Initially, our pool has $10 from each player, so that's $20 in the pool. Our opponent adds another $10, making the pool available to win $30.
If we play, we can win $30 in the pool, making our total money $40, or lose an additional $10 we just put in. So, we just need to win with a \( \frac{10}{40} \) probability to play the increased bet.
One more way to think about this problem is: to play the increased bet, the gain/loss should be less than the gain/loss of not playing the game.
We start the game by putting in $10. If we don't play again, we lose $10 with 100% probability. On the other side, i.e., playing the increased bet, we need our loss to be less than $10, so if the probability of winning is \(x\), we need:
$$20x + (-20)(1-x) \geq -10$$
$$40x - 20 \geq -10$$
$$40x \geq 10$$
$$x \geq \frac{1}{4} \text{ or } 25\%$$
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