Probability Poser

Are Aces high or low? - and to work out exact probability (assuming the "nothing for a pair" rule is in play) would also need to know the exactly what the first three cards were!

But the answer is - crack on (imo) and then ask (Dolly Dealer) Emma Noble out for a drink (Nap).

presumably ace is high and you don't need to know the exact cards. I think the odds of getting two low are 190/1176, but could have made a fu somewhere.

Actually I think Emma was on The Price is Right, but worth a try imo

Presuming that one of the - " first three cards have been 7 (or higher).

Then - There are ...

24 Cards left Numbered 1 to 6
26 Cards left Numbered 8 to Ace (High)
3 Cards left Number 7 (= Loser - Lower or Higher - in this game)

So the Answer is ...

You do not need a Probability figure (down to several decimal points) - as ...

What's Tenner ? - Nowadays.

Are we dealing with a 52 card pack , and we currently have 7 exposed cards ?

Of which we know 4 of those are 7 or higher.

He is being offered 7/1.

Is the guy rational and good at maths ?

Sorry - I don't understand the game - There are 4 exposed cards I understand.

He can't be playing properly - He must be guessing high whilst getting high cards.

No - thats not right either.

TM

Emma Noble was indeed the most gorgeous of all the dolly dealers on the price is right , used to be married to John Major's son.

3 cards exposed all 7 or higher and he is being asked to guess whether the next card will be higher or lower than a 7 and the final card will be higher or lower than the 4th card. I didn't understand it either as I gave the probability of two cards being lower than 7. However, going high the first time is less than evens and if that wins it can't be less more evens for the last one, so £10 must be a bad deal as it is less than £20 which would be the expected amount of two even money shots.

We have an exposed starting card - unknown , and 3 exposed high cards.

There are 48 cards left  - 23 are lower or equal to 7   and 26 are higher than a 7  - You need to adjust this for the starting card - unknown.

Even if this is a high card it still doesn't change the choice to go high on pick 4.  You are slightly odds on to be correct.

So definitely call high and press on - 7/1 is a value play.

Where it gets interesting is when it's  10K or 80K i.e. where the money might mean something.

I can only repeat my original answer (above) - the answer (or the value) is to crack-on - but you have not provided enough information for us to calculate the exact probabilities.

3 cards 7 or higher have been drawn leaving 29 equal to 7 or higher, and 20 cards lower than 7.
49 cards left.

49/29 =1.68
49/20= 2.45


Take the tenner?

or am I miles out

The way this question was initially posed, no mention was made of what the first card was, simply that on the next three turns the contestant guessed correctly, and that on the third turn ( now four cards on the board ) the 7 of hearts appeared; moreover, those three cards were all 7 or higher, but let's assume the first was too, for arguments sake.

I'm assuming the game is being played with a standard deck of 52 cards.

Here's the clincher : by being offered a tenner to pass up guessing the last two turns, the contestant is being offered odds of 7/1, but he has - apparently - a 2/1 chance of succeeding by pressing on. The contestant in question took the £10, and this was said to have been a big error.

I was puzzled how the odds of 2/1 were calculated - and still am ! My thinking - probably wrong - was that there were two even-money shots to win the £80 from the point reached in the narrative, but by my reckoning, that would mean a 25% chance, or 3/1, not 33% and 2/1.

But, has this something to do with the fact that if you're wrong first, the game ends, but if you're right, you then have a 50/50 shot ?!

I tried to rethink the question as follows : four cards are visible on the board, all 7 or higher. At this point, you're offered the chance to bet £10 to win £70 if you can correctly predict the turn of the next two cards, and the last card showing (the fourth ) is the 7 of hearts.

What are the odds of guessing the next two turns correctly ?

the 2/1 will almost def come from the final guess which eg might might be a v low or v high card, making the guess far less than an even money shot

And, does it really matter what cards have come up previously ?

For arguments sake, remove the Ace altogether, so there are only 12 cards, 2 through to King, so you have 2 to 7 'low' and 8 to King 'high'. The first four cards out were two high and two low. As such, this now has little bearing on whether you choose to take the bet, or not. It's simply a case of judging your chances on the next two turns against the 7/1 on offer.

I think... .

ok I see why it is 2-1. If he goes high and wins, the card must be 8 to ace, with approx probability of 1/13 for each. Assuming he then goes low you then take all those 1/13 and work our the p of winning. So Ace would would be 1/13 x 12/13 (approx - obv a bit lower because only three aces left) and 8 would be 1/13 x 6/13. Add all these together and you get much more of a chance that evens by evens.

Turns out 2/1 was a typo ; should've read 3/1.

But, still interesting how the choice whether to go on, or not, doesn't necessarily have much to do with the cards already seen, it seems.

on my rough calcs, it could only be 3-1 if ace was low.

You also need to factor in that the next card could match or the last card could too & on top of that if the next is correct & is an Ace (presuming this is a high card) or a 2 & (then unless a matching card is the final one turned) then this is a no lose scenario for the last decision. An interesting question & I would be curious to see the maths behind getting the correct answer but it is a bit beyond me with the info given as others have asked for clarification.

Wasn't it the case that in the tv series, if the card turned over was the same as the previous one, you simply guessed again ?

Been so long since I saw an episode.

Obviously, going low or high and getting a 2 or an Ace respectively guarantees a win on the fifth card, so whether it's actually bang on 3/1 is questionable. But, the odds are pretty close to 3/1, so why would you take the tenner when you're being offered such terrific value, that's to say 7/1 for a 3/1 chance ? I think much boils down to whether you're being offered a tenner rather than pondering risking a tenner of your own to win £70. The point is that you're obtaining the same value either way, so the decision should be identical, but it often isn't.

In the TV game if the card turned over matched, you lost (nothing for a pair - not in this game!) and the other couple usually ran away with the game through a very kind sequence of card.

I'll say it one final time! - you haven't provided enough information (or even the precise rules) for anybody (even somebody that "does proper maths") to work out the exact mathematical probability. The biggest problem to getting the exact mathematical answer is that we don't even know what the first card was!

All you really need to know is that the value/odds are in your favour to continue. It's a fact that the prize for doing that successfully is 7/1 and an approximation that the true price (via optimal play) is probably somewhere close to about 9/4 imo.

You're absolutely right, The Management, but I'm simply repeating the problem as I found it. No mention was made of the first card, nor what would happen if a pair turned up.

The game was being played by somebody from the winning team at the end of an evening's pub quiz. The person playing wouldn't have had time to work out the probability, even if they'd had the brains to do so. The contestant took the £10, and apparently the pub crowd cheered that decision, but the writer was at pains to point out that it ought to have been obvious to reject the offer and continue going for the £80 given the odds. Part of the dilemma is having last turned a seven, which makes the next decision tricky, thus increasing the temptation to simply take the tenner. But, assuming the first four cards out were 7 or higher, it seems even more logical to crack on, and go low.

A lot of non-mathematical misunderstanding here. It is a simple maths problem. Since 3 cards of 7 or higher are showing there are 24 cards less than a 7 and 25 cards higher or equal to a 7 left. Approximately evens. If you do guess correctly you have no idea of the next card except it is not a 7 and it matches your choice. Again the guess at getting the last card right is AT WORSE evens. Therefore for the 2 guesses the odds against be right is at worse 3/1. You are being offered 7/1 for your bet. No brainer. Play on!

Not just that bluebook. The first of the two "guesses" is a bit daunting and not that far off a coin-flip - but even assuming it's an even-money poke (to a lay-man with no betting/maths knowledge), there is a very good chance that if guessing the first one correctly, the second/final one could well be a very long odds-on chance.

i.e. you guess correctly at lower - there is every chance you will get a 2,3 or 4 - making it long odds-on they can go high for the final card and the payout. Similarly even if they "guess" higher and are correct - there is every chance you will get a card (Ace, King, Queen) that makes going lower for the final card payout, a long-odds-on shot.

In summary, even a moron with no knowledge of betting should (wrongly) be able to guess that it's a 3/1 chance by (wrongly) assuming it's 2 x Even money shots; and that 7/1 about a 3/1 shot is a very good bet.

In reality, while we can't calculate the true odds (due to missing info), it's significantly shorter than 3/1 imo.

As somebody else said, it would be more interesting if more money were involved as although getting 7/1 about a 9/4 chance is a licence to print over time (i.e. if you can do it repeatedly); being able to do it only once, you are still more likely to lose than you are to win. So for £10k, maybe taking the money would be more forgivable if £10k is a lot to the player.

I didn't see your post jimmy - but spot on - the very worst case scenario is that you are getting offered 7/1 about 2 x very slightly odds-on shots, in a double. The more likely scenario is that you are getting offered 7/1 about a slightly odds-on shot and a long odds-on shot in a double.