Sweepstakes probability in which one had to click 3 icons, which would reveal 3 pics

Mattokun

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Jul 20, 2017
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Hello,
I’m new to this forum, I hope someone can give some inputs.
I recently played a sweepstakes in which one had to click 3 icons, which would reveal 3 pictures, either Pic1 or Pic2. If the 3 pics matched, one wins.
Of course, this was such that always, no matter which order clicked, the first 2 icons matched, but the third one did not.
Since this is in reality very unlikely (and, I feel, this is actually a malfeasance), I would like to know what the probability of this happening is.
It would seem to me that it is somehow related to conditional probability (or rather, lack thereof), since the outcome of one icon should be independent of any other icon previously unveiled.
The other thought would be, to take the overall outcome together and determine the likelyhood of its outcome, but that would ignore the "it’s always the last one that misses".
Anyone can share any insights?
Thank you.
 
You'll need an actual number of possible pictures for starters.

How sure are you that Picture#3 is static?
It would be very simple to code the game so that ALL pictures are the same until you have seen Picture #2.
 
I think I have to maybe clarify this a bit more:
There are 3 “doors”, if you will, that you can click on, in any order.
Each “door” has 2 possible outcomes, a “win” or a “dud”.
If all 3 “doors” are clicked, and all 3 show a “win”, then it’s an overall “win”. Else, it’s a “loss”.
The situation, and what I feel like a cheat, is that the first 2 “doors” clicked are ALWAYS “wins”, no matter which ones are clicked in what order.
I guess I call this the “bait”.
Then, of course the third “doors” clicked is ALWAYS a “dud”, no matter which (remaining) “door” that is.
This gives then of course the “oh, you were sooo close!!” message.
My inquiry here is that, I believe it cannot be that, if those “doors” were truly assigned with random outcomes “win” or “dud”, that the outcome described above would be very likely.
In other words, I think they cheat, and their trick is to make people think that it is all random and people are just not lucky enough to get the price.
So, what I want to know is, how can I calculate the probability of the outcome described above, or maybe more clearly, how can I prove that this game is cheating?
 
I think I have to maybe clarify this a bit more:
There are 3 “doors”, if you will, that you can click on, in any order.
Each “door” has 2 possible outcomes, a “win” or a “dud”.
If all 3 “doors” are clicked, and all 3 show a “win”, then it’s an overall “win”. Else, it’s a “loss”.
The situation, and what I feel like a cheat, is that the first 2 “doors” clicked are ALWAYS “wins”, no matter which ones are clicked in what order.
I guess I call this the “bait”.
Then, of course the third “doors” clicked is ALWAYS a “dud”, no matter which (remaining) “door” that is.
If there is indeed no possible chance of winning, then obviously this "game" is a scam. ;)
 
So, what I want to know is, how can I calculate the probability of the outcome described above, or maybe more clearly, how can I prove that this game is cheating?

With only the information available to you as an end user, such a calculation is likely impossible. As you mentioned, you can't ever be certain if the game is rigged or if you're just extremely unlucky. If this game is on a website or something, it might well be possible to examine the code and thus determine for sure if the game's rigged or not.
 
We need to know what odds are offered, that is the value of the prize (for three "wins") compared to the stake.
 
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