six tens before a seven is (1/3)^6 = 1/729.
Etc.

I know what you're thinking. Wait for 4 hits, lay the ten and the odds of losing is 1/243. That's incorrect thinking.

The odds of seeing your 'trigger' of 4 tens without a seven is 1/81.
But once you get to that point, the odds of seeing a 5th ten are back to 1/3.

The 'strategy' you're proposing is a modified version of the Martingale. And because you are waiting for a trigger, you are thinking that the seven is 'due'.

There is no strategy that will shift the odds to the player's advantage. Sure, you can get lucky, but gambling is supposed to fun.

Quote: DJTeddyBear

Therefore the odds of a single ten before a seven is 3/9 = 1/3.
The odds of a two tens before a seven is (1/3) * (1/3) = (1/3)^2 = 1/9.
The odds of a three tens before a seven is (1/3)^3 = 1/27.
The odds of a four tens before a seven is (1/3)^4 = 1/81.
The odds of a five tens before a seven is (1/3)^5 = 1/243.
The odds of a six tens before a seven is (1/3)^6 = 1/729.
Etc.

I know what you're thinking. Wait for 4 hits, lay the ten and the odds of losing is 1/243. That's incorrect thinking.

link to original post

Exactly.
The probability of rolling 4 tens before a 7 when starting counting from when you place your bet is 1/243.
The probability of rolling 4 tens before a 7 when starting counting from "three tens already rolled without a 7" is the same as the probability of rolling 1 ten before a 7, since this what you need to do; this = 1/3.

I rolled five 12's in six rolls last time, so I guess that's a 1/36th of a 2.176 billion chance of that happening.

Quote: MeltTheFeltPA

I came across a Craps strategy]

Like Roulette or betting on a fair coin flip, Craps is a game of chance. 100% chance, so there is no skill or strategy involved.

The only decisions are what bets to make and for how much. I suggest bets with an overall edge < 0.5% and a bet level that gives your bankroll a 90% chance of surviving your trip

Quote: ChumpChange

I rolled five 12's in six rolls last time, so I guess that's a 1/36th of a 2.176 billion chance of that happening.
link to original post

I believe you’re implying that the one non-12 occurred in rolls 2-5. Therefore, the chance of this is 35 * 4 / 36^4 or about 1 in 12,000.

Quote: ChumpChange

I rolled five 12's in six rolls last time, so I guess that's a 1/36th of a 2.176 billion chance of that happening.
link to original post

I got a 1 in ~10.4 million chance using a spreadsheet (for that series of rolls).