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Let say a solitaire game can be solved 1 in 10 plays. (A play is defined as shuffling of cards and trying to solve the solitaire).

If after each successfully solved game we skip 8 games (we shuffle but don't try to solve it), what would be the average number of plays of non-skipped games per 1 solvable game and why?

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    $\begingroup$ It should just be 10:1 right? Shuffling does not alter the probability of the next game being solvable if it did, then the number of times you shuffle the cards would matter. $\endgroup$ – Tony Ruth Jun 13 '16 at 19:07
  • $\begingroup$ OK, taking card game as an example probably is not the best choice, but the question could be "how the ratio would be changed if removing particular number of samples after next win sample?" $\endgroup$ – kamenf Jun 13 '16 at 19:20
  • $\begingroup$ I still don't think there would be any difference. The probability distribution for the 1st sample after a win is exactly the same as the 9th sample after a win, and the 2nd sample has the same probability distribution as the 10th sample, and so on, and so forth. There is something kinda close to this which I think you may be getting at. The average run length between two Heads-Heads flips is longer than the average run length between two Heads-Tails flips. It would be wrong to interpret this as: "the probability of tails after heads is greater than the probability of heads after heads." $\endgroup$ – Tony Ruth Jun 13 '16 at 19:36
  • $\begingroup$ @TonyRuth Agreed. $\endgroup$ – pajonk Jun 13 '16 at 20:04
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    $\begingroup$ Is the ratio (plays : solvable plays) or (plays : solvable shuffles)? $\endgroup$ – kaine Jun 13 '16 at 20:39
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Each game has two possible outcomes. We care about number of played games and number of winnable games (both played and unplayed).

Win

  • $\frac{1}{10}$ chance

  • Add 1 to played games

  • Add 1 to solvable played games

  • Add $\frac{8}{10}$ on average to solvable unplayed games

Loss

  • $\frac{9}{10}$ chance

  • Add 1 to played games

So on average once every 10 played games the number of unplayed solvable games rises by $\frac{8}{10}$ on average.

$10 : \frac{8}{10}$

We calculated ratio of played games to skipped solvable games. We have to add one to the right side to account for the one won played game.

$10 : \frac{8}{10} + 1$

$10 : \frac{18}{10}$

$100 : 18$

$50 : 9$

And that is the answer. Ratio of played games to winnable (both played and unplayed) is $50 : 9$. Average number of played games per solvable game would be $\frac{50}{9}$ or approximately $5.55$.

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The average number of games played per solvable shuffle (given you always successfully solve if you play a solvable shuffle) would be

$\frac{50}9$

Because

If you shuffled for a play and it is not solvable you have made $1$ shuffle and it was unsolvable. (probability $\frac9{10}$: $0$ solvable shuffles)
Whereas if you shuffled for a play and it was solvable you have made a total of $9$ shuffles, the first of which was solvable and the other $8$, which you do not play, each have a probability of $\frac1{10}$ of being solvable. (probability $\frac1{10}$: $1+8\frac1{10}$ solvable shuffles)

The expected value of solvable shuffles per play is therefore $0\frac9{10}+(1+8\frac1{10})\frac1{10}=\frac{18}{100}=\frac9{50}$

The expected value of the number of plays per solvable shuffle is the inverse of this $=\frac{50}9$

In practice

It will be less if you decide ahead of time to play $n$ games and not to shuffle the other $8$ times if your last play is successful (why bother). In this case you expect $\frac9{50}+\frac1{10n}$ solvable shuffles per play and hence $\frac{50n}{9n+5}$ plays per solvable shuffle.

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