Probability of 28 hands in a row without a pair or an ace?

[idrinkcoors]

Does anyone know what this probability is? Happened to me last night, and I thought I remembered reading in Super System about not catching an ace or a pair after 20 hands to be extremely rare.

Thanks in advance.

[The Legend]

probability of an Ace (not including AA) .012*12
probabliity of a pocket pair, .0045*13
probability of neither 1-(.012*12+.0045*13)
probability of neither for 28 hands is last answer to the 28th power

.00177198, or 1/564.34

[ChromePony]

lets see if i can tackle this one,

the probability of getting a single ace is 4/52 + 48/52*3/51 = .13122
the probability of a pocket pair is 3/51 = .05882

so the probability of getting neither for 28 hands:
[ 1 - (.13122 + .05882) ]^28
[ .8099 ]^28 = .002737 or .2737%

so yeah not very likely

there is also a strong chance i messed this up since im new at the probability thing, if bruce wants to step up and put me in my place that would be cool.

EDIT: oops /images/graemlins/frown.gif

[fiskebent]

Probability of not getting an ace as your first card is 48/52. Probability of not getting an ace or a card to match your first card as your second card is 44/41 (7 cards to dodge).
So probability of neither an ace or a pair is 79.638%

To have that happen 28 hands in a row is 79.638% ^ 28 or 0.1703613% (1 in 587).

[shermanash]

to take this a step further, what are the odds that this will happen AT least once (no ace pair in 25 hands) over a, say 500 hand session.. also i don't care about the answer but i would like to how to figure this stuff out.