Math Problem; Similar To Craps Fire Bet
April 28th, 2013 at 2:59:43 AM
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I'm hoping that one of the resident math guys might be able to help me out. I am trying to determine the correct math for a bet similar to the Fire Bet in Craps. To simplify, I have a nine sided die numbered 1 thru 9. I am trying to determine what the probability of rolling all the numbers 1 thru 8 without rolling the 9. Numbers can roll multiple times; the only thing that kills the bet is the roll of the 9. I've looked at the Wizard's spreadsheet for the Fire Bet, but I understand it enough to be dangerous to myself. I'm also trying to determine what the probability of rolling the last 7 after making 1 number, then the last 6 after 2 numbers, etc. Thanks in advance for any help that can be provided; this seemed like the ideal group of people to ask.
April 28th, 2013 at 9:35:50 AM
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I did some reasonably serious number crunching, and then, when what I thought was a non-intuitive answer came up, wrote a Monte Carlo simulation, before banging my head against the table for not noticing that 1/9 is the obvious answer.
Proof: List the numbers in the order of when they first appear. You want the probability that 9 will be last. Since each of the numbers has an equal probability of appearing in any roll (which is not true for the Fire Bet, which is why it is harder to calculate), the 9 is just as likely as the others to be the last one in the list.
Similarly, the probability of rolling the last 7 after making the first one is 1/8 (use the same reasoning, but ignore all rolls of the first number), the probability of rolling the last 6 after making the first two is 1/7, and so on.
Proof: List the numbers in the order of when they first appear. You want the probability that 9 will be last. Since each of the numbers has an equal probability of appearing in any roll (which is not true for the Fire Bet, which is why it is harder to calculate), the 9 is just as likely as the others to be the last one in the list.
Similarly, the probability of rolling the last 7 after making the first one is 1/8 (use the same reasoning, but ignore all rolls of the first number), the probability of rolling the last 6 after making the first two is 1/7, and so on.
April 28th, 2013 at 1:04:32 PM
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Yep, the odds of winning such a bet is 1/9. In fact, the odds of winning any bet is 1/9. That is, if you had a bet that stated "what number of numbers will appear before the 9", the odds of the answer being 0 vs 8 are the same at 1:9.
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April 29th, 2013 at 5:44:15 AM
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Don, thanks for your reply, but intuitively (as you said) 1 in 9 just doesn't seem right. After you replied, I realized that the Fire Bet was a poor comparison. There you have to establish the point then make it; plus only 7's rolled as a 7 out kill the bet. I think a better comparison would have been Bonus Craps where you have to roll all 10 numbers once before any 7 rolls. That pays 175 to 1. I think if I saw the math for that then I could either be convinced of 1 in 9 or look in a different direction.
April 29th, 2013 at 6:55:41 AM
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It's pretty intuitive.
Odds of throwing any number before a 9: 8/9
Odds of throwing another different number before a 9: 7/8
Odds of throwing another number before a 9: 6/7
Odds of throwing another number before a 9: 5/6
Odds of throwing another number before a 9: 4/5
Odds of throwing another number before a 9: 3/4
Odds of throwing another number before a 9: 2/3
Odds of throwing the last number before a 9: 1/2.
It's because the other numbers are out of the equation and don't count as a trial.
Multiply: 8/9 x 7/8 x 6/7 x 5/6 x 4/5 x 3/4 x 2/3 x 1/2 = 1/9
Odds of throwing any number before a 9: 8/9
Odds of throwing another different number before a 9: 7/8
Odds of throwing another number before a 9: 6/7
Odds of throwing another number before a 9: 5/6
Odds of throwing another number before a 9: 4/5
Odds of throwing another number before a 9: 3/4
Odds of throwing another number before a 9: 2/3
Odds of throwing the last number before a 9: 1/2.
It's because the other numbers are out of the equation and don't count as a trial.
Multiply: 8/9 x 7/8 x 6/7 x 5/6 x 4/5 x 3/4 x 2/3 x 1/2 = 1/9
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You want the truth! You can't handle the truth!
April 29th, 2013 at 10:52:52 AM
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First, nice question.Quote: DaGongJiDon, thanks for your reply, but intuitively (as you said) 1 in 9 just doesn't seem right.
There are many things in probability and statistics that do not seem right at first glance or first thought.
IMO, That is the beauty of math.
The correct answer or the correct path to the answer is not always what seems right.
The trick is seeing all the possibilities (the sample space)
(I remember this type of question from high school math)
To me, without any proof, is if I always have a 1 in 9 chance of losing on each roll,
that must be the highest probability of winning over an infinite number of rolls.
It is possible that the game could never end (stuck on 7 numbers)
or take a very long time to end. So we must be converging to some number.
But that is just me.
You are dealing with unequal probabilities for the sum of two dice for those 2 Craps bets.Quote: DaGongJiAfter you replied, I realized that the Fire Bet was a poor comparison. There you have to establish the point then make it; plus only 7's rolled as a 7 out kill the bet. I think a better comparison would have been Bonus Craps where you have to roll all 10 numbers once before any 7 rolls. That pays 175 to 1. I think if I saw the math for that then I could either be convinced of 1 in 9 or look in a different direction.
The Wizard and BruceZ both have solved and showed their math on the Bonus Craps bets.
(aka All Tall Small)
Your example is all about equal probabilities.
Step through each roll and see what can happen.
Place probabilities at each step (each state)
This sure looks like it is leading us to some simple (?) Matrix Algebra (an absorbing transition matrix)
http://people.hofstra.edu/Stefan_Waner/RealWorld/Summary8.html
The matrix looks like this
the probability distribution for the first 30 rolls
and all the probabilities for ending up in each state
You have at the 8th roll about a 61% chance of losing (cumulative) and only a 0.000936657 chance of winning
That means you could be at any other state (# of different numbers hit) 0.388807392
Looks like 5#s has the highest probability at 0.163915
None of this is intuitive.
At least to me. it takes some effort to see where we can end up.
You can easily make many games of this, for example,
by limiting the number of rolls to X and a "winner" could be how many #s were hit before #9
added:
The d6 example: roll a 1 and lose.
get at least 1 of each other #s 2 thru 6 before a 1 and win
5th roll
1#: 0.000643004
2#: 0.0385802
3#: 0.192901
4#: 0.154321
win: 0.0154321
lose: 0.598122
game continues: 0.386445
R / #s 0 1 2 3 4 win lose W on R L on R
1 0 0.8333 0.0000 0.0000 0.0000 0.0000 0.1667 0.0000 0.1667
2 0 0.1389 0.5556 0.0000 0.0000 0.0000 0.3056 0.0000 0.1389
3 0 0.0231 0.2778 0.2778 0.0000 0.0000 0.4213 0.0000 0.1157
4 0 0.0039 0.1080 0.2778 0.0926 0.0000 0.5177 0.0000 0.0965
5 0 0.0006 0.0386 0.1929 0.1543 0.0154 0.5981 0.0154 0.0804
6 0 0.0001 0.0133 0.1157 0.1672 0.0412 0.6625 0.0257 0.0644
7 0 0.0000 0.0045 0.0645 0.1500 0.0690 0.7119 0.0279 0.0494
8 0 0.0000 0.0015 0.0345 0.1215 0.0940 0.7484 0.0250 0.0365
9 0 0.0000 0.0005 0.0180 0.0925 0.1143 0.7747 0.0203 0.0263
10 0 0.0000 0.0002 0.0093 0.0677 0.1297 0.7932 0.0154 0.0185
11 0 0.0000 0.0001 0.0047 0.0482 0.1410 0.8060 0.0113 0.0129
12 0 0.0000 0.0000 0.0024 0.0337 0.1490 0.8149 0.0080 0.0088
13 0 0.0000 0.0000 0.0012 0.0233 0.1546 0.8209 0.0056 0.0060
14 0 0.0000 0.0000 0.0006 0.0159 0.1585 0.8250 0.0039 0.0041
15 0 0.0000 0.0000 0.0003 0.0108 0.1612 0.8277 0.0027 0.0028
16 0 0.0000 0.0000 0.0002 0.0073 0.1630 0.8296 0.0018 0.0019
17 0 0.0000 0.0000 0.0001 0.0049 0.1642 0.8308 0.0012 0.0012
18 0 0.0000 0.0000 0.0000 0.0033 0.1650 0.8317 0.0008 0.0008
19 0 0.0000 0.0000 0.0000 0.0022 0.1656 0.8322 0.0006 0.0006
20 0 0.0000 0.0000 0.0000 0.0015 0.1659 0.8326 0.0004 0.0004
winsome johnny (not Win some johnny)
April 29th, 2013 at 3:34:03 PM
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here is what i think your asking
what are odds you roll 1 2 3 4 5 6 7 8 before the 9
meaning you hit all of those before the 9 is rolled? doubles dont matter.
well...
1 numbers 8/9 or 88.89%
2 numbers 8/9 * 7/9 or 69.14% (69.14% is rolling non double and continuing on to next number)...79.01% you will roll new number or double - safe either way
3 numbers 8/9 * 7/9 * 6/9 or 46.09% rolling non double, 70.23% you roll new number or double but are safe
(safe number found 8/9 * 8/9 * 8/9..ect since you are trying to dodge the 9)
(now ill just show percentages using method above)
4 numbers 25.61%, 62.43% safe
5 numbers 11.38%, 55.49% safe
6 numbers 3.79%, 49.33% safe (also implys you will roll 6 times 49.33% of time without rolling a 9)
7 numbers 0.84%, 43.85% safe
8 numbers 0.094%, 38.97% safe
odds of rolling all 1-8 numbers before the 9 is 1068 to 1 - good luck
hopefully whatever the bet you are making is paying good odds.
what are odds you roll 1 2 3 4 5 6 7 8 before the 9
meaning you hit all of those before the 9 is rolled? doubles dont matter.
well...
1 numbers 8/9 or 88.89%
2 numbers 8/9 * 7/9 or 69.14% (69.14% is rolling non double and continuing on to next number)...79.01% you will roll new number or double - safe either way
3 numbers 8/9 * 7/9 * 6/9 or 46.09% rolling non double, 70.23% you roll new number or double but are safe
(safe number found 8/9 * 8/9 * 8/9..ect since you are trying to dodge the 9)
(now ill just show percentages using method above)
4 numbers 25.61%, 62.43% safe
5 numbers 11.38%, 55.49% safe
6 numbers 3.79%, 49.33% safe (also implys you will roll 6 times 49.33% of time without rolling a 9)
7 numbers 0.84%, 43.85% safe
8 numbers 0.094%, 38.97% safe
odds of rolling all 1-8 numbers before the 9 is 1068 to 1 - good luck
hopefully whatever the bet you are making is paying good odds.
April 29th, 2013 at 3:51:42 PM
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hmmm.....
something doesnt feel right about this.
obvioulsy you have to look at each roll interdependently
88.89% of time you knock off 1 number 11.11% F the Dog and start over
next roll
77.78% of time you roll another number 11.11% copy 11.11% gutter ball (87.5% stay alive and continue) 7 to 1
next roll
66.67% of time you roll another number 22.22% copy 11.11% shit the bed (85.71% stay alive and continue) 6 to 1
....hit......copy......lose
55.56%, 33.33%, 11.11% (83.33% stay alive) 5 to 1
44.44%, 44.44%, 11.11% (80% stay alive) 4 to 1
33.33%, 55.55%, 11.11% (75% stay alive) 3 to 1
22.22%, 66.67%, 11.11% (66.67% stay alive) 2 to 1
11.11%, 77.78%, 11.11% (50% stay alive and win) 1 to 1
something doesnt feel right about this.
obvioulsy you have to look at each roll interdependently
88.89% of time you knock off 1 number 11.11% F the Dog and start over
next roll
77.78% of time you roll another number 11.11% copy 11.11% gutter ball (87.5% stay alive and continue) 7 to 1
next roll
66.67% of time you roll another number 22.22% copy 11.11% shit the bed (85.71% stay alive and continue) 6 to 1
....hit......copy......lose
55.56%, 33.33%, 11.11% (83.33% stay alive) 5 to 1
44.44%, 44.44%, 11.11% (80% stay alive) 4 to 1
33.33%, 55.55%, 11.11% (75% stay alive) 3 to 1
22.22%, 66.67%, 11.11% (66.67% stay alive) 2 to 1
11.11%, 77.78%, 11.11% (50% stay alive and win) 1 to 1
