Joined: Apr 3, 2017
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April 30th, 2017 at 11:42:12 AM permalink
I play Pai gow pretty often as my game of choice. Today the guy two seats to the left of me had a losing streak which I have not seen before. Excluding pushes (6) he lost 9 hands in a row so in theory not a single win in 15 hands. Good news is he was only betting $20 a hand so was stuck for $200 (bet 40 last hand)

Got me wondering and I knew someone here would know the math as to the formula for this and the math behind it?

I have seen people lose 5 hands (not counting pushes before) that I remember but never more than that.
Joined: Feb 18, 2015
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April 30th, 2017 at 1:23:23 PM permalink
If you play reasonably well, the odds of losing any given Pai Gow Poker hand is about 32.6% and the probability of pushing is about 36%. You will win 31.4% of the time.

The odds of NOT winning any of 15 consecutive hands is about 0.0035 or about 1 in 285. Of course, that doesn't quite capture what happened, because even when you don't win in 15 consecutive hands we would have expected fewer than 9 losses.
Joined: Jan 15, 2010
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April 30th, 2017 at 3:08:51 PM permalink
I would look at losses in a row without winning, pushes not counting for anything. So nine losses without winning, 1/512. However, my personal preference with these is to start counting after the first resolution. So you lose one, then you lose 8 more in a row. 1/256.

For the calculations, 32.6%=31.4% - Winning/losing is effectively a coin flip in Pai Gow.
Its - Possessive; It's - "It is" / "It has"; There - Location; Their - Possessive; They're - "They are"

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