A bar near where I work has a game where you pay $5$ dollars which gets you two chances of rolling $5$ dice and if roll results in all of the dice having the same number you win the running pot, effectively its Yahtzee bar game.

I'm trying to figure out when it would make the most sense to start trying to play. Even though its all chance it would seem like if the pot were $20000$ it would be well worth the risk.

The chances of rolling Yahtzee are pretty straightforward:

  • The first die doesn't matter because it can be anything

  • The following dice must all be the same and there are six sides of the die so your chances of rolling Yahtzee are:

    $\dfrac1{1\times6\times6\times6\times6} = \dfrac1{1296} $

And since you get $2$ rolls it would cost $\$2.50$ per roll. So it would only be worth rolling if the pot was $1296\times2.50 = \$3240$ and the pot just hit $\$3300$ yesterday.

Does this make sense or am I doing fuzzy math or fuzzy logic for that matter?

  • $\begingroup$ Wait, how did the pot get that large to start with? Did the pot start out at like \$1000? $\endgroup$ Commented Jul 27, 2013 at 0:11
  • $\begingroup$ Since your money also end up in the pot, you could start trying at any pot size, considering you have enough money to go on until you win. And this makes your calculaction roughly the amount of money you need to have in advance. $\endgroup$
    – sukhmel
    Commented Oct 4, 2018 at 13:26

1 Answer 1


Your logic is basically sound. There is a tiny correction in that your chance of winning the pot is not $\frac 2{1296}$, but $\frac 2{1296}-\frac1{1296^2}$ because you have double counted the chance you win on both rolls. But that doesn't change the threshold. Your odds of winning are then about $1$ in $648.25$ so you should play at $5\cdot \$648.25-\$5=\$3236.25$ (because your $\$5$ gets added to the pot) rounded up to $\$3240$ if your criterion is positive expected value.


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