# The bet

A person wants to get 5,000 euros playing a game of chance that involves betting an amount of money, which must always be a multiple of 1,000, so that, if he wins, he recovers twice as much and if he loses he runs out of money. bet money.

The player starts with 1,000 euros and always plays on each bet in the riskiest way possible to achieve exactly his goal by applying logic. Thus, for example: if you have 2,000 euros, you will play 2,000, while if you had achieved 3,000 euros you would not play them in full, but you would only bet 2,000 euros, since in the case of winning you would get 5,000 euros and if you lost would be with 1,000, with the possibility of playing again.

If we know that the probability of winning or losing in each bet is the same,

How likely are you to get 5,000 euros?

#### Solution

Let's analyze the different options that our player has. Start with 1000 euros and have a 1/2 chance of losing them and 1/2 of getting 2000.

In case you have 2000, you will bet everything, and again you have 1/2 chances of losing everything and 1/2 of getting 4000.

If you have 4000, you will bet only 1,000 and you will have a 1/2 chance of becoming 3000 and 1/2 of successfully finishing the game

Finally, if you play with 3000, you will have 1/2 of probability of winning and 1/2 of having 1000 again.

We can make a system of equations with the probabilities of winning and losing from each of the quantities. We can call P1 the probability of winning if you have 1000, P2 the probability of winning if you have 2000, P3 which you have to win if you have 3000, and P4 if you have 4000. Since all these situations are unstable, except losing everything or win, we know that the probability of winning and losing adds up in any case.

If we imagine that we repeat the experience many times, it is easy to think that half of those who start with 1000 euros lose, while half start to have 2000. Therefore, P1 = P2 / 2. Of those who have 2000, the same happens, so P2 = P4 / 2. In the same way, P4 = 1/2 + P3, and P3 = 1/2 + P1. In this way, we have a system of four equations with four unknowns. Eliminating P4 first, we have that P1 = P2 / 2, P2 = 1/4 + P3 / 4 and P3 = 1/2 + P1 / 2.
After eliminating P3, we have that P1 = P2 / 2 and P2 = 3/8 + P1 / 8, where P1 = 3/16 + P1 / 16, that is, that 16P1 = 3 + P1, where P1 = 1/5.