What are the odds of rolling the same number back to back

I know somewhere on one of the Wizard's sites I saw the answer to this question, but after searching for 2 hours I cannot find the answer.

So... I think I read that the odds(probability) of rolling the same number twice in a row(any combination) was 10.67%...

Any two sixes in a row.... any two nines in a row... etc...

The odds of rolling a pair are one in six.

The odds of rolling the same pair twice in a row is (1/6) * (1/36) = 1/(36*6) = 1/216

The odds of rolling a non pair is 5/6. The odds of rolling the same non-pair twice is (5/6) * (1/18) = 5/(6*18) = 5/108 = 1/21.6

times in a row	Pair (1/x)	Non pair (1/x)
2	216	21.6
3	1,296	388.8
4	46,656	6,998.4
5	1,679,616	125,971.2

Snake Eyes/Midnight: 1/36 * 1/36 * 2 = 0.0015432098765432097

Three/Yo: 2/36 * 2/36 * 2 = 0.006172839506172839

Four/Ten: 3/36 * 3/36 * 2 = 0.013888888888888888

Five/Nine: 4/36 * 4/36 * 2 = 0.024691358024691356

Six/Eight: 5/36 * 5/36 * 2 = 0.03858024691358025

Seven: 6/36 * 6/36 = 0.027777777777777776

Total: 0.11265432098765432 or 11.27%

I think by, "Same number," he meant, "Same outcome," to the extent that a six could be 5-1 or 3-3, doesn't matter. I could be wrong, though, but he did mention, "Two nines," so that's why I interpreted it that way.

You need the combined probability of rolling 2 followed by 2, 3 followed by 3, ..., ..., ..., 12 followed by 12.

1 way to roll a 2 so P(2 followed by 2) = 1/36 X 1/36
2 ways to roll a 3 so P(3 followed by 3) = 2/36 X 2/36
3 ways to roll a 4 ... = 3/36 X 3/36
4 ways to roll a 5 ... = 4/36 X 4/36
5 " " 6
6 " " 7 etc etc etc So you end up with:-

(1x1 + 2x2 + 3x3 + ... ... ... + 6x6 + 5x5 + ... ... 1x1) / (36 x 36)

= (1 + 4 + 9 + 16 + 25 + 36 + 25 + 16 + 9 + 4 + 1) / 1296

= 146/1296 = 11.265432%

edit: Mission beat me to it. At least we got the same answer :-)

It could be an interesting side-bet. The crapless come bet on the hop! If you don't hit the come bet right away, you lose! If you do hit it, you win 11 for 1.

Thank you for the quick replies !!
When I said any number, I meant the same number in ANY combination. Any combination of nines, back to back.
For some reason, I had 10.67% stuck in my brain cells.

I use this knowledge while betting the dark side and laying odds on the FIRST occurrence of a number to stoke the lay-odds. I then regress the odds to a 'comfortable-for-me' amount after the second roll. For example(I am a cheap bettor, playing a $5 table), when the 9 comes up on the first roll, I lay $15. When the next number appears and it is not a 9 or 7, I regress to $9 and leave it for the remainder of that shooter's turn.

I hope that makes sense... If not, PLEASE feel free to lamb-baste me and show me the errors in my ways.
Thanks in advance... Jeffrey

Edit to add :

Given the math above by the fine reply-ers, can I deduce the following odds for each in-play number for the Don't side? :




4 = 16 / 1296 = 0.012345679
5 = 25 / 1296 = 0.0192901235
6 = 36 / 1296 = 0.0277777778
8 = 36 / 1296 = 0.0277777778
9 = 25 / 1296 = 0.0192901235
10 = 16 / 1296 = 0.012345679
TOT = 154

154/1296 = 11.88 %

Quote: jkluv7

I use this knowledge while betting the dark side and laying odds on the FIRST occurrence of a number to stoke the lay-odds. I then regress the odds to a 'comfortable-for-me' amount after the second roll. For example(I am a cheap bettor, playing a $5 table), when the 9 comes up on the first roll, I lay $15. When the next number appears and it is not a 9 or 7, I regress to $9 and leave it for the remainder of that shooter's turn.

It's classic Gambler's Fallacy, but the long-term expectation of any odds bet is $0.00, so you're not really hurting yourself. The probability of rolling a Nine at any given time is 4/36 and the probability of rolling a Seven is 6/36. For every two Nines that are rolled, in the long-term, there should be three Sevens rolled, which is why the Don't Odds pay the way they do and result in an expectation of 0.

The fundamental error in your reasoning is simply the difference between a known result and an unknown result. The probability of Two Nines in a row:

4/36 * 4/36 = 0.012345679012345678

Is based on rolls in which neither of the results are known.

If you have a roll which establishes a Point of Nine, then that first Nine has already happened and has a 100% probability of occuring. Therefore, the probability of rolling two Nines in a row, given the first (known) nine, is simply 4/36, which is the same as the probability of rolling a Nine any other time.