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 08-19-2012, 08:32 PM #1 journeyman   Join Date: Jun 2011 Posts: 210 Help with expected winnings I'm trying to include 'jackpot' prizes in my expected winnings calculations. I play 6-man tournaments. I finish in the top two roughly 40% of the time, (2 games out of 5.) I win a jackpot prize for 6 top two finishes in a row. (Every time this happens, the sequence is reset to zero though, so 7 top twos in a row wouldn't count as two separate jackpot wins for example.) How often should I expect to achieve this jackpot given the figures above? Here was my attempt at calculating the answer but I'm not sure if this is correct: I finish in the top 2 every 2.5 games, but I don't think the answer is as simple as 2.5 to the power of 6= 244, therefore every 244 games. That's not right, is it? I think it might be every 244 new 'attempts,' ie a new attempt starts every time a sequence of top twos has ended. I reckon it's roughly once every 405/410 games ish, but could someone verify this for me please. Thanks for your help.
 08-19-2012, 08:44 PM #2 Carpal \'Tunnel     Join Date: Jun 2005 Location: Psychology Department Posts: 7,430 Re: Help with expected winnings I'm not sure if this is going to answer your question or not, but the probability of you making the jackpot on your next 6 games (i.e, your counter is at 0) is .4^6 ~= .004, or about 4 times in 1000 such situations.
08-19-2012, 08:48 PM   #3
journeyman

Join Date: Jun 2011
Posts: 210
Re: Help with expected winnings

Quote:
 Originally Posted by Sherman I'm not sure if this is going to answer your question or not, but the probability of you making the jackpot on your next 6 games (i.e, your counter is at 0) is .4^6 ~= .004, or about 4 times in 1000 such situations.
That'll be the 1 in 244 figure I quoted, but I'm asking if that means every 244 games or every 244 new attempts.

08-19-2012, 10:30 PM   #4
Carpal \'Tunnel

Join Date: Jun 2005
Location: Psychology Department
Posts: 7,430
Re: Help with expected winnings

Quote:
 Originally Posted by Pocoyo That'll be the 1 in 244 figure I quoted, but I'm asking if that means every 244 games or every 244 new attempts.
Every 244 new attempts.

08-20-2012, 02:18 AM   #5
Carpal \'Tunnel

Join Date: Sep 2002
Posts: 8,907
Re: Help with expected winnings

Quote:
 Originally Posted by Pocoyo I'm trying to include 'jackpot' prizes in my expected winnings calculations. I play 6-man tournaments. I finish in the top two roughly 40% of the time, (2 games out of 5.) I win a jackpot prize for 6 top two finishes in a row. (Every time this happens, the sequence is reset to zero though, so 7 top twos in a row wouldn't count as two separate jackpot wins for example.) How often should I expect to achieve this jackpot given the figures above? Here was my attempt at calculating the answer but I'm not sure if this is correct: I finish in the top 2 every 2.5 games, but I don't think the answer is as simple as 2.5 to the power of 6= 244, therefore every 244 games. That's not right, is it? I think it might be every 244 new 'attempts,' ie a new attempt starts every time a sequence of top twos has ended. I reckon it's roughly once every 405/410 games ish, but could someone verify this for me please.
It's 1 in 244.14 attempts. Multiply that by the average number of games per attempt to get the average number of games. The average number of games per attempt is computed with the partial geometric series:

1 + 0.4 + 0.42 + 0.43 + 0.44 + 0.45

= (1 - 0.46) / (1 - 0.4)

= 1.65984 games.

That is, it always takes at least 1, with probability 0.4 it takes at least 1 more, with probability 0.42 it takes at least 1 more, etc. If you win 5 in a row, the next game completes the series win or lose.

Multiplying 1.65984 games/attempt by 244.14 attempts gives an average of once in 405.2 games as you said.

08-20-2012, 09:55 AM   #6
journeyman

Join Date: Jun 2011
Posts: 210
Re: Help with expected winnings

Quote:
 Originally Posted by BruceZ It's 1 in 244.14 attempts. Multiply that by the average number of games per attempt to get the average number of games. The average number of games per attempt is computed with the partial geometric series: 1 + 0.4 + 0.42 + 0.43 + 0.44 + 0.45 = (1 - 0.46) / (1 - 0.4) = 1.65984 games. That is, it always takes at least 1, with probability 0.4 it takes at least 1 more, with probability 0.42 it takes at least 1 more, etc. If you win 5 in a row, the next game completes the series win or lose. Multiplying 1.65984 games/attempt by 244.14 attempts gives an average of once in 405.2 games as you said.

Thanks BruceZ.

I think I worked it out slightly differently to arrive at the same answer of 405, but maybe my 4 years studying maths at university weren't as much of a waste of time as I thought they were.

08-20-2012, 10:06 AM   #7
Carpal \'Tunnel

Join Date: Sep 2002
Posts: 8,907
Re: Help with expected winnings

Quote:
 Originally Posted by Pocoyo I think I worked it out slightly differently to arrive at the same answer of 405
You mean like this?

1*0.6 +
2*0.4*0.6 +
3*0.4^2*0.6 +
4*0.4^3*0.6 +
5*0.4^4*0.6 +
6*0.4^5*0.6 +
6*0.4^6

= 1.65984 games/attempt

That's equivalent, just more work.

08-20-2012, 11:44 AM   #8
enthusiast

Join Date: Jun 2011
Posts: 78
Re: Help with expected winnings

Quote:
 Originally Posted by BruceZ = (1 - 0.46) / (1 - 0.4) = 1.65984 games. That is, it always takes at least 1, with probability 0.4 it takes at least 1 more, with probability 0.42 it takes at least 1 more, etc. If you win 5 in a row, the next game completes the series win or lose. Multiplying 1.65984 games/attempt by 244.14 attempts gives an average of once in 405.2 games as you said.
That makes the formula:
N = (1 - 0.4^6) / (1 - 0.4) * (1 / 0.4^6)
N = (1 - P^run) / (1 - P) * (1/P^run)

which is equivalent to
N = (1 / P^run - 1) / Q

Number of Attempts to Get n Successes in a Row

The median looks to be, from the distribution, 282

Last edited by sallymustang; 08-20-2012 at 12:06 PM. Reason: typo

08-20-2012, 01:46 PM   #9
enthusiast

Join Date: Jun 2011
Posts: 78
Re: Help with expected winnings

Quote:
 Originally Posted by sallymustang The median looks to be, from the distribution, 282
I like to add the median value in these type of questions as most just think that the average, median and the mode are about the same. We were taught this in school and seem to think it applies to all distributions.
It is far from the truth when we talk about runs or streak where the mode is the number of trials equal to the run length and the median is about 69% of the average.

Bruce likes these kind of Qs.
In the link below he estimates the median.

for this particular problem we can estimate the answer

My estimation for the median is
log0.5 / log (1-(1/average number of trials))

example:
average number of trials = for run of 6 or more
=(1/(P^run))*(1/(Q)) = 406.9010417

log0.5 / log (1-(1/406.9010417))=281.70

My BF came up with this, he says it should be a very good approximation, and actually used calculus for an exact answer that I do not understand
so I wonder how accurate this is at different values of P and run
I know longer runs need an extra trial added.
I need more time to work with this concept

Sally

Last edited by sallymustang; 08-20-2012 at 01:48 PM. Reason: quote

 10-16-2012, 11:11 AM #10 journeyman   Join Date: Jun 2011 Posts: 210 Re: Help with expected winnings Now a slightly different angle on the original question; Given the figures used in the original post, after how many games WITHOUT six consecutive top two finishes, would you start to be concerned that something may be wrong?
 10-16-2012, 11:38 AM #11 Carpal \'Tunnel     Join Date: Jun 2005 Location: Psychology Department Posts: 7,430 Re: Help with expected winnings This is given by the binomial distribution. In excel type: =binomdist(0,10,.4,FALSE) The result is the probability of winning 0 games out of 10 where you have a .4 chance of winning each game. This gives an answer of .006047, or about 6 times in 1000 sets of 10 games. You can replace the 10 with however many trials (games) you want. At what point you decide a) You suck or b) You are being cheated is up to you. Many people use a conventional level of .05, but if you are already suspicious of data you have gathered it might not be appropriate here. Better to collect a new sample. Further, a level of .05 means that 5% of the time you will "think something is wrong" even when it isn't. Further still, if you conduct the test on multiple sets of 10 games, the probability of "thinking something is wrong when it isn't" increases by 5% each time (if you use the .05 cutoff). Personally, if I missed 10 games in a row I would think it is time for a break and come back later. If I am still not successful later it is probably best just to quit the game because either I suck, the games are too tough, or I am being cheated. Either way, that is not the kind of game I am looking for.
10-16-2012, 02:17 PM   #12
Carpal \'Tunnel

Join Date: Sep 2002
Posts: 8,907
Re: Help with expected winnings

Quote:
 Originally Posted by Sherman This is given by the binomial distribution. In excel type: =binomdist(0,10,.4,FALSE) The result is the probability of winning 0 games out of 10 where you have a .4 chance of winning each game. This gives an answer of .006047, or about 6 times in 1000 sets of 10 games. You can replace the 10 with however many trials (games) you want.
ARRRGGHHH!!! You don't use a binomial distribution. You use a streak calculator. We don't want to know that a streak of 10 happens an average of 6 times in 1000 independent sets of 10 games. We want to know that it happens an average of about 24 times in 10,000 games. A streak of 6 will happen an average of over 196 times in 10,000 games. You can use the streak calculator available here. The bottom output box tells you how often to expect a streak.

I was reading about this guy Wittgenstein who massively influenced many diverse fields including psychology and the philosophy of mathematics. He used to beat children when they made math errors. Just thought I'd throw that out there. LOL

 10-16-2012, 03:44 PM #13 Carpal \'Tunnel     Join Date: Jun 2005 Location: Psychology Department Posts: 7,430 Re: Help with expected winnings Actually, re-reading the question I was answering a different question. I missed the part about "without six consecutive." It is pretty obvious that a streak calculator is needed for that type of question. I thought he was asking how rare it is to finish outside the top 2 in all of X games (i.e. what is the probability of 0 successes in X games where X changes). Which is why I answered the question with the binomial. You can correct me if I am wrong Bruce, but I believe the binomial does answer the question that I thought he was asking. In any case, my answer above is also given by p(success)^X where X is the number of games. So I'm pretty sure I have the right answer...just to the wrong question. ;-)
 10-16-2012, 03:53 PM #14 Carpal \'Tunnel     Join Date: Sep 2002 Posts: 8,907 Re: Help with expected winnings Either p(lose)^X or the binomial gives the probability of losing all X games. But multiplying that by 1000 just gives you the average number of times you lose all X games in 1000 X game matches, not how many times you lose X games in a row in 1000*X games. Nobody wants the first number because the 1000*X games are always continuous not individual matches.
10-16-2012, 03:55 PM   #15
Carpal \'Tunnel

Join Date: Jun 2005
Location: Psychology Department
Posts: 7,430
Re: Help with expected winnings

Quote:
 Originally Posted by BruceZ Either p(lose)^X or the binomial gives the probability of losing all X games. But multiplying that by 1000 just gives you the average number of times you lose all X games in 1000 X game matches, not how many times you lose X games in a row in 1000*X games.
Yeah, I wasn't recommending multiplying the resulting probability times 1000. I was just saying that if you do that sort of test multiple times the probability of finding a result below a given probability value is the given probability value * the number of tests (assuming the tests are independent).

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