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Odds of winning these bets
| November 5th, 2011 at 2:04:26 AM permalink | |
| roony672 Member since: Nov 5, 2011 Threads: 2 Posts: 3 | If I have four bets and I will win the first bet 42% of the time, the second bet 42% of the time, the third bet 48% of the time and the fourth bet 15% of the time. Then what are the odds that I lose all four bets? How about winning just one of them (doesn't matter which one)? Two? How about winning all four of them? Thank you. |
| November 5th, 2011 at 4:22:26 AM permalink | |
| SOOPOO Member since: Aug 8, 2010 Threads: 49 Posts: 1324 | Rather than just give you an answer, think of it this way... there are 16 total possibilities... WWWW WWWL WWLW WWLL WLWW WLWL WLLW WLLL LWWW LWWL LWLW LWLL LLWW LLWL LLLW LLLL The first one (WWWW) will happen .42 x .42 x .48 x .15= .0127008, or about 1.3% of the time. Why don't you try and figure out your other answers now.... |
| November 5th, 2011 at 8:47:22 AM permalink | |
| EdCollins Member since: Oct 21, 2011 Threads: 3 Posts: 24 | Hi Roony. Welcome to the forum. You will lose all four bets 14.86888% of the time. You will win exactly one bet approximately 37.89% of the time. You will win exactly two bets approximately 33.89% of the time. You will win all four bets 1.27008% of the time. Good luck with your "system." |
| November 5th, 2011 at 10:39:03 AM permalink | |
| FleaStiff Member since: Oct 19, 2009 Threads: 75 Posts: 4829 | Can you "show your work"? Or as I used to have to do in math class... draw a picture of me craning my neck to see the paper of the person sitting in front of me. Or, better yet, give it to us textually rather in these formulas that don't make much sense to me anyway. |
| November 5th, 2011 at 11:25:46 AM permalink | |
| EdCollins Member since: Oct 21, 2011 Threads: 3 Posts: 24 |
Well, for what it's worth, here is my work: winning all four bets: 42/100 * 42/100 * 48/100 * 15/100 losing all four bets: 58/100 * 58/100 * 52/100 * 85/100 Notice for my other two answers I said "approximately." My work for this is a computer simulation, for 1 billion trials. RANDOMIZE TIMER max_sims = 1000000000 FOR x = 1 TO max_sims win = 0 rand_number = RND(1, 100) IF rand_number <= 42 THEN INCR win rand_number = RND(1, 100) IF rand_number <= 42 THEN INCR win rand_number = RND(1, 100) IF rand_number <= 48 THEN INCR win rand_number = RND(1, 100) IF rand_number <= 15 THEN INCR win IF win = 4 THEN INCR won_all_four IF win = 0 THEN INCR lost_all_four IF win = 2 THEN INCR won_two_of_them IF win = 1 THEN INCR won_one_of_them NEXT PRINT "Won all four:"; won_all_four PRINT "Probability:";won_all_four/max_sims PRINT "Lost all four:"; lost_all_four PRINT "Probability:";lost_all_four/max_sims PRINT "Won just one of them:"; won_one_of_them PRINT "Probability:";won_one_of_them/max_sims PRINT "Won just two of them:"; won_two_of_them PRINT "Probability:";won_two_of_them/max_sims Program Output: Won all four: 12,700,208 Probability: .012700208 Lost all four: 148,687,737 Probability: .148687737 Won just one of them: 378,828,971 Probability: .378828971 Won just two of them: 338,971,187 Probability: .338971187 |
| November 5th, 2011 at 12:11:38 PM permalink | |
| Doc Member since: Feb 27, 2010 Threads: 21 Posts: 2825 | I don't see any reason it would require a simulation. On the other hand, I'm too lazy to do all the calculations manually, so I started with the guidance from SOOPOO and used a spreadsheet. Below is a copy of the spreadsheet, and I suspect you can figure out the formulas for each cell.  |
| November 5th, 2011 at 2:03:02 PM permalink | |
| EdCollins Member since: Oct 21, 2011 Threads: 3 Posts: 24 |
Well, it doesn't REQUIRE a simulation. A simulation is just one of the many ways to arrive at the answer. |
| November 5th, 2011 at 2:34:48 PM permalink | |
| Doc Member since: Feb 27, 2010 Threads: 21 Posts: 2825 | Understood. I just wouldn't use a simulation when an explicit calculation of the answer is so straight forward. Do you ordinarily simulate a million rolls of a pair of dice to determine the probability of a roll that totals 7? It's easier, and much clearer, just to calculate the EV as 1/6, I think. The OP's 4-bet question is not quite that simple, but it isn't hard to calculate, and you can see every calculation that is needed represented in the small image of a spreadsheet above. On the other hand, there are many probability questions where explicit calculation becomes impractical or maybe impossible, and that is were most people would use simulations. No? |
| November 5th, 2011 at 4:25:03 PM permalink | |
| MathExtremist Member since: Aug 31, 2010 Threads: 46 Posts: 2521 | I usually draw the line at anything I can't actually do in Excel. For computations O(<10^5), I'll do it in Excel. For O(10^6 .. 10^10) or so, I'll write an iterator. For O(>10^11), that's usually too big for an iterator to run through in a reasonable amount of time (on my machine) so I'll simulate. This problem, being O(2^4), is definitely Excel fodder. "In my own case, when it seemed to me after a long illness that death was close at hand, I found no little solace in playing constantly at dice."
-- Girolamo Cardano, 1563 |
| November 5th, 2011 at 5:24:22 PM permalink | |
| roony672 Member since: Nov 5, 2011 Threads: 2 Posts: 3 |
Thank you for your assistance. It's not actually a system however. |
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