27

The average payout for each ticket is So Thus


20

I think the answer is Proof


19

Number of qualifying times: Probability:


17

To start off, I believe... So, we can consider the odds as: If a zero is scratched, the game ends and you get nothing, otherwise Therefore you can consider


15

Most people here have zeroed in on the underdefined statement of "90% good". I would try to be a good Bayesian here and start with a few of my personal observations about the real world as priors. From these two, we can conclude that Which we can combine with another set of priors: The above observations point in one direction: the metric he ...


12

They win with probability


12

I think the answer is as follows. Case One: Case Two: Case Three: Bonus:


11

Suppose Thelma asks her mother on the first day. I originally misread the question, thinking that the rule was that if you ask one parent on two consecutive days, they won't say yes twice. This is a pretty interesting question too, so I'll keep the answer for that question below: Clearly she must alternate which parent she asks, because if she asks the same ...


11

The answer would simply be: because example


11

I'm going to assume that when the partner says that they are 90% good at vetting they mean they accept an unqualified candidate 10% of the time, and reject a qualified one 10% of the time.


10

Suppose we have $n$ dice. Then When $n=15$ Just for fun, here's a smartass combinatorial way to prove the identity I used above: Perhaps there's a smarter-ass way to do it a bit more briefly.


10



9

The expected value is What you can do here is Then, Next, Therefore


9

This is an example of the Using that,


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Explanation: The first step is to determine the optimal strategy. Now imagine It is only interesting to scratch another square if Note: Hence the best strategy is: How much are you winning with the best strategy ? Conclusion Since the ticket costs $10$, the lottery will distribute: of its income.


9



9

Their best strategy is Their chances with this strategy are Optimality


8

There are two simplifications we should make first: Then, it's a matter of calculation.


8

I think the answer is ... because


7

A strategy: This works as long as This is optimal because:


7

Here is one simple way to solve the problem in a single round of tests. This is not optimal, but easy to explain. At the end I'll describe an improved version. When the results come back, you can easily find which candidate is the successful one, because If there is more than one successful candidate, The above method is not optimal, but can be improved ...


7

As @Gareth McCaughan was asking for a more smarty pants solution:


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This solution was derived independently but the result is equivalent to the answer already given by Mark Tilford and the confident tone of a suggestion by Greg Martin led to the realization that this does provide the highest possible probability of winning the car. Imagining a repeating cycle of doors, . . . Left - Middle - Right - Left - Middle - Right - ...


6

I think the answer is... Analysis:


6

Revised to include a recursive path to a perspective of the result from a situation that did not occur. Did either of them make a mistake? That question is more interesting than the rest of this sentence but, simply put, . . . . . . . both Jack (J) and Kirby (K) aren’t perfect logicians and both had suboptimal strategies.   As such, . . . . . . . which ...


6

How can Ash win? Looking at the game tree: Formalising the above, we get the following relationship for $p_A$, the probability that Ash wins: Ash and Bree desire for $p_A = \frac{1}{2}$. Substituting and rearranging, we get: which can be solved Now, to finish up: But, there's a catch:


5

If you plot the "sales so far this year" line against the "days so far this year" line (which will be a straight line with a slope of 1), the lines will or In the first case, Note that


5

Alice can achieve a victory probability of Proof: Proof that there is no better strategy:


5

In a group of 25, you might have good odds of duplicate birthdays, as the linked question says, but you have almost good odds of not having them, too. For all ten groups of 25 to have duplicates is unlikely. But this is not a group of 250 random people. It's an exam, and since it's teenagers, it is perhaps a high-school exam -- a situation in which people ...


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