To irchans: your question of 3 years ago answered

[BruceZ]

This question asked for the number of hands needed before there was a 50% or better probability of seeing all 1326 hold'em hands. At the time I approximated the answer at 10,016 hands. Irchans said that Sklansky posted the answer of 10,019.8 hands on rgp, and asked how to get his answer, but no one could it figure it out. Now using my "new toy" shown to me by Pzhon, we know that the correct answer to the nearest tenth is 10,019.0 hands. And to think that the new toy is just the inclusion-exclusion principle which I have been using all along for other problems. It doesn't give the answer directly, you have to run it over the range of possible values, or use the goal-seek function of Excel to solve this non-linear equation for N. The number of terms also had to be found empirically, as 6 and 7 terms both ensure the result is < 0.5, while decreasing N by a tenth would make both terms > 0.5.

1326*(1325/1326)^N - C(1326,2)*(1324/1326)^N + C(1326,3)*(1323/1326)^N - C(1326,4)*(1322/1326)^N + C(1326,5)*(1321/1326)^N - C(1326,6)*(1320/1326)^N + C(1326,7)*(1319/1326)^N <= 0.5.

[uuDevil]

So, would this "new toy" be considered "hi falootin math," or not? Is this the method Sklansky used? If not, did he leave any notes in the margins? [img]/images/graemlins/wink.gif[/img]

[BruceZ]

I made a change the above post. The answer to the nearest tenth is actually 10,019.0. You have to compute 7 terms, and find the minimum value that make both 6 and 7 terms less than 0.5, while decreasing it by a tenth makes both both 6 and 7 terms greater than 0.5.