Probability is relative to what we know ... what information we have.
OK, so with this game, everyone has a probability of $\frac{1}{10}$ of winning. To be precise: If we know that 10 cards are being randomly distributed, and we know that one of those is a winner, and we know nothing else related to the game, then each of those 10 players has a probability of having the winning card of $\frac{1}{10}$.
Now, once we also know that the first card is not the winner, then yes, the probability of the second player being the winner is now $\frac{1}{9}$, and the same is true for the other 8 remaining players. To be precise: If we know that 10 cards are being randomly distributed, and we know that one of those is a winner, and we know that the first player's card is not the winner, then each of the remaining nine players has a probability of having the winning card of $\frac{1}{9}$.
However, this does not mean that the game is somehow unfair if the players only reveal their cards one by one and that the game is only fair (i.e. that everyone has equal chance of being the winner) fair if they reveal their cards all at once.
It is intuitive that if all players reveal their cards at once, they all have the same probability of having the winning card of $\frac{1}{10}$. OK, fine. But if we are told that the players unfold their cards one by one, will that effect the probability of each of them having the winning card? No! They still have a probability of having the winning card of $\frac{1}{10}$ each.
For example, consider player 2: There is a $\frac{9}{10}$ probability that player 1 does not have the winning card, and we al;ready established that in that case, the probability of player 2 having the winning card is $\frac{1}{9}$. But if player 1 reveals the winning card, the probability of player 2 having the winning card beomes $0$. In total, this means that the probability of player 2 having then winning card is $\frac{9}{10} \times \frac{1}{9} + \frac{1}{10} \times 0 = \frac{1}{10}$.
Notice that what we are dealing with here are conditional probabilities. The conditional probability that player 2 wins given that player 1 does not win is $P(2|1') = \frac{1}{9}$. And the conditional probability that player 2 wins given that player 1 wins is $P(2|1) = 0$. But what we are interested in is the probability player 2 winning ... period. That is: not knowing whether player 1 is going to reveal a winning card or not, what is the probability of player 2 winning? Well, that is $P(2) = P(2|1')\times P(1') + P(2|1)\times P(1) = \frac{9}{10} \times \frac{1}{9} + \frac{1}{10} \times 0 = \frac{1}{10}$.
You can do similar calculations for any other player, but every time you will find that the unconditional probability (i.e. without knowing whether any of the previous players have a wining card or not) is $\frac{1}{10}$.
And yes, the probability of any player winning does change once we do know what cards the other players have (e.g. if we know that player 6 does not have a wining card, then the probability of player 1 having the winning card becomes $\frac{1}{9}$ ... but knowing how the players reveal their hand (all at the same time or one by one, or in groups of two, or ...) does not provide us with any such knowledge, and hence the a priori, unconditional probability remains $\frac{1}{10}$.
In sum, if we know that 10 cards are being randomly distributed, and we know that one of those is a winner, and we know that the players will all reveal their cards at the same time, and we know nothing else related to the game, then each of those 10 players has a probability of having the winning card of $\frac{1}{10}$. But it is also true that if we know that 10 cards are being randomly distributed, and we know that one of those is a winner, and we know that the players will reveal their cards one at a time, and we know nothing else related to the game, then each of those 10 players has a probability of having the winning card of $\frac{1}{10}$.
With this, we can now make sense of your scenarios:
S1: You're right, we don't know what the first player sees when peeking, and so to us, the probability for each player winning is still $\frac{1}{10}$ each.
S2: Until you get to the parenthetical remark, we don't know what Peter sees, and so the probability remains $\frac{1}{10}$. But once you tell us in the parenthetical remark that the card was not the winner, then the probability of player 2 winning becomes $\frac{1}{9}$. So: Peter looking at the cards does nothing to the probability. It is only when we know what it is that Peter sees that the probability changes. The parenthetical remark changes everything: without it, the probability woukld remain $\frac{1}{10}$ .. but with it, the probability becomes $\frac{1}{9}$.
S3: Same as above. Alice seeing the cards does not change anything. But us knowing what Alice sees does. And if know that Peter did not see a winning card, and we also know that Alice did not see a winning card, then yes, the probability of the third player winning is now $\frac{1}{8}$.
S4: Once again, a bacteria seeing the card of the first player does not change anything as far as our knowledge is concerned. It is only when we know that the bacteria was looking at a winning or non-winning card that the probability changes. And bacteria not being able to perceive the winning card as the winning card (since bacteria can;t read) is only important in terms of how our knowledge is somehow effected by that. For example, if we try to ask the bacteria what they saw we get no response, and so we will gain no knowledge... and hence the probability remains $\frac{1}{10}$ for any of the other players.
