Here's a counter-intuitive puzzle:

Post solutions below!

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Here's a counter-intuitive puzzle:

Tom and Harry are trapped on a desert island, playing a game of dice for the last banana. They roll two dice, and agree that if the biggest number rolled is a one, two, three, or four, Tom wins, but if the biggest number rolled is a five or six, Harry wins. Who has the best probability of winning the banana?

Post solutions below!

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Puzzle

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2 comments
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Let's call the rolls $r_1, r_2 \in {1,2,3,4,5,6}$ and the sets of rolls $A = {1,2,3,4}, B = {5,6}$ for convenience.

Harry wins if both dice are less than $5$. That is if $\mathbb{P}(r_1 \in A, r_2 \in A) = \frac{4 \times 4}{6 \times 6} = \frac{16}{36} < \frac{1}{2}$. Thus, Tom is more likely to win.

Another way is to visualize it as a matrix in Python:

```
>>> import numpy as np
>>> wins = np.zeros((6, 6)) # r_1 is rows, r_2 is columns
>>> wins[[4, 5], :] = 1 # Tom wins if the first die is 5 or 6
>>> wins[:, [4, 5]] = 1 # Tom wins if the second die is 5 or 6
>>> wins
array([[0., 0., 0., 0., 1., 1.],
[0., 0., 0., 0., 1., 1.],
[0., 0., 0., 0., 1., 1.],
[0., 0., 0., 0., 1., 1.],
[1., 1., 1., 1., 1., 1.],
[1., 1., 1., 1., 1., 1.]])
>>> wins.mean() # Tom's probability of winning: no. of ways he could win / no. of ways the dice could fall
0.5555555555555556
```

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Looks good kl2792! I'm currently adding markdown support for posts and comments to make code look nicer.

jdry1729
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For each individual die, the probability of coming up with $\leq 4$ eyes is $\frac{4}{6} = \frac{2}{3}$. It therefore follows that the probability of two independent rolls both coming up $\leq 4$ is equal to

$\left( \frac{2}{3} \right)^2 = \frac{4}{9}$. Hence the probability of Tom winning is $\frac{4}{9} < \frac{4}{8} = \frac{1}{2}$, i.e. the odds are in Harry's favour who has a $\frac{5}{9} \approx 56\%$ chance of winning.

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Nice solution @KvanteKat! 👏👏👏

jdry1729
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