Suppose you are playing a video game. Every time you win a game you have an $x\%$ chance of winning a prize. After the $n^{th}$ win you have a $x\%=(2+2.3(n-1))\%$ chance of winning a prize. Obviously on the $22^{nd}$ win we have about a $50\%$ chance of winning a prize, but intuitively I feel like we can expect to win a prize before then because although we have a $50\%$ chance on the $22^{nd}$ win, we have already had $21$ other chances to win.

So I guess my real question is: For what value of $n$ can we expect to have won a prize with a $50\%$ probability? Is it just $n=22$ or is my intuition correct and how would I go about making this assertion rigorous?

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    $\begingroup$ Do you mean after the $n^{th}$ game the probabilities increase, as opposed to wins? Because then how can you say you expect to have won a prize by the time when $n=22$, which means by your $22^{nd}$ win? Or alternatively, do you mean to find the value of $m$ (where $m$ is number of games played) for which we can expect (all that stuff)? Also in this case, what happens to the chance of winning if you lose? Does it stay the same? $\endgroup$ – John Doe Jun 2 '17 at 21:52
  • $\begingroup$ Hmmmm, maybe I wasn't very clear. You can assume I win every game and so my chance of winning a prize goes up every game. So $n$ is effectively games played and we can disregard the concept of whether or not I won a game. $\endgroup$ – user451847 Jun 2 '17 at 21:59
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    $\begingroup$ @ErickWong I got confused by this too, but the comment clears it up, and also careful reading - you may win the game but you only have $x\%$ chance of winning a prize - and this is what we care about. $\endgroup$ – John Doe Jun 2 '17 at 22:05
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    $\begingroup$ @user451847 Perhaps you should try improving the wording of your question. For instance by using "complete" instead of "win" when referring to a game. $\endgroup$ – Erick Wong Jun 2 '17 at 22:07
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    $\begingroup$ I see where there was confusion. I went and made the problem less confusing and more explicit. $\endgroup$ – user451847 Jun 2 '17 at 22:08

The probability of not winning the prize on the nth game win, in percentage terms is $$100 - \left[2+2.3\left(n-1\right)\right] = 100.3-2.3n$$ In fractional terms, this is $$1.003 - 0.23n$$ Let's ignore the fact the you have a less than 0 chance of winning the prize when n=0, and assume the probability is just 0 then.

The probability of not winning any prize at all from any wins up to and including the nth win is $$\Pi_{k=1}^{n}(1.003-0.023k)$$ We can calculate this for different values of n:

n Probability of not getting the prize

1 0.98

2 0.93786

3 0.87596124

4 0.79800068964

5 0.7086246124

6 0.612960289726

7 0.51611256395

8 0.422696189875

9 0.33646616714

10 0.260088347199

11 0.1950662604

12 0.14181317131

13 0.0998364726026

14 0.0679886378424

15 0.0447365237003

16 0.0284076925497

17 0.0173855078404

18 0.010240064118

19 0.00579587629078

20 0.0031471608259

21 0.00163652362947

22 0.000813352243845

23 0.000385528963582

24 0.000173873562576

25 7.44178847824e-05

So you probably get the prize by the 8th win, and are very unlikely to have to wait until the 22nd.

  • $\begingroup$ This is what I just tried to work out on my own. This makes perfect sense. $\endgroup$ – user451847 Jun 2 '17 at 22:12

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