Out of pure boredom during the corona period back in 2020 I 'invented' a drinking game which seemed to have a fair outcome, but I am not sure. We were sitting on the couch and after a lot of beer, and some games of Perudo (which is played with a lot of dice) I came up with the following game.

  • There are 24 dice on the table.
  • One person throws all the dice.
  • Every '1' and '6' that is rolled is removed from the table. (So if you roll 5 times a '6' and 4 times a '1', your next throw contains 15 dice)
  • Repeat this process for a total of 8 times.
  • If you are able to remove all the dice in 8 throws or less, you win the game and the rest of the room has to drink a X amount of sips.

That night we a had a lot of fun playing this very simple and rather stupid drinking game, and it seemed to have a fair outcome: the chances for winning this game felt like a 50/50. The next day I woke up with a hangover and I was curious about the actual probability of winning this game. I calculated the expected number dice you would be left with after 8 throws:

The number in the third row is calculated as follows; 24 - (1/3*24) etc.

The number in the third row is calculated as follows; 24 - (1/3*24) etc. So, after 8 throws the expected number of dice you are left with is .936. I am probably wrong but; On the one hand, as this is less than 1, you might argue that the probability of winning this game is therefore greater than fifty percent. On the other hand you might argue that it is more than .5 dice and therefore it could be that the probability of winning this game is smaller than fifty percent.

What is the probability of winning this game? There is probably one 'easy' answer which is simulating the game for a million times. However, I am not familiar with coding, is there anyone who could help me? Furthermore, I am very curious how you could interpret this .936. Is there any way to express this as a percentage of winning the game? Or is there any calculation that gives you the percentage of winning this game?

I am very curious! If something is not clear let me know.

Thanks in advance,


  • $\begingroup$ Just a note about being precise about what you want: please be aware that odds and probability are different things. en.wikipedia.org/wiki/Odds ... if you want probability, don't say odds. $\endgroup$
    – Glen_b
    Commented Jan 23, 2023 at 6:55
  • $\begingroup$ The problem is equivalent to rolling each dice 8 times (even when it had a 1 or 6 in the meantime) and see whether any had zero times a 1 or 6 (that's a dice that never got removed in your version). $\endgroup$ Commented Jan 23, 2023 at 7:06
  • $\begingroup$ @Glen_b Thanks a lot for your clear reply! I understand it now :) $\endgroup$
    – Pim
    Commented Jan 23, 2023 at 7:11
  • $\begingroup$ @SextusEmpiricus Oops, it was a typo I guess... I've edited it :) $\endgroup$
    – Pim
    Commented Jan 23, 2023 at 7:11

1 Answer 1


The chance a single fair die doesn't show a $1$ or a $6$ in eight rolls is $(\frac46)^8$ $\approx$ $0.039$.

The chance that all 24 dice do show a $1$ or a $6$ within 8 rolls is therefore $[1-(\frac46)^8]^{^{24}}$ $\approx$ $0.3847$. That is, the probability of a win is just a little below 40%.

[In case you did want odds, the odds in favour of a win are therefore $0.3847/0.6153 \approx 0.625$ or about $5:8$.]

  • $\begingroup$ Very clear! Thanks a lot :) $\endgroup$
    – Pim
    Commented Jan 23, 2023 at 7:20

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