I came across a casino-like game a few days ago where there is a playing field of face-down cards of 5x5 cards. Out of these 25 cards, 5 are 'bombs'. The player bets a fixed amount of money at the start of the round, and then can uncover as much cards as he wishes. Uncovering a 'good'(any non-bomb) card gives him a little bit of money. The amount of money he gets is determined from the no. of cards he already has uncovered. - When a player decides to stop, he gets back his fixed bet as well as all the money won from the uncovered good cards. - When a player however hits a bomb, he doesn't get any money and also loses the money he put in at the start of the round.
Now my question is, how big is the chance of hitting a bomb on your first turn, and then on your second turn, ...and then on your 20th turn(when there are only bombs left).
In other words: How do the cards you already uncovered affect the probability of hitting a bomb in the current round?
I'm not sure if this is what you are looking for, but I guess that it could be interesting as well, besides knowing what is the next probability of hitting a bomb.
Probability of not hitting a bomb after uncovering
ncards = $\dfrac{20Pn}{25^n}$So that, the probability of uncovering 1 card (in a row) without losing = $\dfrac{20}{25}$
So that, the probability of uncovering 2 cards in a row without losing = $\dfrac{20\times19}{25\times25}$
etc.
The probability of winning the big prize (i.e. uncovering all cards but the bombs) becomes = $\dfrac{20P20}{25^{20}}$ or $\dfrac{20!}{25^{20}}$