Expected values for dicecoin - competitions answers

Question 1: is there a winning strategy for this gambling game? What is it?

Both competition entrants iopq and 3phase put a lot more work into this than I expected anyone to make! Good job both of you and I'll make some comments about your results after I give the answer.

I was actually assuming that entrants would simply try the game and see what happens, as I mentioned in one of the hints. If you had tried that you would have obtained something like the following 100 games, which took quite a while using the online dice generator I linked to:

Number of throws per game:4, 3, 5, 13, 52, 26, 23, 2, 5, 5, 3, 4, 6, 3, 14, 8, 12, 15, 6, 2, 14, 39, 5, 1, 13, 5, 1, 12, 3, 6, 5, 18, 17, 8, 3, 4, 3, 2, 1, 9, 4, 3, 5, 3, 2, 3, 7, 1, 2, 4, 2, 2, 8, 20 ,17 ,2 ,2, 9, 9, 1, 1, 12, 6, 5, 9, 17, 24, 7, 2, 2, 2, 8, 11, 9, 29, 12, 14, 27, 17, 3, 4, 7, 4, 8, 14, 8, 6, 4, 24, 15, 6, 6, 3, 28, 5, 40, 14, 12, 1, 25

If you weren't going to use probability theory to solve this, the first intuitive leap you had to make was that assessing one throw per game

**before** the costs and prizes were distributed amongst the players seems much easier to assess than trying to work out a per player result (which would depend on how many players were present). If you accept this for now, I'll derive the result mathematically later.

To work out the value of the first throw, we have to find all of the games which contain one or more throws - i.e. all of them. The profit for the first throw of a game before costs and winnings are split amongst players is:

Winnings for the throw minus cost of the throw = $100/(number of throws in the game)-$10

Average profit for the first throw of each game:

= mean($100/4-$10, $100/3-$10, $100/5-$10, $100/13-$10, …. , $100/25-$10)

= mean($100/(all throws per game) - $10)

= mean(25.24 - $10) = $15.24

So our average profit (when we define profit as winnings minus cost) for playing only the first throw in a game for this series of games is $15.24

To estimate the value of the second throw, we have to find all of the games which contain two or more throws -

Number of throws per game, for games>=2 throws: 4, 3, 5, 13, 52, 26, 23, 2, 5, 5, 3, 4, 6, 3, 14, 8, 12, 15, 6, 2, 14, 39, 5, 13, 5, 12, 3, 6, 5, 18, 17, 8, 3, 4, 3, 2, 9, 4, 3, 5, 3, 2, 3, 7, 2, 4, 2, 2, 8, 20, 17, 2, 2, 9, 9, 12, 6, 5, 9, 17, 24, 7, 2, 2, 2, 8, 11, 9, 29, 12, 14, 27, 17, 3, 4, 7, 4, 8, 14 ,8, 6, 4, 24, 15, 6, 6, 3, 28, 5, 40, 14, 12, 25

Calculating the average profit as above results in $9.62 per game for the second throw of each game.

Keep in mind that for a constant player, the expected profit is $0 per game.

If we continue calculating the expected profits for a throw as above, we can generate the following results:

When the expected or average value of the throw is below $0, the throw is more likely to reduce your overall winnings than increase them. So for this dataset the strategy would be to stay on for the first 5 throws. I would have accepted this as an answer. It turns out (interestingly) that when the expected values are more accurately calculated the strategy would still be a winner if the 4th throw or the 5th throw is used as the final throw.

Both iopq and 3phase came up with similar results. Although 3phase was first with

The best number is given by the strategy of playing 5 throws and then leaving.

he later added that any consecutive 5 throws would be a winning strategy (although I can't find that quote in the post anymore, only in the discussion between iopq & 3phase - can you confirm that for me 3phase?)

So, contingent on 3phase confirming his "any consecutive 5 throws" statement,

**I'm awarding iopq the huge 1btc prize!** Congratulations iopq, and well done to both iopq and 3phase.

So I think that's enough for the moment. A little later I'll post a proof for the above, then a discussion of both sets of results from iopq and 3phase, and finally I explain how to calculate the expected profit from the game if you use this strategy.

iopq, please PM me with a bitcoin address to send your prize to.