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If you were to start out with a 2% chance of winning a prize at the casino but each time you lost you were granted an additional 2% chance of winning, what would be the odds that you lost enough to reach a 70% chance of winning.

So say you start gambling with a 2% chance to win and you lose, so now you have a 4% chance to win. now with that 4% chance of winning you lose again, so now you have a 6% chance. but you lose again so now you have an 8% chance to win. and you lose again so now you have a 10% chance, and so fourth.

What would be the odds that you lost so many times that you were able to reach a 70% chance of winning.

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  • $\begingroup$ what happens when you win, it resets back to 2% or stays at the current level? $\endgroup$
    – karakfa
    Jan 25, 2018 at 19:56
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    $\begingroup$ The odds are 100%, eventually you will reach it. Maybe you wanted to ask something else? $\endgroup$
    – orlp
    Jan 25, 2018 at 19:58
  • $\begingroup$ My guess is, it's supposed to ask how many rounds you're expected to play. $\endgroup$
    – karakfa
    Jan 25, 2018 at 20:04
  • $\begingroup$ when you win it resets, but that is besides the point. im just asking what are the odds that you could lose so many times that eventually you would have a 70% chance of winning, the odds could not be 100% because after 25 rolls or so you will start rolling to win with over 50% chances of winning, and still losing over 10 times in a row. $\endgroup$
    – Michael Mitchell
    Jan 25, 2018 at 20:05
  • $\begingroup$ say you're at a slot machine, and each time you pull the lever and lose you are granted an extra 2% chance. so starting at 0% chance you pull and lose, so now you have a 2% chance and you pull and lose, so now you have a 4% chance and you pull and lose, so now there is a 6% chance of winning, but you lose so you keep losing and with each pull having this percent chance of winning so then you move onto having as such 8% 10% 12% 14% 16% 18% 20% 22% 24% 26% 28% 30% 32% 34% 36% 38% 40% 42% 44% 46% 48% 50% 52% 54% 56% 58% 60% 62% 64% 66% 68% 70%. what would be the odds that this happened. $\endgroup$
    – Michael Mitchell
    Jan 25, 2018 at 20:11

1 Answer 1

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So first I hope you are talking about an increase in percentage points and second that the winning probability is set back to 2% once you win before hitting the 70% winning chance. I guess your question is how likely it to loose n times in a row to directly hit the 70% winning probability right?

In such a case you need to loose 34 times in a row to reach a 70% chance of winning since $$2 + 2\cdot n \stackrel{!}{=} 70 \Leftrightarrow n = 34$$

So the odds to actually reach this scenario is $$P_1(loose) \cdot P_2(loose) \cdots \cdot P_{34}(loose) = 0.98 \cdot 0.96 \cdot \cdots \cdot 0.32 = 0.0000000024%$$

Therefore the probability that this happens is 0.00000024%

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