[Stewart]: The exact probability of at least one run of r consecutive successes in n independent trials, each with success probability p, is given by de Moivre's formula, which I quote from Uspensky's (1937) Intro to Mathematical
Probability:

In TeX symbolism,

$$
P=1-\sum_{k=0}^{[n/(r+1)]}(-1)^k{n-kr\choose k}(qp^r)^k
+p^r\sum_{k=0}^{[(n-r)/(r+1)]}(-1)^k{n-r-kr\choose k}(qp^r)^k.
$$

[Don]: Wow! I like my way better! :-)

[Stewart]: In your case, p=2*8*32/(104*103), r=4, and n=5000, so it suffices to consider the first few terms in each sum (the remaining terms are negligible). I get P=0.0245 (rounded). So your approximation was not too
bad.

[Don]: Surprisingly close. Doesn't make my way right, but didn't expect to be within one-tenth of the correct answer.

[Stewart]: In practice, the dealer doesn't necessarily shuffle between hands,

[Don]: This was foolish of me. Don't know what I was thinking. See below, for what I hope is a much better approximation to the true correct answer.

[Stewart]: so the true probability will be smaller than 2.45 percent.

[Don]: Shockingly smaller. Consider that, after three straight naturals, there are only five aces left, and the prob. of getting a snapper is dramatically decreased (by almost half, when one considers the reduced tens, as well).

[Stewart]: For example, if the dealer has had three straight naturals and is still dealing from the same deck, the fourth natural will be somewhat less likely, because three of the aces will be gone.

[Don]: MUCH less likely! Suppose I were to deal nothing but four straight hands of two cards each (but with no replacement), to simplify matters. Then the prob. of four straight naturals becomes:

[(2*8*32)/(104)(103)][(2*7*31)/102)(101)][(2*6*30)/(100)(99)][(2*5*29)/(98)(97)] = 1/1,442,515.

Compare this result to the 1/191,603 that we got with replacement! The latter is 7.53 times more likely than the former! So, if I now do it my way, I get a new probability of 1 - 0.99654 = 0.00346, or once in every 289 attempts, which, of course, is much rarer than once in 39. What do you get with Uspensky's formula, replacing p with the new 0.000000693?

Thanks very much Stewart; I appreciate your help.

Don