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thank you very much for any help

that is 141 attempts ((244/495)^7) and each attempt is not 1 betQuote:clay32just a couple questions, for ya guys. to my understanding that winning 7 pass lines happends 1 in 141.

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?

I say close but not exactly the same as the dpass can push and the pass can not push.Quote:clay32would 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.

It can be calculated

you must be after something!

Sally

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

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 244/495Quote:clay32just a couple questions, for ya guys. to my understanding that winning 7 pass lines happends 1 in 141.

roughly.Quote:would the odds be the same if i made a pass, dont pass, pass, dont pass, pass, dont pass, pass?

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.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

I'd start here,

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

and if interested in Craps, here

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

https://wizardofvegas.com/member/oncedear/blog/4/#post1370

actually the pass for 7 in a row would beQuote:clay32thats 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

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

and many have and still do!Quote:clay32one last question if i could. if someone was to use the martingale system.

so each win would add $5 to your starting bankroll of 127 units (127*$5)Quote:clay32starting 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

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

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