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In a city, the official lottery consists of scratch cards that cost 10 bucks each. Each scratch card contains 36 squares with a hidden number on each square : 21 of them are “zeroes”, 9 of them are “ones” and 6 of them are “tens”. To proceed, one can scratch as many squares as one wants. The total revenue consists of the product of all the scratched numbers. For example, if someone scratches 2 “ones” and 1 “ten”, the total revenue would be 10 bucks. If everyone in the city plays with the optimal strategy, what percentage of the total sum that the lottery company would make would be given back to the players? Please don’t use computers.

This problem is adapted from a statistics exam for AP Statistics.

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To start off, I believe...

We can ignore all ones in this case, since they have no affect the total score and I can't think of any case in which you are in a situation when scratching again is optimal, you scratch and get a one, and then it is optimal to stop. I have a feeling may be incorrect in this assumption however so perhaps I need to revisit this.

So, we can consider the odds as:

21/27 "Zero" and 6/27 "10"

If a zero is scratched, the game ends and you get nothing, otherwise

The "$10" odds reduce to 5/26, 4/25, 3/24, and 2/23, the last of which is no longer above 1/10 and therefore it is not in the player's interest to continue drawing.

Therefore you can consider

6/27 * 5/26 * 4/25 * 3/24 is the odds in order to get a total product of \$10,000. Finding the product of these probabilities gets 360/421200, or roughly 0.0008547, multiplying by \$10,000 gets you an average winnings of \$8.547, divided by \$10 and multiplied by 100 to get a percentage gets us that 85.47% of the money is given back to the players

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    $\begingroup$ Very clever method. $\endgroup$ – Display maths Jun 8 at 13:54
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    $\begingroup$ To see that scratching ones does not affect your stopping criteria, consider that payout multiplier for scratching one more is $10T/(T+Z)$ for $T$ tens and $Z$ zeros, which changes to $(10T+O)/(T+Z+O)$. Adding the same number to the numerator and denominator is just moving the weighted average closer to $1$, but cannot possibly make it cross unity in either direction. $\endgroup$ – obscurans Jun 9 at 3:04
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The average payout for each ticket is

Less than $10$ for whatever number of scratched squares.

So

Since "everyone in the city plays with the optimal strategy" as stated in the question, the "optimal strategy" is not to buy the ticket.

Thus

No percentage of the total sum is given back to the players.

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    $\begingroup$ Clever clever. Although I'm not sure choosing not to buy a ticket can be interpreted as "playing" with the optimal strategy. $\endgroup$ – Michael Moschella Jun 8 at 12:50
  • $\begingroup$ @MichaelMoschella I know, this is more a [lateral-thinking] answer $\endgroup$ – melfnt Jun 8 at 12:50
  • $\begingroup$ It is amusing to upvote answers which disagree $\endgroup$ – Ross Millikan Jun 8 at 21:12
  • $\begingroup$ well, if nobody plays, no money is collected! So the percent return is (0/0)*100 = undefined. Not zero percent. $\endgroup$ – Ross Presser Jun 8 at 21:35
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    $\begingroup$ "A strange game.The only winning move is not to play. How about a nice game of chess?" $\endgroup$ – zovits Jun 9 at 8:00
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85.47 %

Explanation:

The first step is to determine the optimal strategy.

First notice that squares containing $1$'s are irrelevant. The gain is only affected by discovering a $0$ or a $10$ : So we can dismiss all $1$'s ; if you discover one of them, just scratch another square. The game is now considered to have $27$ squares, $21$ of them zeroes and $6$ of them ten's.

Now imagine

That you are in a situation were the open squares reward you with a gain of $G$, and there are $z$ zeroes and $t$ tens still hidden. You expected gain by scratching one more square (that is, scratching until you discover something else than a one) is $EG=\frac{10*t*G}{t+z}$.

It is only interesting to scratch another square if

if $\frac{EG}{G}>1$, ie if $z<9t$.

Note:

(since $\frac{EG}{G}$ is a decreasing function of $t$, you cannot hope to compensate a loss of expected value at one step by a higher multiplicator later on).

Hence the best strategy is:

With z fixed at $21$, it means you should play when there are still at least $3$ ten's hidden, and stop when there are only $2$ of them left.

How much are you winning with the best strategy ?

You win $10000$ with a probability of $\frac{6*5*4*3}{27*26*25*24}=\frac{23*22}{2} * {{27}\choose{6}}$
and $EG\approx 8.547$

Conclusion

Since the ticket costs $10$, the lottery will distribute:

$85.47$%

of its income.

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  • $\begingroup$ Very clever method. $\endgroup$ – Display maths Jun 8 at 13:54

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