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-   -   Drawing a card probability (http://mymathforum.com/probability-statistics/331033-drawing-card-probability.html)

 lenks April 27th, 2016 01:56 PM

Drawing a card probability

This is just something I became curious about, not homework or anything, but I don't understand the result I'm getting.

My attempt at a problem statement:
A contestant is presented with a game where the goal is to draw 1 winning card from a collection of cards. Each time they draw, they must place the card back into the pile and the cards are randomized.

The number of times the contestant can attempt to draw the winning card is equal to the total number of cards.

For example, 1,000 tries for a pile of cards containing 999 losing and 1 winning card.

The question is: how does the probability of winning change as the number of cards increases by multiples of 10, starting at 10 and going to 10^20.

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When I graph the result it's very unexpected to me and I'm asking here in hopes I can understand. At first it holds steady at about 36% chance of losing for 10^1 through 10^13, but around here it starts to fluctuate wildly until by the time we're at 10^17 the probability of losing shoots to nearly 1 and stays there (?!)

Is this an error with the calculator?

Here is the equation I used for probability of drawing a losing card every time:
y=[((10^x)-1)/(10^x)]^(10^x), from x=1 to x=20

e.g. (99/100)^100 to represent the probability of drawing a losing card 100 times in a row.

 mathman April 27th, 2016 03:56 PM

Your analysis is correct. I presume the calculator is limited by precision. For large n, $\displaystyle (1-\frac{1}{n})^n \approx \frac{1}{e}$.

 lenks April 27th, 2016 05:12 PM

So as the number of cards gets very large, the probability of winning approaches 1-1/e This is a very clean answer and what I was initially curious about, thanks.

I feel like if I understood e better this would be an intuitive result... I'll have to think about this a bit in the future.

Also, looking again, I feel a bit silly for not doubting the calculator more confidently. Thanks again.

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