# Probability of hitting a 'bomb' in a memory-like game

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?

-
Where did you get stuck? – Keep these mind Mar 27 '13 at 19:55
Well, I can imagine that the probability for hitting a bomb the first round is 5/25 (or 1/5). But on the second round you already have an uncovered square. Is the probability then 5/24? (not sure about this) And where I seriously get stuck: How to calculate how big the chance is of missing the bombs instead of hitting them for multiple rounds in a row? So Where I get stuck: How does the outcome of the last round(s) affect the next round? – Qqwy Mar 27 '13 at 20:12
This is a simple conditional probability questin. Why don't you start by writing down the first few. and whatever the expected pay off is, it is less than how much you put in, so expect to lose money over a long run, though it must be case there is an optimal time you flip this thing to minimise your expected loss. – Lost1 Mar 27 '13 at 20:14
@Qqwy Yup, $5/25$ and $5/24$ are correct. As I understand it you don't get to turn more cards after hitting a bomb, therefore, whenever you're allowed to turn a card, the numerator always stays $5$. – Keep these mind Mar 27 '13 at 20:24
@Qqwy This could be an interesting question, but you would have to create a new question then and leave this one as it is. Note that you would have to clarify how much you get for a card that's not a bomb. Is it always the same? Is it the number on the card? Etc. Also, this time don't use the game-theory tag, that's only if there are more than 1 players. Tip: If you have a picture of the game, that would always help getting good answers. – Keep these mind Mar 27 '13 at 21:50

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 n cards = $\dfrac{20Pn}{25^n}$

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 $\dfrac{20!}{25^{20}}$

-