The following solution, bet three, always win 0.5 3.
The bet is 5, winning 4 games and winning the last game, 4C3*0.5^5.
The bet is 7, winning 5 and losing 2, the number of games is not 5, and the last game is won, (1-4c3 * 0.5 5) * 6c4 * 0.5 7.
Bet 9, win 6 and lose 3, the number of games is not 7, 5 and 3, and the last game wins, (1-4c3 * 0.55-(1-4c3 * 0.5) * 6c4 * 0.57) * 8c5 * 0.5.
The following 1 1 bureau, 13 bureau or even infinity are calculated in turn, and then all of them are summed, and then the limit is found. Take your time, some of you have already suffered.
If you lose everything, you must lose five games, and the number of gambling games must be 5, 7, 9, 1 1 ...
Calculated in a similar way as above.
Note that this is to calculate the probability of winning 8 or losing all. It includes all the possibilities of gambling. If you are sure that there are only these two results, you need to find out the relative percentage of the two probabilities obtained above. Normalize with 1 and find their respective probabilities.
It should be noted that the probability itself is relatively small, because you must win 8 or lose all, and you must normalize. This is an important probability. However, it is difficult to predict the number of games. I mean, it may take a lot of money to get the result, maybe dozens or hundreds. In this way, the result before finding the relative percentage is more meaningful, because that is the natural probability.
Some people upstairs have no brains. If they win the first three games, there will be gambling in the fourth and fifth games!