What are the odds of losing 16 straight Texas hold ‘em poker hands when bets and folds are nonexistent and hands are completed without prejudice.

2 Answers 2


You don't say how many players there are, so let's say there are n players.

For simplicity, ignore "chopped pots", which make only a small difference to the outcome; in other words, assume that every hand has exactly one winner. Assume also that the game is fair.

In the situation you describe, each player is then equally likely to win each hand (with probability 1/n).

This also means the probability of you (or any player) losing a given hand is (n - 1)/n. For example, in a 'full ring' 9-handed game, your odds of losing a given hand are 8/9.

So the probability of your losing k hands in a row is ((n - 1)/n)^k.

In your example, k = 16, so the probability is ((n - 1)/n)^16.

If n = 9, this gives (8/9)^16 = 0.152. This means there is just over a 15% chance you'll lose 16 hands in a row - so not particularly unlikely.

If n = 6, the equivalent number is (5/6)^16 = 0.0541 = 5.4%, or around 1 in 19 - slightly worse than the odds of being dealt a pocket pair.

Even with three players (n = 3), the equivalent number is (2/3)^16 = 0.00152 = 0.152%, or roughly 1/660. This is unlikely, but (for comparison) still more likely than flopping a full house with an arbitrary starting hand, so far from impossible.

(By the way, it is pure coincidence that the significant figures of the numbers above for n = 9 and n = 3 both begin "152".)

  • Okay I’ll be more clear I suppose I didn’t mention it’s just myself and the dealer. This game at the card club is called ultimate Texas hold em. There was no other people at the table and you do not play against other players only the house. So 1/2 is the ratio Commented Oct 5, 2023 at 9:50
  • Then n = 2, so the probability is 0.00001525 = 0.00153%, or 1/65536. Perhaps you should try a different casino...
    – Neil T
    Commented Oct 5, 2023 at 10:46
  • For n=6 I believe you should use 5/6 (instead of 6/9) for about 5.4% chance of losing 16 in a row.
    – TTT
    Commented Nov 7, 2023 at 3:22
  • @TTT - you are of course completely correct. Fixed in my answer above. Thank you for pointing out my silly error.
    – Neil T
    Commented Nov 7, 2023 at 9:25
  • 1
    @TTT - also fixed; thank you! (This was just a typo; the calculated values were correct.) Since OP has already commented on this answer I think I'll leave it as is; the other answer by Nij is probably more relevant to OP's original question. Hopefully the maths will help someone, now you've corrected it!
    – Neil T
    Commented Nov 8, 2023 at 9:54

Unless the payouts are also 1:1, you should not be expecting anything close to 50:50 odds. If it is like any other common casino poker-based game, your odds of winning any hand are almost nothing, but you get paid big when you do hit one.

The payouts table for Ultimate Texas Holdem shows the lowest hand that gets paid out is a straight, at 1:1, and the dealer does not open at all on less than a paired board. For comparison, your odds of getting that straight in the first place are worse than 19:1. Even if you win, you are being paid only a few percent of what you should be.

It is not just common to lose any given hand, but in fact, losing multiple hands in a row is expected. 16 hands in a row is unlikely, but when enough people play enough hands, even unlikely things will be almost sure to happen. You just got the bad luck this time, instead of the good.

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