Why do you have a client for such silliness?
p(first card is good) = 5/6
p(second card is good) = 4/5
p(third card is good) = 3/4
etc.
Hi,
Lately I have been facing quite complex (complex for me) calculation that my client is expecting me to provide the most accurate results.
Basically, my problem is defining probability result for one of the simple computer game I do.
There are 6 cards with numbers 1 - 6 facing down. Only one card is randomly selected by computer.
My task is to reveal all the 5 cards but not the one chosen by device.
My question is, what is the probability of choosing correctly 5 cards?
I have been banging my head against the brick wall to get the right results and however I calculate I am really not sure of being correct.
I believe the answer will be very simple.
I would really be grateful for your help.
Thanks a lot and have a good day.
Really thanks for reply.
I didn't want to go in details so much.
I am working for company making casino games.
The example above is very simplified, but the concept is the same.
I actually thought the same way but the entire idea is to give probability for occurrence of 5 correct reveals only.
What is the average number of trials, player must commit to reveal 5 cards correctly ?
That shown above is each draw considered separately.
Standard defininition.
P(first card is bad) = 1/6
If the first card is bad, there is no second card.
P(2nd card is bad) = (5/6)*(1/5)
P(3rd card is bad) = (5/6)*(4/5)*(1/4)
p(4th card is bad) = (5/6)*(4/5)*(3/4)*(1/3)
p(5th card is bad) = (5/6)*(4/5)*(3/4)*(2/3)*(1/2)
p(6th card is bad) = (5/6)*(4/5)*(3/4)*(2/3)*(1/2)*(1)
This wonderfully complicated list simplifies nicely to a uniform distribution. The probability that a bad draw will occur on any numbered draw is 1/6. This can be seen by slightly redesigning the experiment. Just throw six cards into the air. Line them up in a logical sequence after they land. The probability that the bad card will land in any of the six possible positions is the same, 1/6. Expected number on which bad card falls: 21/6 = 3½
Hello, plastic!
There are 6 cards with numbers 1 - 6 facing down.
Only one card is randomly selected by computer.
My task is to reveal all the 5 cards but not the one chosen by device.
My question is, what is the probability of choosing correctly 5 cards?
First, guess which card is the undesirable one.
. . Then reveal the other five cards.
What is the probability that you guessed correctly?
Ah, Captain Black said this first . . .
.
It may not be quite the same as the correct answer we have so far, since we've spoken only of the initial distribution. The rules may change a bit if the payoff or bid increases or decreases between cards. If you can avoid the temptation to switch after your initial selection, that would be fine. If the dealer knows the bad card and raises the stakes to entice you to switch, that's a different problem.