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I was trying to come up with a dice game which makes the players' intuition fail.

The game has two dice. Player $A$ and player $B$ have different victory conditions. Player $A$ and $B$ alternate tossing the dice. Player $A$ needs to throw two consecutive $6$ to win. Player $B$ throws both dice together and needs to roll double sixes.

The expected number of rolls for player $B$ before winning is $36$. For player $A$, it is $42$. Should this be enough to convince someone to play as player $B$?

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What is the probability that player A wins?

Well, if has not yet rolled a 6 (as is the case at the beginning), $p = \frac{35}{36}(\frac{1}{6}p'+\frac{5}{6}p)$, and if he has rolled a 6, $p'= \frac{1}{6}+\frac{5\cdot 35}{216}p$. Solving this yields

$$p=\frac{1260}{2731}\approx.4614$$

Note that this calculation assumes that A gets to roll first. Otherwise, multiply by an extra factor of $\frac{35}{36}$ to obtain

$$p\approx.4486$$

So in either case B is clearly better off.

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  • $\begingroup$ So it seems rational to take a $(8,10)\%$ chance edge in the game? Is this really enough to "run a gambling circle and make profit"? $\endgroup$ – Cehhΐro Oct 16 '16 at 21:40
  • $\begingroup$ If you play it enough times, yes. $\endgroup$ – ruadath Oct 16 '16 at 21:49

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