the chances of "how many times will any 2 field numbers be rolled consecutively" in 100 is much harder math

That is a weird way of asking the question. It happens 19.56 times out of 100 rolls. However, it only happens 1.78 times out of 10 rolls. It happens It happens 0.198 times out of 2 rolls. It happens It happens zero times out of 1 roll.Quote:CasinoinvestorIn 100 rolls of the dice how many times will any 2 field numbers be rolled consecutively ( 2 times in a row ).

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The ratio is different for every number of rolls because the last roll cannot be part of two consecutive rolls. For a large number of rolls, the ratio converges on 0.19753 consecutives per roll because this end effect becomes insignificant.

EDIT: odiousgambit explains it correctly. He gets a different answer for the long run due to rounding.

Quote:CasinoinvestorIn 100 rolls of the dice how many times will any 2 field numbers be rolled consecutively ( 2 times in a row ).

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Questions:

If your first three rolls are all field numbers, does this count as 1 time, or two (rolls 1 & 2, and rolls 3 & 4)?

What about if your first four rolls are all field numbers?

I find the HA on the field bet to be far too much for this type of progression (same for roulette). I'd try this progression on the PL or the DC or the PB 6 or 8 separately ($12, $18, $18, $24 and onwards).

I'd use a $15, $20, $25 and onwards progression on BJ (and the PB 5 or 9 in craps) because it's harder to put together 2 wins in a row and sometimes the double downs, & splits can come back to bite you if you're betting too high.

If you don't count three consecutive field numbers as 2 separate sets of consecutive rolls: 21.97 times

If you do: 43.44 times (this includes 4 in a row as 3 sets of consecutive rolls, 5 in a row as 4 consecutive sets, and so on)

Quote:CasinoinvestorIn 100 rolls of the dice how many times will any 2 field numbers be rolled consecutively ( 2 times in a row ).

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If I roll 6 in a row, do you count that as 5 separate instances of rolling two in a row? Or would you count that as 3 instances?

If you do it the first way, it’s around 19-20 times in 100 rolls.

Quote:CasinoinvestorJust checking my math. So, 16/36=0.44 so divide that by 2 or multiply by 2?

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Multiply .44 x .44.

The answer has to be less than 100*(16/36)^2. Both these number are too high.Quote:ThatDonGuySimulating gets this:

If you don't count three consecutive field numbers as 2 separate sets of consecutive rolls: 21.97 times

If you do: 43.44 times (this includes 4 in a row as 3 sets of consecutive rolls, 5 in a row as 4 consecutive sets, and so on)

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Quote:MentalThe answer has to be less than 100*(16/36)^2. Both these number are too high.Quote:ThatDonGuySimulating gets this:

If you don't count three consecutive field numbers as 2 separate sets of consecutive rolls: 21.97 times

If you do: 43.44 times (this includes 4 in a row as 3 sets of consecutive rolls, 5 in a row as 4 consecutive sets, and so on)

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Correct. Change the 100 to 99.

Quote:MentalThe answer has to be less than 100*(16/36)^2. Both these number are too high.Quote:ThatDonGuySimulating gets this:

If you don't count three consecutive field numbers as 2 separate sets of consecutive rolls: 21.97 times

If you do: 43.44 times (this includes 4 in a row as 3 sets of consecutive rolls, 5 in a row as 4 consecutive sets, and so on)

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Say, here's a crazy idea: how about if I remember to set the "number of times in a row" back to zero when I roll a non-point number?

Now, it's:

If you reset after each pair: 13.58

If you count 3 in a row as 2, 4 in a row as 3, and so on: 19.555

The 13.58 one is confirmed by spreadsheet:

Let P(n, 0) be the probability of the "zero in a row" state after n rolls

P(n, 1) be the probability of the "one in a row" state after n rolls

E(n) be the expected number of pairs after n rolls

P(n,0) = 1; P(n,1) = E(n) = 0

P(n+1, 1) = P(n, 0) x 4/9

P(n+1, 0) = P(n, 0) x 5/9 + P(n,1)

E(n+1) = E(n) + P(n, 1) x 4/9

Note that, after n = 25, P(n,0) stabilizes at about 0.6923, so each E(n) value increases by about 0.13675 over the previous one.

It turns out that, for the second one, the expected number of pairs is 16 / 81 x the number of rolls.

p^2 ((N - 1) + p (1 - (-p)^(N-1)) / (1 + p)) / (1 + p)

where p is the probability of rolling a field number, which is 4/9, and N is the total number of rolls, which is 100.

The solution is 16/117 * (99 + 4/13 * (1-(-(4/9))^99)), or about 13.58054.