What are my chances of winning?
Author : Johnny
Submitted : 20180115 10:13:46 Popularity:
Tags: chances winning
I entered a competition in which there are 8 items that will be given to 8 individuals randomly. I have 35 entries, and there are 108 total entries (including mine). Each time a name is drawn, that individual gets one of the items. What are my chances of
107.3%
Let's find the probability of winning NONE of the items, then subtract that from 1 to find the probability that you win at least 1 item (possible to win all 8!)
If you have 35 tickets of the 108 entries, the probability of you winning the first prize is 35/108. So the probability of you NOT winning that prize is (108  35) / 108 or 73/108.
Now that you didn't win the first prize, you still have 35 tickets in, but there are only 107 tickets left. So the probability of you NOT winning prize 2 is (107  35) / 107 or 72/107
So we continue this pattern until all 8 prizes are drawn and you don't win any of them. Multiply all of these ratios together to get the probability that you win NONE of the eight prizes:
73/108 * 72/107 * 71/106 * 70/105 * 69/104 * 68/103 * 67/102 * 66/101 = 541986522295680/14193673376238720 ≈ 0.038185
So there is only about a 3.8% chance of you winning NOTHING. Which means the probability of you getting at least one prize is 1 subtracted from that, or:
1  0.038185 = 0.961815 or about 96.2% chance of you winning at least one prize.
When you own almost 1/3 of the tickets from the start, with 8 prizes drawn, yeah. you're almost guaranteed to win something. :)
It easier to figure out the probability that none of your tickets are drawn.
For the first prize, there are 73 out of 108 tickets that aren't yours.
For the second prize, there are 72 out of 107.
For the third prize, 71/106
etc.
If you multiply that all together, you get the probability you don't win anything. Then subtract from 1 to get the opposite probability that you will at least one prize.
1  (73/108 * 72/107 * 71/106 * 70/105 * 69/104 * 68/103 * 67/102 * 66/101)
Answer:
~0.962 (or 96.2%)
Whenever you see "probability of at least one," the easiest solution is almost always to calculate "probability of none," then subtract that from 1.
There are 10835= 73 entries that are not yours.
There are 73C8 ways for all 8 picks to not include yours.
There are 108C8 ways in all to pick 8.
Your probability of winning at least 1 is 1  73C8/108C8
= 1  73x72x71x70x69x68x67x66 / (108x107x106x105x104x103x102x101)
= 0.961814922...


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