Odds of a particular "Dont' Four" series

Wizard/Forum Contributors:

Forgive my simple question if asked by others previously--

What are the odds of the following:

1) two fours before a seven (not necessarily back to back, just two fours being rolled before a seven is rolled)

2) three fours before a seven

3) three fours before a seven

4) four fours before a seven


The purpose of my question is simply that I would like to know my risk profile of increasing my subsequent "dont four" bet if my initial "dont four" wager is lost due to a seven being rolled.

Thanks in adavance for your reply. I'm headed to vegas Friday Oct 14 and would love to be equipped with this knowledge before arriving there.

Much appreciated!

There are 3 ways to roll a 4 and 6 ways to roll a 7. Since you're unconcerned about anything else, ignore the 27 other rolls. There are 3 chances out of 9, or 1/3 chance, that a roll of either 4 or 7 will be a 4.

For your specific question, the chance of 2 fours before any seven = ( 1 / 3 ) ^ 2 = 1 / 9 = 11.11%

3 fours before any seven = ( 1 / 3 ) ^ 3 = 1 / 27 = 3.7%
4 fours before any seven = ( 1 / 3 ) ^ 4 = 1 / 81 = 1.2%
5 fours before any seven = ( 1 / 3 ) ^ 5 = 1 / 243 = 0.4%
6 fours before any seven = ( 1 / 3 ) ^ 6 = 1 / 729 = 0.1%

The odds of hitting a 4 before a 7 is 1/3
The odds of hitting n 4s before a 7 is (1/3)^n

Your no-4 bet will win, not lose, if a seven is rolled. But more importantly, the risk profile does not change after a four is rolled one or more times. If you make a no-4 bet and the 4 shows, you still have exactly a 2/3 chance of winning when you make the next one. Remember that you're not betting on "two fours before a seven" or "four fours before a seven". At the moment you make the bet, only two numbers matter -- the *next* four and the *next* seven. What happens after that is irrelevant because by then that bet is over.

Quote: DJTeddyBear

There are 3 ways to roll a 4 and 6 ways to roll a 7. Since you're unconcerned about anything else, ignore the 27 other rolls. There are 3 chances out of 9, or 1/3 chance, that a roll of either 4 or 7 will be a 4.

For your specific question, the chance of 2 fours before any seven = ( 1 / 3 ) ^ 2 = 1 / 9 = 11.11%

3 fours before any seven = ( 1 / 3 ) ^ 3 = 1 / 27 = 3.7%
4 fours before any seven = ( 1 / 3 ) ^ 4 = 1 / 81 = 1.2%
5 fours before any seven = ( 1 / 3 ) ^ 5 = 1 / 243 = 0.4%
6 fours before any seven = ( 1 / 3 ) ^ 6 = 1 / 729 = 0.1%

These are for at least x fours before any seven. For exactly x fours before a seven its (1/3)^x * (2/3)

Thank you very, very much! You are VERY KIND to reply.