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# Probability Improvement in Sharing Tournament Prize

Posted on 2012-09-05
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My son is going to a poker tournament which pays out equal amounts to the top 10%.

Asssuming that all the players have equal ability, then the odds of my son winning is 1 in 10. My son is trying to help decide whether he and two friends should share evenly their prizes. If they do so, how do the odds improve that they will all take home a prize?

Tournament begins in 3 hours so please answer quickly; explaining your results will help me understand the results better.

Do the number of entries in the tournament affect the results, whether it be 100 or 1000 players?

Thanks,
Paul
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Question by:phoffric
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LVL 27

Expert Comment

ID: 38369834
The odds are three times higher to win something.
If he is the only one of the group to win, that something will be three times smaller.

This is essentially a fair bet.
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LVL 27

Expert Comment

ID: 38369859
You may be able to improve the team odds even more:

Assume 100 players.
If the player who is first loses enough in the last hand to move his teammate from 12th to 9th, he doubles his winnings.

Be aware of it and beware of it: This is cheating.  Ask the poor fellow who would have been 10th.

Team play of the sort you are describing may in fact be prohibited.  I would check.
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LVL 32

Author Comment

ID: 38370001

>> The odds are three times higher to win something
Originally, the odds of my son winning is 10%. Now, I think you are saying that by sharing the prize pool with his two friends, the odds is now triple or 30%. That would be nice, but if I extrapolate and if he had 9 friends, then it couldn't be that the odds of winning would be 10 times higher (or 100%), which would be very nice; but not realizable. Maybe I am misunderstanding what you mean.

By team play, I understand that you are talking about collaboration which is not allowed. They have never collaborated in any way. But thanks for the cautionary advice!

They are not team playing. They are just considering sharing any prizes amongst themselves to try to at least take home something. It is a form of chopping, which is legit.
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LVL 27

Expert Comment

ID: 38370268
If there are lots of people and the probability of winning is 10%, the possible outcomes for three people are:

p =  0.729    LLL
P =  0.081    LLW      LWL    WLL
p =  0.009    LWW     WLW   WWL
P =  0.001    WWW

Probability of winning something would be 3*(0.081 + 0.009) + 0.001 ==>  0.271
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LVL 27

Assisted Solution

d-glitch earned 1148 total points
ID: 38370289
The expected value of the prize for the above case is

EV  =  0.081 + 2*0.009 + 3*0.001  =  0.1

This is the same as for a single player, which is why I said this is a fair bet.
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LVL 17

Assisted Solution

Thibault St john Cholmondeley-ffeatherstonehaugh the 2nd earned 852 total points
ID: 38370293
Agree with d_glitch who works faster than me and has reached the same result with a tree very much like I have drawn on this envelope. Odds of no wins are 9*9*9 /10*10*10 or 729/1000. Meaning that there is a 271/1000 or 2.71 in 10 chance of a shared prize.
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LVL 27

Expert Comment

ID: 38370299
For a smaller number of players, you would probably need to look at permutations.
It there are fewer than 30 players, there may only be two winners.
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Author Comment

ID: 38370341
To summarize what I gathered for the individual:
no sharing:
the EV for an individual is 0.10 with a 10% chance of winning
with 3-way sharing:
the EV for an individual is 0.0333 with a 27.1% chance of winning
Thanks.
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LVL 32

Author Comment

ID: 38370363
Suppose they play this tournament 100 times, and the payout for the top 10% is, say, \$1000.
Is the following individual take-home correct?

No Sharing:
10*\$1000 = \$10000

Sharing:
27.1*\$1000/3 = \$9033

Even bet?
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LVL 17

Expert Comment

ID: 38370403
I would prefer to stay with my calculation that when they share there is a 72.9% chance of winning nothing at all. You have the odds of different sized wins above, using those to calculate the value if a prize might bring false hope.
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Expert Comment

ID: 38370408
*value OF a prize
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Author Comment

ID: 38370503
>> You have the odds of different sized wins above, using those to calculate the value if a prize might bring false hope.
I do not understand why there is false hope.

And I am not sure if I am doing something wrong with the 100 tournament scenario. I thought "even bet" meant that the outcome would be the same over a large number of tournaments; yet, the "not sharing" scenario appears to pay out better.

(btw, to correct typos, you can now hit the Edit Comment button.)
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Assisted Solution

Thibault St john Cholmondeley-ffeatherstonehaugh the 2nd earned 852 total points
ID: 38370593
false hope because if I was entering a competition and told I was likely to win \$9033, it would be wrong. Being told that I had a 1/10 (or a 2.71/10) chance of winning something it would be more accurate.
I think  d-glitch showed how the expected values for each possibility worked out, and that the total prize taken from the prize fund did not vary whether it was shared or not. This was to show that sharing gave no disadvantage to the other players.
The odds listed for each of the possible outcomes show that for instance  there is a .009 chance of just two of the three winning, and so the prize would be 2/3 each (Two wins divided between three people).
To extrapolate this into a total chance of winning a certain amount loses a lot of its meaning, a bit like saying I have a chance of winning \$1000 on a lottery ticket, when I really only have a slight chance of winning millions. The maths aren't incorrect, but they are describing a near impossible event. There is only that single multi-million prize and a very slim chance of winning it. Offering to share (syndicate) with others does increase my chances of winning, and assuming the shares are even and there are enough people, then my chances are increased, but certainly not so much that I can be assured of a profit on my gamble.

(Thanks for the typo tip, but it doesn't work on my phone.)
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Assisted Solution

d-glitch earned 1148 total points
ID: 38370700
Sharing:
27.1*\$1000/3 = \$9033

That is not correct.   You are missing the probability of multiple wins in a particular round.
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LVL 27

Accepted Solution

d-glitch earned 1148 total points
ID: 38372403
Suppose they play this tournament 1000 times (to make the probabilities work out), and the payout for the top 10% is \$100.

``````A single player will win 100 times for a total of \$10,000

A team of three can expect the following:

729    No winners
243    One winner               24,300
27    Two winners               5,400
1    Three winners               300
------------------------------------------
\$30,000 which must be shared three ways.
``````
The Expected Value for each player is identical in both scenarios.
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LVL 27

Assisted Solution

d-glitch earned 1148 total points
ID: 38372453
>>  That is not correct.   You are missing the probability of multiple wins in a particular round.

That is not correct either.  You were missing the benefit of multiple wins in some rounds.
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LVL 17

Assisted Solution

Thibault St john Cholmondeley-ffeatherstonehaugh the 2nd earned 852 total points
ID: 38372801
I thought it was misuse of values rather than missing something.
The 27.1 is the chances of winning a part of any prize. The actual value of the prize is not included in that figure, in fact it is a mix of three values in varying quantities.
The chances of winning an exact value were shown above and separately shiwn was the 27.1 chance of receiving any of those values. You can't just divide the odds of winning any prize into the total prize pool and expect to calculate your win.
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LVL 18

Expert Comment

ID: 38375335
the chance of a player getting no prize is 0.9

the chance of all 3 players getting no prize is 0.9^3 = .729

the chance of at least 1 player getting a prize is 1 - chance of no prizes = 1-.9^3 = .271

*******************************************************************

the chances of getting a prize is higher, but expected winnings will be unaltered
expected winnings is always 10% of the prize
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LVL 32

Author Comment

ID: 38378000

@deighton: It seems that your post summarizes nicely what was said before, or am I missing something?
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LVL 32

Author Comment

ID: 38381780
>> You can't just divide the odds of winning any prize into the total prize pool and expect to calculate your win.
>> \$30,000 which must be shared three ways
Ok, thanks. I believe I may have mixed up probability of winning with expectation. I will straighten this out and try to post a correction.

>> ... false hope ... lottery ... The maths aren't incorrect, but they are describing a near impossible event.
Not sure I agree (or maybe just not following you - I follow better with concrete math examples). If there is a slim chance of winning, but the game or lottery has a positive expectation, then you can be confident of winning if participating in a large number of events. Most state lotteries have a negative expectation; and therefore, it is a losing proposition. But not always! Although each ticket had a slim chance of winning, the group purchase large buckets of tickets and won millions, as they predicted for their school project. I think they got an 'A' for the project. I also read once (maybe this was the case) that there was a lottery promotion - buy one get one free - and a group computed a positive expectation and also won millions.
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Author Closing Comment

ID: 38544163
Thanks for this discussion.
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