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Let's say you make three bets... first one has 30% chance of winning, second has 40%, and last has 50%. What are the odds that you win at least two of them? How is that calculated?

So far I know that I can't just add them together. I can't multiply them together. I tried manually doing each pair (.3*.4 , .4*.5, .3*.5) to calculate the odds of those pairs both winning... but I'm not sure how to add them together.

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  • $\begingroup$ What have you tried? People are much more likely to help you if you show them where you are stuck. $\endgroup$
    – Leonhard Euler
    Aug 9, 2018 at 15:51
  • $\begingroup$ Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments. $\endgroup$
    – José Carlos Santos
    Aug 9, 2018 at 15:52

2 Answers 2

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Hint: Let $A_i$ the probability that you win bet $i$ with the following probabilities:

$P(A_1)=0.3, P(A_2)=0.4, P(A_3)=0.5$ and $P(\overline A_i)=1-P(A_i)$

Then the probability to win at least 2 bets is

$P(A_1)\cdot P(A_2) \cdot P(\overline A_3)+P(A_1)\cdot P(\overline A_2) \cdot P( A_3)+P(\overline A_1)\cdot P( A_2) \cdot P( A_3)$ $+P( A_1)\cdot P( A_2) \cdot P( A_3)$

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Hint. Draw the binary tree with three levels, one for each bet. Put probabilities on the edges. Multiply along paths to get the probabilities of the $8$ possible outcomes (leaves). Those are disjoint events. Add the probabilities of the ones you care about (i.e. at least two wins).

Your profile says you're a programmer, so this should work for you.

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