Probability of higher or lower?!?
I want to know how to work out the odds of the following game using normal ace to king playing cards (ace is high).
1) the first card is revealed
2) the player must guess whether the next card is higher or lower
3) if he is correct he can continue
4) if he is wrong the game ends.
He must get to the end of all the cards before he can win any prize. (for example 15 cards)
What are the odds? If we knew what each card was going to be the calculation would be easier, but since each card is random, is the probability of getting it right 50%?
Answers:
What do you do with a tie? For example a Jack showing and another Jack is next. Also, the odds depend on the player "playing perfectly." Do they always make the mathematically correct choice? To do this they would have to be keeping track of which cards had already appeared. If so. it isn't much of a game.
Aloha
YOU tell us and your teacher
Depends what the card is, if it is something like a queen then you could guess lower and you've got a good chance of being right. If its a 7 then it is 50%.
Once the first card is known (eg a Q) then the maths is:
(Cards lower than a Q) x 4 suits DIVIDED BY (Cards higher than a Q) x 4 suits.
thereafter, its the same maths, except that this time you have to minus the cards already in play.
I'd say odds get better after each try, as there would be less cards and you would know what previous ones were.
As you don't know which cards have been selected, you have to work on the basis that the chances are the same as if you had the whole pack.
The 8 is the middle card. If the last card shown is less than 8, you should guess higher, if it is higher you should guess lower. If it is 8, take your pick.
The chances of getting 1 guess right is not 50%. If a 2 is showing, the chances of a greater guess being correct is more than 50%, it is 12/13 on the first run. On average, if you play the game correctly, your chances of being right at each guess is more than 50%, it is more like 75%(roughly). However, these probabilities have to be multiplied together to get the probability of being correct every time with more cards.
1 75%
2 56%
3 42%
4 32%
5 24%
6 18%
7 13%
8 10%
9 8%
10 6%
11 4%
12 3%
13 2%
14 1%
15 1%
I suspect a lot of games finish quite quickly, but this does seem to make the probability of winning rather lower than I expected.
short answer is yes, its about 50%, but if you do the math its not, for example if there is 52 cards in a pack shuffled randomly and the first card revealed is high then the next card has a 26/51 chance of being low, which is actually 51% chance of getting a low card. this kind of thing continues, if the next card that is taken is high again, then the chance of the next card being low is even greater now 26/50 = 52% however if your 2nd card you pick is low, then the chances are balanced, if you remember all the cards that have gone and try to and always try to balance the probabilties, e.g if you have had 10 highs and 9 lows it is slightly more likely that the next card is low, if you correctly get to the end then you can be always sure what the last card is. however this is only increasing his chances to very slightly over 50%, thats it, it by no way gaurantees you to win sorry. the most important thing for you to do is to decide what constituetes as a low and high card, and stick to it unless what you remember from what has already gone changes the chances enough, wnhat you remember to have gone is unlike to effect the game unless, what has already gone is unbalanced e.g more highs than lows, but then you must decide by how much what has already gone should weight your desicion e.g is a 8 now a high card or a low card while before it was in the dead centre what you know from what has gone may mean that you would count 8 as high then predicting the next card would be low. so its not quite as simple as it first looks, but to be honest its 95% luck and 5% not being stupid. thanks for listening,
The odds of getting the first one right won't be 50% because of the possible of the first to cards being equal. But it should be
(1 - 3/51 ) / 2 = 47%.
The probability of get the next question right is much more difficult to work out. Firstly will the player follow a random selection of highs and lows or will he follow a certain strategy? Either way there isn't a simple formula to calculate the probability, you need to work out the probability for each possible scenario individually taking into account duplicate cards (ie. 1,2,3 of four occurrences of the same card).
A much simpler way would be to write a computer program to run through all possible combinations (or a random sample) and use that to calculate the probability
To clarify your question. the probability without knowing the card is not necessarily 50%. In fact it should be higher. To actually calculate it without much work assuming a tie is not a loss, find the following sum. (which is
(53/53 + 49/53 + 45/53 + 41/53 + 37/53 + 33/53 + 29/53 + 33/53 + 37/53 + 41/53 + 45/53 + 49/53 + 53/53) / 13
= 0.7910015 approximatly
This is the probability using the optimal strategy given a tie is a win of winning on the first card. It is 1/2 only if you GUESS RANDOMLY.
If you guess randomly then the probability of winning is always 1/2. Then you can simply find the probabiliyt of winning the game as
1 - (1/2)^(n-1) ([n-1] since the first card is determined you only play n-1 times)
HOWEVER
I suggest simulation. essentially if you had a fair 52 card deck and used n of them, then the probability would only depend on n.
You could get a very accurate answer with little real work. You could make a table that gave you the probability for values of n.
IF you assume that a tie is a loss. then the guessing probability is smaller than 1/2 (unless all ties are out already i.e., all remaining card numbers are unique)
EDIT Reading up on the game from wikipedia a tie is a loss.
We can still find the first probability with some modification
(53/53 + 49/53 + 45/53 + 41/53 + 37/53 + 33/53 + 29/53 + 33/53 + 37/53 + 41/53 + 45/53 + 49/53 + 53/53) / 13 - 4/53
= 0.7155298 (only slightly lower)
The game was played with 5 cards.. not 15. and since im interested i will write code now to calculate the proability.
EDIT2 i just realized the optimal solution requires more coding than i want to do :P but it is possible..
do you get anything for a pair? and can i be a dolly dealer?
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1) the first card is revealed
2) the player must guess whether the next card is higher or lower
3) if he is correct he can continue
4) if he is wrong the game ends.
He must get to the end of all the cards before he can win any prize. (for example 15 cards)
What are the odds? If we knew what each card was going to be the calculation would be easier, but since each card is random, is the probability of getting it right 50%?
Answers:
What do you do with a tie? For example a Jack showing and another Jack is next. Also, the odds depend on the player "playing perfectly." Do they always make the mathematically correct choice? To do this they would have to be keeping track of which cards had already appeared. If so. it isn't much of a game.
Aloha
YOU tell us and your teacher
Depends what the card is, if it is something like a queen then you could guess lower and you've got a good chance of being right. If its a 7 then it is 50%.
Once the first card is known (eg a Q) then the maths is:
(Cards lower than a Q) x 4 suits DIVIDED BY (Cards higher than a Q) x 4 suits.
thereafter, its the same maths, except that this time you have to minus the cards already in play.
I'd say odds get better after each try, as there would be less cards and you would know what previous ones were.
As you don't know which cards have been selected, you have to work on the basis that the chances are the same as if you had the whole pack.
The 8 is the middle card. If the last card shown is less than 8, you should guess higher, if it is higher you should guess lower. If it is 8, take your pick.
The chances of getting 1 guess right is not 50%. If a 2 is showing, the chances of a greater guess being correct is more than 50%, it is 12/13 on the first run. On average, if you play the game correctly, your chances of being right at each guess is more than 50%, it is more like 75%(roughly). However, these probabilities have to be multiplied together to get the probability of being correct every time with more cards.
1 75%
2 56%
3 42%
4 32%
5 24%
6 18%
7 13%
8 10%
9 8%
10 6%
11 4%
12 3%
13 2%
14 1%
15 1%
I suspect a lot of games finish quite quickly, but this does seem to make the probability of winning rather lower than I expected.
short answer is yes, its about 50%, but if you do the math its not, for example if there is 52 cards in a pack shuffled randomly and the first card revealed is high then the next card has a 26/51 chance of being low, which is actually 51% chance of getting a low card. this kind of thing continues, if the next card that is taken is high again, then the chance of the next card being low is even greater now 26/50 = 52% however if your 2nd card you pick is low, then the chances are balanced, if you remember all the cards that have gone and try to and always try to balance the probabilties, e.g if you have had 10 highs and 9 lows it is slightly more likely that the next card is low, if you correctly get to the end then you can be always sure what the last card is. however this is only increasing his chances to very slightly over 50%, thats it, it by no way gaurantees you to win sorry. the most important thing for you to do is to decide what constituetes as a low and high card, and stick to it unless what you remember from what has already gone changes the chances enough, wnhat you remember to have gone is unlike to effect the game unless, what has already gone is unbalanced e.g more highs than lows, but then you must decide by how much what has already gone should weight your desicion e.g is a 8 now a high card or a low card while before it was in the dead centre what you know from what has gone may mean that you would count 8 as high then predicting the next card would be low. so its not quite as simple as it first looks, but to be honest its 95% luck and 5% not being stupid. thanks for listening,
The odds of getting the first one right won't be 50% because of the possible of the first to cards being equal. But it should be
(1 - 3/51 ) / 2 = 47%.
The probability of get the next question right is much more difficult to work out. Firstly will the player follow a random selection of highs and lows or will he follow a certain strategy? Either way there isn't a simple formula to calculate the probability, you need to work out the probability for each possible scenario individually taking into account duplicate cards (ie. 1,2,3 of four occurrences of the same card).
A much simpler way would be to write a computer program to run through all possible combinations (or a random sample) and use that to calculate the probability
To clarify your question. the probability without knowing the card is not necessarily 50%. In fact it should be higher. To actually calculate it without much work assuming a tie is not a loss, find the following sum. (which is
(53/53 + 49/53 + 45/53 + 41/53 + 37/53 + 33/53 + 29/53 + 33/53 + 37/53 + 41/53 + 45/53 + 49/53 + 53/53) / 13
= 0.7910015 approximatly
This is the probability using the optimal strategy given a tie is a win of winning on the first card. It is 1/2 only if you GUESS RANDOMLY.
If you guess randomly then the probability of winning is always 1/2. Then you can simply find the probabiliyt of winning the game as
1 - (1/2)^(n-1) ([n-1] since the first card is determined you only play n-1 times)
HOWEVER
I suggest simulation. essentially if you had a fair 52 card deck and used n of them, then the probability would only depend on n.
You could get a very accurate answer with little real work. You could make a table that gave you the probability for values of n.
IF you assume that a tie is a loss. then the guessing probability is smaller than 1/2 (unless all ties are out already i.e., all remaining card numbers are unique)
EDIT Reading up on the game from wikipedia a tie is a loss.
We can still find the first probability with some modification
(53/53 + 49/53 + 45/53 + 41/53 + 37/53 + 33/53 + 29/53 + 33/53 + 37/53 + 41/53 + 45/53 + 49/53 + 53/53) / 13 - 4/53
= 0.7155298 (only slightly lower)
The game was played with 5 cards.. not 15. and since im interested i will write code now to calculate the proability.
EDIT2 i just realized the optimal solution requires more coding than i want to do :P but it is possible..
do you get anything for a pair? and can i be a dolly dealer?
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