DougD
DougD
  • Threads: 2
  • Posts: 21
Joined: Aug 15, 2022
March 24th, 2024 at 2:24:32 AM permalink
I was using the Wizard's Pai Gow calculator and entered a player hand of Teen Bo ( 6-6, 6-6) and Day Bo (1-1, 1-1).
With this player hand the dealer cannot win. He/she can only lose and push.The dealer will push when the dealer hand contains the Gee pair.

According to the calculator the player's best hand returns 20,178 loses and 297 pushes for a return of 0.936220.
With the above player tiles removed there are 20,475 dealer hands. Of these hands, how many dealer hands will have the Gee pair?

I think there are 275 dealer Gee pairs which will cause a push but the calculator produces 297 pushes.
Why are there more than 275 pushes? Even if the dealer splits the Gee pair it's still a loss against the player's Teen Bo and Day Bo.

Also, what's the formula for calculating the number of the dealer's Gee pairs when the Teen Bo and Day Bo are removed?

The calculator is a great too for learning the game!!. I use it all the time and recommended it to many friends.
Thanks in advance for your help.
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