expectation
February 15th, 2011 at 7:24:11 AM
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I have $5 bet on DP4 and $5 bet on DC5. What is the expectation for this combination? Thank You. HAK
February 16th, 2011 at 12:47:20 PM
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This is what I came up with, but someone definitely needs to check my calculations.
In the scenario you describe, the most risky part of the process for don't bettors has been successfully completed. The point and come numbers have been established, and you did not lay the odds. In this case, there are three ways to make the four (1-3, 2-2, 3-1), and four ways to make the five (1-4, 2-3, 3-2, 4-1), for a combined seven ways to lose. You will win both bets if a seven is rolled. There are six ways to make a seven (1-6, 2-5, 3-4, 4-3, 5-2, 6-1). At this point, no other combinations matter to you. On the next throw you could:
Win both bets: Probability: 6 - 13 (46.1%): pays $10, value: $4.62
Lose the 5: Probability: 4 - 13 (30.8%): pays -$5, value: -$1.54
Lose the 4: Probability: 3 - 13 (23.1%): pays -$5, value: -$1.15
Expectation from your $10 action on the next roll of the dice: $1.92
Note that the most likely outcome overall, is that you will lose one then win the other, for a net gain of $0.
In the scenario you describe, the most risky part of the process for don't bettors has been successfully completed. The point and come numbers have been established, and you did not lay the odds. In this case, there are three ways to make the four (1-3, 2-2, 3-1), and four ways to make the five (1-4, 2-3, 3-2, 4-1), for a combined seven ways to lose. You will win both bets if a seven is rolled. There are six ways to make a seven (1-6, 2-5, 3-4, 4-3, 5-2, 6-1). At this point, no other combinations matter to you. On the next throw you could:
Win both bets: Probability: 6 - 13 (46.1%): pays $10, value: $4.62
Lose the 5: Probability: 4 - 13 (30.8%): pays -$5, value: -$1.54
Lose the 4: Probability: 3 - 13 (23.1%): pays -$5, value: -$1.15
Expectation from your $10 action on the next roll of the dice: $1.92
Note that the most likely outcome overall, is that you will lose one then win the other, for a net gain of $0.
Simplicity is the ultimate sophistication - Leonardo da Vinci
February 16th, 2011 at 3:27:43 PM
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1) What happens if you lose the 5 and win the 4? or 2) lose the 4 and win the 5? It seems to me these 2 scenarios need to be taken into account. Thanks for your post. HAK
February 16th, 2011 at 3:55:17 PM
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win both (6/13) 46.2%
win 5 lose 4 (3/13)*(6/10) 13.8%
win 4 lose 5 (4/13)*(6/9) 20.5%
lose both (3/13)*(4/10)+(4/13)*(3/9) 19.5%
You will average a profit of $2.67
win 5 lose 4 (3/13)*(6/10) 13.8%
win 4 lose 5 (4/13)*(6/9) 20.5%
lose both (3/13)*(4/10)+(4/13)*(3/9) 19.5%
You will average a profit of $2.67
“Man Babes” #AxelFabulous
February 16th, 2011 at 4:18:51 PM
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I agree with Miplet's calculation, and here's another way to get the same result:
P(win 5) = 6/(4+6) = 0.600
P(win 4) = 6/(3+6) = 0.667
E = 5(0.600)-5(1-0.600)+5(0.667)-5(1-0.667) = $2.67
P(win 5) = 6/(4+6) = 0.600
P(win 4) = 6/(3+6) = 0.667
E = 5(0.600)-5(1-0.600)+5(0.667)-5(1-0.667) = $2.67
February 17th, 2011 at 10:03:13 AM
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Quote: ChesterDogI agree with Miplet's calculation, and here's another way to get the same result:
P(win 5) = 6/(4+6) = 0.600
P(win 4) = 6/(3+6) = 0.667
E = 5(0.600)-5(1-0.600)+5(0.667)-5(1-0.667) = $2.67
I also get $2.67.
Cheers,
Alan Shank
Woodland, CA
Cheers,
Alan Shank
"How's that for a squabble, Pugh?" Peter Boyle as Mister Moon in "Yellowbeard"
February 17th, 2011 at 3:23:50 PM
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Quote: hakoelliI have $5 bet on DP4 and $5 bet on DC5. What is the expectation for this combination? Thank You. HAK
So you can understand and calculate these in the future.
expectation is EV(per decision) times the $ bet amount.
| Type | EV | $ wager | expectation | $ expectation |
|---|---|---|---|---|
| DP4 | 1/3 (.33) | $5 | 1 2/3 | $1.67 |
| DP10 | 1/3 (.33) | $5 | 1 2/3 | $1.67 |
| DP5 | 1/5 (.20) | $5 | 1 | $1.00 |
| DP9 | 1/5 (.20) | $5 | 1 | $1.00 |
| DP6 | 1/11 (.0909) | $5 | 5/11 | $0.45 |
| DP8 | 1/11 (.0909) | $5 | 5/11 | $0.45 |
How to calculate the EV?
DP4 as an example.
6 ways to win.
3 ways to lose.
We add the win probability times $1 to the loss probability times $1 to arrive at the EV.
win probability: 1*(6/(6+3))... 1*(6/9)... (6/9) or 2/3
loss probability: -1*(3/(6+3))... -1*(3/9)... (3/9) or -1/3
2/3 PLUS -1/3 = 1/3 (or .3333)
expectation is EV(per decision) times the $ bet amount.
So a $5 DP4 = $5 * 1/3 = $5/3 or $1.67(rounded)
I just did the math for my wife with Baccarat, so I share with you.
Correct me if I made any mistakes.
