Probability Problem, Quick question...

[Minotar]

Ok, here is the quick question for any who remember anything from Problem Stats... Two people are playing a game that has 8 possible "hands" (1 through 8), and hand 8 is the best "hand" that can be achieved, and "hand" 1 is the worst... If each of these 8 "hands" have a probability to come up, how would you find the probability that Player 1 will get the better hand than Player 2? How about vice versa (P2 > P1)? Does anyone have any idea how to approach this problem? Thanks for any help!

 

[interchange]

(P2-8 + P2-7 + ... + P2-2) * (P1-1) +
(P2-8 + P2-7 + ... + P2-3) * (P1-2) +
.... +
(P2-8) * (P1-7)

I don't see why that wouldn't work.

 

[DrPizza]

Maybe I'm misunderstanding your problem. But, wouldn't the probability simply be 50/50?
Deal the hand to players A and B, face down. Before either of them looks at their cards, have them switch seats. It makes no difference.

 

[Minotar]

Originally posted by: DrPizza
Maybe I'm misunderstanding your problem. But, wouldn't the probability simply be 50/50?
Deal the hand to players A and B, face down. Before either of them looks at their cards, have them switch seats. It makes no difference.

That would work if it was a simple, clear-cut case, but it won't work here... With this problem, assume that the probabilities for each hand, for each player are different. In other words, assume you have already calculated the probability of coming up with a certain hand, and it is different for each player... Approaching the problem is very different in this case. It should be close to 50/50, but not quite.

 

[Minotar]

Originally posted by: interchange
(P2-8 + P2-7 + ... + P2-2) * (P1-1) +
(P2-8 + P2-7 + ... + P2-3) * (P1-2) +
.... +
(P2-8) * (P1-7)

I don't see why that wouldn't work.

Hmmm...This seems interesting... Just wondering, is there any theorem or reasoning you used to come up with this?

 

[Bobthelost]

Odds of Player 1 losing with a known hand = Probability of Player 2 getting a higher score.

Thus if Player 1 has a hand "X" and Player 2 has a hand "Y" the odds are as follows.

P8 + P7 +P6 + .... P(X+1) Or in other words the sum of all the chances that player 2 gets a hand of greater value. So if you know the odds of player 2 getting each hand then you can work out the odds of winning if you know player 1's hand.

 

[Minotar]

Originally posted by: Bobthelost
Odds of Player 1 losing with a known hand = Probability of Player 2 getting a higher score.

Thus if Player 1 has a hand "X" and Player 2 has a hand "Y" the odds are as follows.

P8 + P7 +P6 + .... P(X+1) Or in other words the sum of all the chances that player 2 gets a hand of greater value. So if you know the odds of player 2 getting each hand then you can work out the odds of winning if you know player 1's hand.

This seems similar to interchanges idea, but the only problem with what you have listed, is that the probs don't add up to "1". The probs for each hand are actually very low, except hand 1, in which P(Player one gets hand 1)=1 and P(Player two gets hand 1)=1. All other Probs are very low. I can post the table if everyone wishes to see it. Also, what about the case of a tie, i.e. NO player wins because both have the same hand? Ugggg...This is confusing

 

[Bobthelost]

Originally posted by: Minotar

Originally posted by: Bobthelost
Odds of Player 1 losing with a known hand = Probability of Player 2 getting a higher score.

Thus if Player 1 has a hand "X" and Player 2 has a hand "Y" the odds are as follows.

P8 + P7 +P6 + .... P(X+1) Or in other words the sum of all the chances that player 2 gets a hand of greater value. So if you know the odds of player 2 getting each hand then you can work out the odds of winning if you know player 1's hand.

This seems similar to interchanges idea, but the only problem with what you have listed, is that the probs don't add up to "1". The probs for each hand are actually very low, except hand 1, in which P(Player one gets hand 1)=1 and P(Player two gets hand 1)=1. All other Probs are very low. I can post the table if everyone wishes to see it. Also, what about the case of a tie, i.e. NO player wins because both have the same hand? Ugggg...This is confusing

You're not doing a great job of explaining this one.

What do you want to be able to do exactly, and if you could stick the table up so we'd have some real numbers to walk through as an example...

 

[Minotar]

_______________Player1_______Player 2
Straight Flush_______0____________0
4 of a kind_______1/741_________1/741
Full House________5/741_________5/741
Flush___________15/247__________0
Straight___________0___________7/741
3 of a kind_______43/741________43/741
2 pair__________170/741_______170/741
1 pair_____________1____________1

EDIT: I had to use underscores to separate the columns!

These are the probabilities for each type of hand. By defult, the worse hand a player can have is 1 pair. What is the P(Player 1 wins) and what is the P(Player 2 wins) and what is the P(both tie)? I hope this clarifies what I am looking for. I am just not sure how to do this final calculation.... Thanks for any help!

 

[Bobthelost]

I see, in which case you want to work out a realistic stats:

Hand 1 Player 1 Probablity: 477/741 (1 - All the other possiblities)
Hand 2 Player 1 Probablity: 170/741
Hand 3 Player 1 Probablity: 43/741
Hand 4 Player 1 Probablity: 0/741
Hand 5 Player 1 Probablity: 45/741
Hand 6 Player 1 Probablity: 5/741
Hand 7 Player 1 Probablity: 1/741

Then you should be able to work it out.

 

[Minotar]

Originally posted by: Bobthelost
I see, in which case you want to work out a realistic stats:

Hand 1 Player 1 Probablity: 477/741 (1 - All the other possiblities)
Hand 2 Player 1 Probablity: 170/741
Hand 3 Player 1 Probablity: 43/741
Hand 4 Player 1 Probablity: 0/741
Hand 5 Player 1 Probablity: 45/741
Hand 6 Player 1 Probablity: 5/741
Hand 7 Player 1 Probablity: 1/741

Then you should be able to work it out.

I don't follow what you are saying here???? Well, I know what you did, but I don't know where to go from there! I know there must be a theorem or rule that lets you compute the prob from my table, but I am not sure what it would be... To me, my data looks more like a distribution. I need some help, if anyone has any idea

 

[Matthias99]

Originally posted by: Minotar

Originally posted by: Bobthelost
I see, in which case you want to work out a realistic stats:

Hand 1 Player 1 Probablity: 477/741 (1 - All the other possiblities)
Hand 2 Player 1 Probablity: 170/741
Hand 3 Player 1 Probablity: 43/741
Hand 4 Player 1 Probablity: 0/741
Hand 5 Player 1 Probablity: 45/741
Hand 6 Player 1 Probablity: 5/741
Hand 7 Player 1 Probablity: 1/741

Then you should be able to work it out.

I don't follow what you are saying here???? Well, I know what you did, but I don't know where to go from there! I know there must be a theorem or rule that lets you compute the prob from my table, but I am not sure what it would be... To me, my data looks more like a distribution. I need some help, if anyone has any idea

It's not hard to work it out mechanically, but it will take a little while. What you would do is: for each possible hand for Player 1, figure out the probability of a win/loss/draw (which you can do, since you know the frequency distribution for Player 2's hands). Then you can use those probabilities, combined with the probability of getting each hand, to get the overall win/loss/draw figure. This is a zero-sum game, so Player 2 will win when Player 1 loses and vice versa (IOW, you don't need to work it out twice).

I don't want to give you too much more, since this sounds like homework.

 

[Minotar]

I still am not sure how to proceed:/ And, no, this is not homework. It is a project that I have been working on for some time.

 

[DrPizza]

I understand what you mean now.

Easy way to proceed. (lets assume ties are possible?)
there are 8*8 possible combinations of hands.
Figure out how many of those 64 times player 1 will win.

_______________Player1_______Player 2
Straight Flush_______0____________0
4 of a kind_______1/741_________1/741
Full House________5/741_________5/741
Flush___________15/247__________0
Straight___________0___________7/741
3 of a kind_______43/741________43/741
2 pair__________170/741_______170/741
1 pair_____________1____________1

So, with a pair, player 1 can't win.
with 2 pair, he can win 1 way (player 2 has 1 pair.)
with 3 of a kind, he can win 2 ways
with a straight, 3 ways,
... 1 + 2 + 3 + 4 + 5 + 6 + 7 ways out of the 64 to win.
Find the probability of each of these occuring, add them together.
(Sorry, it seems like a pita to do so)

Also, shouldn't each person's probabilities sum to 741/741? And, what do you mean by the probability of a pair is 1? Did you mean that it's the complement of the sum of the other probabilities?

 

[Matthias99]

Originally posted by: Minotar
I still am not sure how to proceed:/ And, no, this is not homework. It is a project that I have been working on for some time.

For instance, assume that Player 1 has, say, three-of-a-kind.

There is a (1/741 + 5/741 + 7/741) chance of Player 2 having a better hand, a (43/741) chance of Player 2 having the same hand, and a (685/741) chance of Player 2 having a worse hand. If you do this for each possible hand that Player 1 can get, you can then work out the overall win-loss-tie probabilities based on how frequently each hand will come up.

 

[Minotar]

Originally posted by: DrPizza
I understand what you mean now.

Easy way to proceed. (lets assume ties are possible?)
there are 8*8 possible combinations of hands.
Figure out how many of those 64 times player 1 will win.

_______________Player1_______Player 2
Straight Flush_______0____________0
4 of a kind_______1/741_________1/741
Full House________5/741_________5/741
Flush___________15/247__________0
Straight___________0___________7/741
3 of a kind_______43/741________43/741
2 pair__________170/741_______170/741
1 pair_____________1____________1

So, with a pair, player 1 can't win.
with 2 pair, he can win 1 way (player 2 has 1 pair.)
with 3 of a kind, he can win 2 ways
with a straight, 3 ways,
... 1 + 2 + 3 + 4 + 5 + 6 + 7 ways out of the 64 to win.
Find the probability of each of these occuring, add them together.
(Sorry, it seems like a pita to do so)

Also, shouldn't each person's probabilities sum to 741/741? And, what do you mean by the probability of a pair is 1? Did you mean that it's the complement of the sum of the other probabilities?

Thanks for the helpful thoughts, Doc! This might just be it?! Just to tell you, the probs will not add up to 1, though. They will only add up to 1 if you include all of the non useful card combinations as well (which we are not interested in). The reason the P(1 pair)=1 is kinda hard to explain, but this is why I didn't want to explain the actual game being played here. The card game scenario is actually very comlicated, and I just can't explain this scenario... Too much! LOL I will try this thought... any other ideas?

 

[eLiu]

It should be 50/50 by symmetry. Look at it this way:
I can give player A hand 1, and player B hand 8.
I could also give player A hand 8 and B hand 1.

^^Those 2 events have equal probability.

Thus for any combination of hands like (2,4) or (8,3) or whatever, you can turn the tables and hence change the winner.

 

[Minotar]

Originally posted by: eLiu
It should be 50/50 by symmetry. Look at it this way:
I can give player A hand 1, and player B hand 8.
I could also give player A hand 8 and B hand 1.

^^Those 2 events have equal probability.

Thus for any combination of hands like (2,4) or (8,3) or whatever, you can turn the tables and hence change the winner.

No, this will not work. You cannot think of it that way at all. The reason is because the probabilities are NOT equal. Please view the chart. If they are not equal then you cannot possibly say that there is a 50/50 chance of winning.