Thread: Hold'em Probability Puzzler View Single Post
#13
 PseudoPserious Senior Member Join Date: Oct 2002 Posts: 151 Re: I like your problem more lol.

Heya Huh,

I think I have a solution to the 169-hand problem. Let's see if I can explain it clearly [img]/forums/images/icons/wink.gif[/img]

When you're dealt a 2-card hand, it can either be a pair, an off-suit non-pair (NPo), or a suited non-pair (NPs).

There are 13 different pairs and 6 ways to deal each pair. As there are 1326 total ways to deal two cards, the odds of you getting a pair are thus Pp = 13*6/1326 = 78/1326.

There are 78 different NPo hands and 12 ways to deal each of them. Thus Pnpo = 78*12/1326 = 936/1326.

There are 78 different NPs hands and 4 ways to deal each of them. Thus Pnps = 78*4/1326 = 312/1326.

Consider only the times you are dealt a pair. Using the method outlined in the above posts, it will take you, on average, Sum[13/n, n=13 to 1] = 41.3 times being dealt a pair before you have seen all of the pairs.

Since you are dealt a pair on roughly 6% of all deals (78/1326), it will take you on average Sum[13/n, n=13 to 1]/Pp = 702.8 deals before you've seen all the pairs.

Likewise, it will take you Sum[78/n, n=78 to 1] = 385.3 times being dealt NPo before you've seen all of the NPo. You're dealt 385.3 NPo hands after 545.9 total deals.

Finally, it will take you Sum[78/n, n=78 to 1] = 385.3 times being dealt NPs before you've seen all of the NPs. You're dealt 385.3 NPs hands after 1637.7 total deals.

So, on average, it will take you 1637.7 hands to have seen the 78 NPs. Since by this time you've also seen all of the pairs and NPo, I'd say that the average number of deals to see all 169 is 1637.7 hands.

Does this make sense?
PP

P.S. Once I get back to my computer, I'll modify the little code I wrote to sim this problem a few times.