Winning the Lottery: Probability & Odds

AI Thread Summary
The discussion focuses on calculating the probability of winning the lottery by choosing 2 numbers from a set of 18 without replacement. Participants explore the correct method for combining combinations, debating whether to add or multiply the outcomes. It is clarified that the solution represents a fraction of total outcomes rather than the actual probability of winning. The correct approach involves multiplying combinations from different ranges to arrive at the final calculation. Overall, the conversation emphasizes understanding the mathematical principles behind lottery odds.
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thanks for helping :)
 
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Think about this as follows: how many ways are there to choose 2 numbers from 18? (no replacement, order doesn't matter)
 
oleador said:
Think about this as follows: how many ways are there to choose 2 numbers from 18? (no replacement, order doesn't matter)

i was thinking about this:

numerator: 18C2, 18C2, 19C2 --- but I'm not sure if you ADD or MULTIPLY these
denominator: 55C6
 
<br /> \frac{C^{2}_{18} \, C^{2}_{18} \, C^{2}_{19}}{C^{6}_{55}}<br />

Wolfram Alpha

This looks terribly high?!
 
i was thinking of that.
if you add it on the other hand, its too low

i'm not quite sure with my solution
:(
 
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No, but this not a probability ot win! This is just a fraction of the total number of outcomes that satisfy your condition. It doesn't mean that if you fill A ticket satisfying this condition that you have this probability to win.
 
Dickfore said:
No, but this not a probability ot win! This is just a fraction of the total number of outcomes that satisfy your condition. It doesn't mean that if you fill A ticket satisfying this condition that you have this probability to win.

so you think the solution is right? :)
 
Yes. The solution is correct.
 
Dickfore said:
Yes. The solution is correct.

thanks! :)
 
  • #10
The intuition for why you have to multiply the combinations is that for every possible combination, say, in [1,18] you can have 18C2 of combinations in [19,36] and 19C2 in [37,55].
 
  • #11
oleador said:
The intuition for why you have to multiply the combinations is that for every possible combination, say, in [1,18] you can have 18C2 of combinations in [19,36] and 19C2 in [37,55].

now i understand. thanks! :)
 
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