7 wins in a row at craps

just a couple questions, for ya guys. to my understanding that winning 7 pass lines happends 1 in 141. would the odds be the same if i made a pass, dont pass, pass, dont pass, pass, dont pass, pass? im asumeing the same but just have to ask. i was just wondering if there would be a pattern that would create higher odds than 1 in 141.
thank you very much for any help

Quote: clay32

just a couple questions, for ya guys. to my understanding that winning 7 pass lines happends 1 in 141.

that is 141 attempts ((244/495)^7) and each attempt is not 1 bet
attempts is great for a martingale bet system

for avg # of bets, that = about 276.92 (say 277)

Thread: On average, how many bets will it take?

Quote: clay32

would the odds be the same if i made a pass, dont pass, pass, dont pass, pass, dont pass, pass? im asumeing the same but just have to ask.

I say close but not exactly the same as the dpass can push and the pass can not push.
It can be calculated

you must be after something!
Sally

thank you very much for your response mustangsally. i thought i was pretty good at math, but this is another deminsion for me.. so for example useing the martingale system. what would be the correct odds that say player one makes 7 pass lines in a row. and would it not reset for player 2 and so forth? awww i think it just hit me sence player 1 can make the pass line. and then keep the dice, we would have to figure in number of rolls?

so what if player one plays dont pass, looses then player 2 plays dont pass. and so fourth.. arrg we would run into the same problem, wow thanks you really opend my understanding more.

one last question if i could. if someone was to use the martingale system. starting at 5 dollers and doubling till 320 dollers.. played untill they won 1 hundred dollers, quit and repeated day after day.. how many times out of 100 would they loose the total, or how many times will they win the 100. sorry forgot to add.. playing the dont pass

thats what i was kind of getting to, if there was a way to increas the odds of a certain pattern not happening
pass, pass, pass, pass, pass, pass, pass #of bets 277
pass, dont, pass, dont, pass, dont, pass sounds like it would be pretty close
pass, pass, dont, dont, pass, pass, dont probley about the same

thank you very much for opening my mind up

Quote: clay32

just a couple questions, for ya guys. to my understanding that winning 7 pass lines happends 1 in 141.

using (244/495)^7 I get the same answer depending on how you want to round it off, but this raises an eyebrow about your statement downthread about being math savvy ... OK, well, maybe you haven't investigated enough about your subject to run into 244/495

Quote:

would the odds be the same if i made a pass, dont pass, pass, dont pass, pass, dont pass, pass?

roughly.

Quote:

im asumeing the same but just have to ask. i was just wondering if there would be a pattern that would create higher odds than 1 in 141.
thank you very much for any help

Forget this quest. Reading between the lines here, you seem to be investigating the Martingale. I make that assumption since you are trying to figure out the odds of so many in a row of something. You really need to read up on gambling fallacies or you are heading for trouble.

I'd start here,

https://wizardofodds.com/gambling/betting-systems/

and if interested in Craps, here

https://wizardofodds.com/games/craps/basics/

Whatever the game, whatever the betting strategy, you need this to answer your basic question.
https://wizardofvegas.com/member/oncedear/blog/4/#post1370

Quote: clay32

thats what i was kind of getting to, if there was a way to increase the odds of a certain pattern not happening
pass, pass, pass, pass, pass, pass, pass #of bets 277
pass, dont, pass, dont, pass, dont, pass sounds like it would be pretty close
pass, pass, dont, dont, pass, pass, dont probley about the same

thank you very much for opening my mind up

actually the pass for 7 in a row would be
1 / (244/495)^7 = 141.4189499
so you got to say 142

7 dpass in a row - not counting the pushes
1 / (949/1925)^7 = 141.303087
so you got to say 142

4 pass in any order and 3 dpass in any order
1 / (244/495)^4 * (949/1925)^3 = 141.3692828
so you got to say 142

Quote: clay32

one last question if i could. if someone was to use the martingale system.

and many have and still do!

Quote: clay32

starting at 5 dollers and doubling till 320 dollers..
played untill they won 1 hundred dollers, quit and repeated day after day.. how many times out of 100 would they loose the total, or how many times will they win the 100. sorry forgot to add.. playing the dont pass

so each win would add $5 to your starting bankroll of 127 units (127*$5)

the prob of winning $5 would be 1 - (244/495)^7 = 0.991380548 = WinItMan
and you need to win 20 times in a row
=WinItMan^20

you can take the math from there for days out of 100.

I like to see the probability to double that 127 unit bankroll. cuz if you cantz
Marty 7 goes away

that probability = 0.460956669 (say 46%) (used a Markov chain solution and one has to bet all if bankroll ends below 127 units)
that is smaller than 49.3% if you just bet all 127 units at 1 time.
close but really not close enough

the probability to double that 127 unit bankroll B4 losing the Marty 7 the 1st time
=WinItMan^127 = 0.333065729 (OUCH!!)

imagine making that 7th bet!
rush!!

Sally

How long would it be before I bust out? Acknowledging that all systems fail.

Red chip units. Given that most shooters go out on the 3rd roll. Wait until after the second roll and wager one unit on the DC, using up to a three step monty if I lose the first and second time I bet. Wagers go like this if I lose my first DC wager, 1 chip, if I lose repeat one chip wager on the DC after the second roll [assuming point is established, and not wagering on the CO roll] If I lose the first bet, I bet two units, If I lose that bet I bet 4 units. If I lose, I return to beginning.

How long before I would either bust out [assuming a 3 hundred dollar buy in] or die of boredom?

edited; I meant DC not DP, so changed it